Module 13: Systems of Equations and Inequalities

Systems of linear equations: two variables, learning outcomes.

  • Solve systems of equations by graphing, substitution, and addition.
  • Identify inconsistent systems of equations containing two variables.
  • Express the solution of a system of dependent equations containing two variables using standard notations.

A skateboard manufacturer introduces a new line of boards. The manufacturer tracks its costs, which is the amount it spends to produce the boards, and its revenue, which is the amount it earns through sales of its boards. How can the company determine if it is making a profit with its new line? How many skateboards must be produced and sold before a profit is possible? In this section we will consider linear equations with two variables to answer these and similar questions.

Skateboarders at a skating rink by the beach.

(credit: Thomas Sørenes)

Introduction to Solutions of Systems

In order to investigate situations such as that of the skateboard manufacturer, we need to recognize that we are dealing with more than one variable and likely more than one equation. A system of linear equations consists of two or more linear equations made up of two or more variables such that all equations in the system are considered simultaneously. To find the unique solution to a system of linear equations, we must find a numerical value for each variable in the system that will satisfy all equations in the system at the same time. Some linear systems may not have a solution and others may have an infinite number of solutions. In order for a linear system to have a unique solution, there must be at least as many equations as there are variables. Even so, this does not guarantee a unique solution.

In this section, we will look at systems of linear equations in two variables, which consist of two equations that contain two different variables. For example, consider the following system of linear equations in two variables.

[latex]\begin{align}2x+y&=15\\[1mm] 3x-y&=5\end{align}[/latex]

The solution to a system of linear equations in two variables is any ordered pair that satisfies each equation independently. In this example, the ordered pair [latex](4,7)[/latex] is the solution to the system of linear equations. We can verify the solution by substituting the values into each equation to see if the ordered pair satisfies both equations. Shortly we will investigate methods of finding such a solution if it exists.

[latex]\begin{align}2\left(4\right)+\left(7\right)&=15 &&\text{True} \\[1mm] 3\left(4\right)-\left(7\right)&=5 &&\text{True} \end{align}[/latex]

In addition to considering the number of equations and variables, we can categorize systems of linear equations by the number of solutions. A consistent system of equations has at least one solution. A consistent system is considered to be an independent system if it has a single solution, such as the example we just explored. The two lines have different slopes and intersect at one point in the plane. A consistent system is considered to be a dependent system if the equations have the same slope and the same y -intercepts. In other words, the lines coincide so the equations represent the same line. Every point on the line represents a coordinate pair that satisfies the system. Thus, there are an infinite number of solutions.

Another type of system of linear equations is an inconsistent system , which is one in which the equations represent two parallel lines. The lines have the same slope and different y- intercepts. There are no points common to both lines; hence, there is no solution to the system.

A General Note: Types of Linear Systems

There are three types of systems of linear equations in two variables, and three types of solutions.

  • An independent system has exactly one solution pair [latex]\left(x,y\right)[/latex]. The point where the two lines intersect is the only solution.
  • An inconsistent system has no solution. Notice that the two lines are parallel and will never intersect.
  • A dependent system has infinitely many solutions. The lines are coincident. They are the same line, so every coordinate pair on the line is a solution to both equations.

Below is a comparison of graphical representations of each type of system.

Graphs of an independent system, an inconsistent system, and a dependent system. The independent system has two lines which cross at the point seven-fifths, negative eleven fifths. The inconsistent system shows two parallel lines. The dependent system shows a single line running through the points negative one, negative two and one, two.

How To: Given a system of linear equations and an ordered pair, determine whether the ordered pair is a solution.

  • Substitute the ordered pair into each equation in the system.
  • Determine whether true statements result from the substitution in both equations; if so, the ordered pair is a solution.

Example: Determining Whether an Ordered Pair Is a Solution to a System of Equations

Determine whether the ordered pair [latex]\left(5,1\right)[/latex] is a solution to the given system of equations.

[latex]\begin{align}x+3y&=8\\ 2x-9&=y \end{align}[/latex]

Substitute the ordered pair [latex]\left(5,1\right)[/latex] into both equations.

[latex]\begin{align}\left(5\right)+3\left(1\right)&=8 \\[1mm] 8&=8 &&\text{True} \\[3mm] 2\left(5\right)-9&=\left(1\right) \\[1mm] 1&=1 &&\text{True} \end{align}[/latex]

The ordered pair [latex]\left(5,1\right)[/latex] satisfies both equations, so it is the solution to the system.

Analysis of the Solution

We can see the solution clearly by plotting the graph of each equation. Since the solution is an ordered pair that satisfies both equations, it is a point on both of the lines and thus the point of intersection of the two lines.

A graph of two lines running through the point five, one. The first line's equation is x plus 3y equals 8. The second line's equation is 2x minus 9 equals y.

Determine whether the ordered pair [latex]\left(8,5\right)[/latex] is a solution to the following system.

[latex]\begin{align}5x-4y&=20\\ 2x+1&=3y\end{align}[/latex]

Not a solution.

Solving Systems of Equations by Graphing

There are multiple methods of solving systems of linear equations. For a system of linear equations in two variables, we can determine both the type of system and the solution by graphing the system of equations on the same set of axes.

Example: Solving a System of Equations in Two Variables by Graphing

Solve the following system of equations by graphing. Identify the type of system.

[latex]\begin{align}2x+y&=-8\\ x-y&=-1\end{align}[/latex]

Solve the first equation for [latex]y[/latex].

[latex]\begin{align}2x+y&=-8\\ y&=-2x-8\end{align}[/latex]

Solve the second equation for [latex]y[/latex].

[latex]\begin{align}x-y&=-1\\ y&=x+1\end{align}[/latex]

Graph both equations on the same set of axes:

A graph of two lines running through the point negative 3, negative 2. The first line's equation is y equals minus 2x minus 8. The second line's equation is y equals x+1.

The lines appear to intersect at the point [latex]\left(-3,-2\right)[/latex]. We can check to make sure that this is the solution to the system by substituting the ordered pair into both equations.

[latex]\begin{align}2\left(-3\right)+\left(-2\right)&=-8 \\[1mm] -8=-8 &&\text{True} \\[3mm] \left(-3\right)-\left(-2\right)&=-1 \\[1mm] -1&=-1 &&\text{True} \end{align}[/latex]

The solution to the system is the ordered pair [latex]\left(-3,-2\right)[/latex], so the system is independent.

Solve the following system of equations by graphing.

[latex]\begin{gathered}2x - 5y=-25 \\ -4x+5y=35 \end{gathered}[/latex]

The solution to the system is the ordered pair [latex]\left(-5,3\right)[/latex].

Two lines that cross at the point negative five, three. One line's equation is y equals four-fifths x plus 7. The other line's equation is y equals two-fifths x plus 5.

Can graphing be used if the system is inconsistent or dependent?

Yes, in both cases we can still graph the system to determine the type of system and solution. If the two lines are parallel, the system has no solution and is inconsistent. If the two lines are identical, the system has infinite solutions and is a dependent system.

Plot the three different systems with an online graphing tool. Categorize each solution as either consistent or inconsistent. If the system is consistent determine whether it is dependent or independent. You may find it easier to plot each system individually, then clear out your entries before you plot the next. 1) [latex]5x-3y = -19[/latex] [latex]x=2y-1[/latex]

2) [latex]4x+y=11[/latex] [latex]-2y=-25+8x[/latex]

3) [latex]y = -3x+6[/latex] [latex]-\frac{1}{3}y+2=x[/latex]

  • One solution – consistent, independent
  • No solutions, inconsistent, neither dependent nor independent
  • Many solutions –  consistent, dependent

Solving Systems of Equations by Substitution

Solving a linear system in two variables by graphing works well when the solution consists of integer values, but if our solution contains decimals or fractions, it is not the most precise method. We will consider two more methods of solving a system of linear equations that are more precise than graphing. One such method is solving a system of equations by the substitution method , in which we solve one of the equations for one variable and then substitute the result into the second equation to solve for the second variable. Recall that we can solve for only one variable at a time, which is the reason the substitution method is both valuable and practical.

How To: Given a system of two equations in two variables, solve using the substitution method.

  • Solve one of the two equations for one of the variables in terms of the other.
  • Substitute the expression for this variable into the second equation, then solve for the remaining variable.
  • Substitute that solution into either of the original equations to find the value of the first variable. If possible, write the solution as an ordered pair.
  • Check the solution in both equations.

Example: Solving a System of Equations in Two Variables by Substitution

Solve the following system of equations by substitution.

[latex]\begin{align}-x+y&=-5 \\ 2x-5y&=1 \end{align}[/latex]

First, we will solve the first equation for [latex]y[/latex].

[latex]\begin{align}-x+y&=-5 \\ y&=x - 5 \end{align}[/latex]

Now we can substitute the expression [latex]x - 5[/latex] for [latex]y[/latex] in the second equation.

[latex]\begin{align}2x - 5y&=1 \\ 2x - 5\left(x - 5\right)&=1 \\ 2x - 5x+25&=1 \\ -3x&=-24 \\ x&=8 \end{align}[/latex]

Now, we substitute [latex]x=8[/latex] into the first equation and solve for [latex]y[/latex].

[latex]\begin{align}-\left(8\right)+y&=-5 \\ y&=3 \end{align}[/latex]

Our solution is [latex]\left(8,3\right)[/latex].

Check the solution by substituting [latex]\left(8,3\right)[/latex] into both equations.

[latex]\begin{align}-x+y&=-5 \\ -\left(8\right)+\left(3\right)&=-5 && \text{True} \\[3mm] 2x - 5y&=1 \\ 2\left(8\right)-5\left(3\right)&=1 && \text{True} \end{align}[/latex]

[latex]\begin{align}x&=y+3 \\ 4&=3x - 2y \end{align}[/latex]

[latex]\left(-2,-5\right)[/latex]

Can the substitution method be used to solve any linear system in two variables?

Yes, but the method works best if one of the equations contains a coefficient of 1 or –1 so that we do not have to deal with fractions.

The following video is ~10 minutes long and provides a mini-lesson on using the substitution method to solve a system of linear equations.  We present three different examples, and also use a graphing tool to help summarize the solution for each example.

Solving Systems of Equations in Two Variables by the Addition Method

A third method of solving systems of linear equations is the addition method,  this method is also called the  elimination method . In this method, we add two terms with the same variable, but opposite coefficients, so that the sum is zero. Of course, not all systems are set up with the two terms of one variable having opposite coefficients. Often we must adjust one or both of the equations by multiplication so that one variable will be eliminated by addition.

How To: Given a system of equations, solve using the addition method.

  • Write both equations with x – and y -variables on the left side of the equal sign and constants on the right.
  • Write one equation above the other, lining up corresponding variables. If one of the variables in the top equation has the opposite coefficient of the same variable in the bottom equation, add the equations together, eliminating one variable. If not, use multiplication by a nonzero number so that one of the variables in the top equation has the opposite coefficient of the same variable in the bottom equation, then add the equations to eliminate the variable.
  • Solve the resulting equation for the remaining variable.
  • Substitute that value into one of the original equations and solve for the second variable.
  • Check the solution by substituting the values into the other equation.

Example: Solving a System by the Addition Method

Solve the given system of equations by addition.

[latex]\begin{align}x+2y&=-1 \\ -x+y&=3 \end{align}[/latex]

Both equations are already set equal to a constant. Notice that the coefficient of [latex]x[/latex] in the second equation, –1, is the opposite of the coefficient of [latex]x[/latex] in the first equation, 1. We can add the two equations to eliminate [latex]x[/latex] without needing to multiply by a constant.

[latex]\begin{align} x+2y&=-1 \\ -x+y&=3 \\ \hline 3y&=2\end{align}[/latex]

Now that we have eliminated [latex]x[/latex], we can solve the resulting equation for [latex]y[/latex].

[latex]\begin{align}3y&=2 \\ y&=\dfrac{2}{3} \end{align}[/latex]

Then, we substitute this value for [latex]y[/latex] into one of the original equations and solve for [latex]x[/latex].

[latex]\begin{align}-x+y&=3 \\ -x+\frac{2}{3}&=3 \\ -x&=3-\frac{2}{3} \\ -x&=\frac{7}{3} \\ x&=-\frac{7}{3} \end{align}[/latex]

The solution to this system is [latex]\left(-\frac{7}{3},\frac{2}{3}\right)[/latex].

Check the solution in the first equation.

[latex]\begin{align}x+2y&=-1 \\ \left(-\frac{7}{3}\right)+2\left(\frac{2}{3}\right)&= \\ -\frac{7}{3}+\frac{4}{3}&= \\ \-\frac{3}{3}&= \\ -1&=-1&& \text{True} \end{align}[/latex]

We gain an important perspective on systems of equations by looking at the graphical representation. See the graph below to find that the equations intersect at the solution. We do not need to ask whether there may be a second solution because observing the graph confirms that the system has exactly one solution.

A graph of two lines that cross at the point negative seven-thirds, two-thirds. The first line's equation is x+2y=negative 1. The second line's equation is negative x + y equals 3.

Example: Using the Addition Method When Multiplication of One Equation Is Required

Solve the given system of equations by the addition method .

[latex]\begin{align}3x+5y&=-11 \\ x - 2y&=11 \end{align}[/latex]

Adding these equations as presented will not eliminate a variable. However, we see that the first equation has [latex]3x[/latex] in it and the second equation has [latex]x[/latex]. So if we multiply the second equation by [latex]-3,\text{}[/latex] the x -terms will add to zero.

[latex]\begin{align}x - 2y&=11 \\ -3\left(x - 2y\right)&=-3\left(11\right) && \text{Multiply both sides by }-3 \\ -3x+6y&=-33 && \text{Use the distributive property}. \end{align}[/latex]

Now, let’s add them.

[latex]\begin{align}3x+5y&=−11 \\ −3x+6y&=−33 \\ \hline 11y&=−44 \\ y&=−4 \end{align}[/latex]

For the last step, we substitute [latex]y=-4[/latex] into one of the original equations and solve for [latex]x[/latex].

[latex]\begin{align}3x+5y&=-11\\ 3x+5\left(-4\right)&=-11\\ 3x - 20&=-11\\ 3x&=9\\ x&=3\end{align}[/latex]

Our solution is the ordered pair [latex]\left(3,-4\right)[/latex]. Check the solution in the original second equation.

[latex]\begin{align}x - 2y&=11 \\ \left(3\right)-2\left(-4\right)&=3+8 \\ &=11 && \text{True} \end{align}[/latex]

A graph of two lines that cross at the point 3, negative 4. The first line's equation is 3x+5y=-11. The second line's equation is x-2y=11.

Solve the system of equations by addition.

[latex]\begin{align}2x - 7y&=2\\ 3x+y&=-20\end{align}[/latex]

[latex]\left(-6,-2\right)[/latex]

Example: Using the Addition Method When Multiplication of Both Equations Is Required

Solve the given system of equations in two variables by addition.

[latex]\begin{align}2x+3y&=-16 \\ 5x - 10y&=30\end{align}[/latex]

One equation has [latex]2x[/latex] and the other has [latex]5x[/latex]. The least common multiple is [latex]10x[/latex] so we will have to multiply both equations by a constant in order to eliminate one variable. Let’s eliminate [latex]x[/latex] by multiplying the first equation by [latex]-5[/latex] and the second equation by [latex]2[/latex].

[latex]\begin{align} -5\left(2x+3y\right)&=-5\left(-16\right) \\ -10x - 15y&=80 \\[3mm] 2\left(5x - 10y\right)&=2\left(30\right) \\ 10x - 20y&=60 \end{align}[/latex]

Then, we add the two equations together.

[latex]\begin{align} −10x−15y&=80 \\ 10x−20y&=60 \\ \hline −35y&=140 \\ y&=−4 \end{align}[/latex]

Substitute [latex]y=-4[/latex] into the original first equation.

[latex]\begin{align}2x+3\left(-4\right)&=-16\\ 2x - 12&=-16\\ 2x&=-4\\ x&=-2\end{align}[/latex]

The solution is [latex]\left(-2,-4\right)[/latex]. Check it in the other equation.

[latex]\begin{align} 5x - 10y&=30\\ 5\left(-2\right)-10\left(-4\right)&=30\\ -10+40&=30\\ 30&=30\end{align}[/latex]

A graph of two lines that cross the point -2,-4. The first line's equation is 2x+3y=-16. The second line's equation is 5x-10y=30.

Example: Using the Addition Method in Systems of Equations Containing Fractions

[latex]\begin{align}\frac{x}{3}+\frac{y}{6}&=3 \\[1mm] \frac{x}{2}-\frac{y}{4}&=1 \end{align}[/latex]

First clear each equation of fractions by multiplying both sides of the equation by the least common denominator.

[latex]\begin{align}6\left(\frac{x}{3}+\frac{y}{6}\right)&=6\left(3\right) \\[1mm] 2x+y&=18 \\[3mm] 4\left(\frac{x}{2}-\frac{y}{4}\right)&=4\left(1\right) \\[1mm] 2x-y&=4 \end{align}[/latex]

Now multiply the second equation by [latex]-1[/latex] so that we can eliminate  x .

[latex]\begin{align}-1\left(2x-y\right)&=-1\left(4\right) \\[1mm] -2x+y&=-4 \end{align}[/latex]

Add the two equations to eliminate  x  and solve the resulting equation for y .

[latex]\begin{align} 2x+y&=18 \\ −2x+y&=−4 \\ \hline 2y&=14 \\ y&=7 \end{align}[/latex]

Substitute [latex]y=7[/latex] into the first equation.

[latex]\begin{align}2x+\left(7\right)&=18 \\ 2x&=11 \\ x&=\frac{11}{2} \\ &=7.5 \end{align}[/latex]

The solution is [latex]\left(\frac{11}{2},7\right)[/latex]. Check it in the other equation.

[latex]\begin{align}\frac{x}{2}-\frac{y}{4}&=1\\[1mm] \frac{\frac{11}{2}}{2}-\frac{7}{4}&=1\\[1mm] \frac{11}{4}-\frac{7}{4}&=1\\[1mm] \frac{4}{4}&=1\end{align}[/latex]

[latex]\begin{align}2x+3y&=8\\ 3x+5y&=10\end{align}[/latex]

[latex]\left(10,-4\right)[/latex]

in the following video we present more examples of how to use the addition (elimination) method to solve a system of two linear equations.

Classify Solutions to Systems

Now that we have several methods for solving systems of equations, we can use the methods to identify inconsistent systems. Recall that an inconsistent system consists of parallel lines that have the same slope but different [latex]y[/latex] -intercepts. They will never intersect. When searching for a solution to an inconsistent system, we will come up with a false statement, such as [latex]12=0[/latex].

Example: Solving an Inconsistent System of Equations

Solve the following system of equations.

[latex]\begin{gathered}&x=9 - 2y \\ &x+2y=13 \end{gathered}[/latex]

We can approach this problem in two ways. Because one equation is already solved for [latex]x[/latex], the most obvious step is to use substitution.

[latex]\begin{align}x+2y&=13 \\ \left(9 - 2y\right)+2y&=13 \\ 9+0y&=13 \\ 9&=13 \end{align}[/latex]

Clearly, this statement is a contradiction because [latex]9\ne 13[/latex]. Therefore, the system has no solution.

The second approach would be to first manipulate the equations so that they are both in slope-intercept form. We manipulate the first equation as follows.

[latex]\begin{gathered}x=9 - 2y \\ 2y=-x+9 \\ y=-\frac{1}{2}x+\frac{9}{2} \end{gathered}[/latex]

We then convert the second equation expressed to slope-intercept form.

[latex]\begin{gathered}x+2y=13 \\ 2y=-x+13 \\ y=-\frac{1}{2}x+\frac{13}{2} \end{gathered}[/latex]

Comparing the equations, we see that they have the same slope but different y -intercepts. Therefore, the lines are parallel and do not intersect.

[latex]\begin{gathered}y=-\frac{1}{2}x+\frac{9}{2} \\ y=-\frac{1}{2}x+\frac{13}{2} \end{gathered}[/latex]

Writing the equations in slope-intercept form confirms that the system is inconsistent because all lines will intersect eventually unless they are parallel. Parallel lines will never intersect; thus, the two lines have no points in common. The graphs of the equations in this example are shown below.

A graph of two parallel lines. The first line's equation is y equals negative one-half x plus 13 over two. The second line's equation is y equals negative one-half x plus 9 over two.

Solve the following system of equations in two variables.

[latex]\begin{gathered}2y - 2x=2\\ 2y - 2x=6\end{gathered}[/latex]

No solution. It is an inconsistent system.

Expressing the Solution of a System of Dependent Equations Containing Two Variables

Recall that a dependent system of equations in two variables is a system in which the two equations represent the same line. Dependent systems have an infinite number of solutions because all of the points on one line are also on the other line. After using substitution or addition, the resulting equation will be an identity, such as [latex]0=0[/latex].

Example: Finding a Solution to a Dependent System of Linear Equations

Find a solution to the system of equations using the addition method .

[latex]\begin{gathered}x+3y=2\\ 3x+9y=6\end{gathered}[/latex]

With the addition method, we want to eliminate one of the variables by adding the equations. In this case, let’s focus on eliminating [latex]x[/latex]. If we multiply both sides of the first equation by [latex]-3[/latex], then we will be able to eliminate the [latex]x[/latex] -variable.

[latex]\begin{align}x+3y&=2 \\ \left(-3\right)\left(x+3y\right)&=\left(-3\right)\left(2\right) \\ -3x - 9y&=-6 \end{align}[/latex]

Now add the equations.

[latex]\begin{align} −3x−9y&=−6 \\ +3x+9y&=6 \\ \hline 0&=0 \end{align}[/latex]

We can see that there will be an infinite number of solutions that satisfy both equations.

If we rewrote both equations in the slope-intercept form, we might know what the solution would look like before adding. Let’s look at what happens when we convert the system to slope-intercept form.

[latex]\begin{align}\begin{gathered}x+3y=2 \\ 3y=-x+2 \\ y=-\frac{1}{3}x+\frac{2}{3} \end{gathered} \hspace{2cm} \begin{gathered} 3x+9y=6 \\9y=-3x+6 \\ y=-\frac{3}{9}x+\frac{6}{9} \\ y=-\frac{1}{3}x+\frac{2}{3} \end{gathered}\end{align}[/latex]

Look at the graph below. Notice the results are the same. The general solution to the system is [latex]\left(x, -\frac{1}{3}x+\frac{2}{3}\right)[/latex].

A graph of two lines that overlap each other. The first line's equation is x+3y=2. The second line's equation is 3x-9y=6.

Writing the general solution

In the previous example, we presented an analysis of the solution to the following system of equations:

After a little algebra, we found that these two equations were exactly the same. We then wrote the general solution as [latex]\left(x, -\frac{1}{3}x+\frac{2}{3}\right)[/latex]. Why would we write the solution this way? In some ways, this representation tells us a lot.  It tells us that x can be anything, x is x .  It also tells us that y is going to depend on x , just like when we write a function rule.  In this case, depending on what you put in for x , y will be defined in terms of x as [latex]-\frac{1}{3}x+\frac{2}{3}[/latex].

In other words, there are infinitely many ( x , y ) pairs that will satisfy this system of equations, and they all fall on the line [latex]f(x)-\frac{1}{3}x+\frac{2}{3}[/latex].

[latex]\begin{gathered}y - 2x=5 \\ -3y+6x=-15 \end{gathered}[/latex]

The system is dependent so there are infinitely many solutions of the form [latex]\left(x,2x+5\right)[/latex].

Using Systems of Equations to Investigate Profits

Using what we have learned about systems of equations, we can return to the skateboard manufacturing problem at the beginning of the section. The skateboard manufacturer’s revenue function is the function used to calculate the amount of money that comes into the business. It can be represented by the equation [latex]R=xp[/latex], where [latex]x=[/latex] quantity and [latex]p=[/latex] price. The revenue function is shown in orange in the graph below.

The cost function is the function used to calculate the costs of doing business. It includes fixed costs, such as rent and salaries, and variable costs, such as utilities. The cost function is shown in blue in the graph below. The x -axis represents quantity in hundreds of units. The y -axis represents either cost or revenue in hundreds of dollars.

A graph showing money in hundreds of dollars on the y axis and quantity in hundreds of units on the x axis. A line representing cost and a line representing revenue cross at the point (7,33), which is marked break-even. The shaded space between the two lines to the right of the break-even point is labeled profit.

The point at which the two lines intersect is called the break-even point . We can see from the graph that if 700 units are produced, the cost is $3,300 and the revenue is also $3,300. In other words, the company breaks even if they produce and sell 700 units. They neither make money nor lose money.

The shaded region to the right of the break-even point represents quantities for which the company makes a profit. The shaded region to the left represents quantities for which the company suffers a loss. The profit function is the revenue function minus the cost function, written as [latex]P\left(x\right)=R\left(x\right)-C\left(x\right)[/latex]. Clearly, knowing the quantity for which the cost equals the revenue is of great importance to businesses.

Example: Finding the Break-Even Point and the Profit Function Using Substitution

Given the cost function [latex]C\left(x\right)=0.85x+35{,}000[/latex] and the revenue function [latex]R\left(x\right)=1.55x[/latex], find the break-even point and the profit function.

Write the system of equations using [latex]y[/latex] to replace function notation.

[latex]\begin{align} y&=0.85x+35{,}000 \\ y&=1.55x \end{align}[/latex]

Substitute the expression [latex]0.85x+35{,}000[/latex] from the first equation into the second equation and solve for [latex]x[/latex].

[latex]\begin{gathered}0.85x+35{,}000=1.55x\\ 35{,}000=0.7x\\ 50{,}000=x\end{gathered}[/latex]

Then, we substitute [latex]x=50{,}000[/latex] into either the cost function or the revenue function.

[latex]1.55\left(50{,}000\right)=77{,}500[/latex]

The break-even point is [latex]\left(50{,}000,77{,}500\right)[/latex].

The profit function is found using the formula [latex]P\left(x\right)=R\left(x\right)-C\left(x\right)[/latex].

[latex]\begin{align}P\left(x\right)&=1.55x-\left(0.85x+35{,}000\right) \\ &=0.7x - 35{,}000 \end{align}[/latex]

The profit function is [latex]P\left(x\right)=0.7x - 35{,}000[/latex].

The cost to produce 50,000 units is $77,500, and the revenue from the sales of 50,000 units is also $77,500. To make a profit, the business must produce and sell more than 50,000 units.

A graph showing money in dollars on the y axis and quantity on the x axis. A line representing cost and a line representing revenue cross at the break-even point of fifty thousand, seventy-seven thousand five hundred. The cost line's equation is C(x)=0.85x+35,000. The revenue line's equation is R(x)=1.55x. The shaded space between the two lines to the right of the break-even point is labeled profit.

We see from the graph below that the profit function has a negative value until [latex]x=50{,}000[/latex], when the graph crosses the x -axis. Then, the graph emerges into positive y -values and continues on this path as the profit function is a straight line. This illustrates that the break-even point for businesses occurs when the profit function is 0. The area to the left of the break-even point represents operating at a loss.

A graph showing dollars profit on the y axis and quantity on the x axis. The profit line crosses the break-even point at fifty thousand, zero. The profit line's equation is P(x)=0.7x-35,000.

Writing a System of Linear Equations Given a Situation

It is rare to be given equations that neatly model behaviors that you encounter in business, rather, you will probably be faced with a situation for which you know key information as in the example above. Below, we summarize three key factors that will help guide you in translating a situation into a system.

How To: Given a situation that represents a system of linear equations, write the system of equations and identify the solution.

  • Identify the input and output of each linear model.
  • Identify the slope and y -intercept of each linear model.
  • Find the solution by setting the two linear functions equal to another and solving for x , or find the point of intersection on a graph.

Now let’s practice putting these key factors to work. In the next example, we determine how many different types of tickets are sold given information about the total revenue and amount of tickets sold to an event.

Example: Writing and Solving a System of Equations in Two Variables

The cost of a ticket to the circus is $25.00 for children and $50.00 for adults. On a certain day, attendance at the circus is 2,000 and the total gate revenue is $70,000. How many children and how many adults bought tickets?

Let c = the number of children and a = the number of adults in attendance.

The total number of people is 2,000. We can use this to write an equation for the number of people at the circus that day.

[latex]c+a=2{,}000[/latex]

The revenue from all children can be found by multiplying $25.00 by the number of children, [latex]25c[/latex]. The revenue from all adults can be found by multiplying $50.00 by the number of adults, [latex]50a[/latex]. The total revenue is $70,000. We can use this to write an equation for the revenue.

[latex]25c+50a=70{,}000[/latex]

We now have a system of linear equations in two variables.

[latex]\begin{gathered}c+a=2,000\\ 25c+50a=70{,}000\end{gathered}[/latex]

In the first equation, the coefficient of both variables is 1. We can quickly solve the first equation for either [latex]c[/latex] or [latex]a[/latex]. We will solve for [latex]a[/latex].

[latex]\begin{gathered}c+a=2{,}000\\ a=2{,}000-c\end{gathered}[/latex]

Substitute the expression [latex]2{,}000-c[/latex] in the second equation for [latex]a[/latex] and solve for [latex]c[/latex].

[latex]\begin{align} 25c+50\left(2{,}000-c\right)&=70{,}000 \\ 25c+100{,}000 - 50c&=70{,}000 \\ -25c&=-30{,}000 \\ c&=1{,}200 \end{align}[/latex]

Substitute [latex]c=1{,}200[/latex] into the first equation to solve for [latex]a[/latex].

[latex]\begin{align}1{,}200+a&=2{,}000 \\ a&=800 \end{align}[/latex]

We find that 1,200 children and 800 adults bought tickets to the circus that day.

Meal tickets at the circus cost $4.00 for children and $12.00 for adults. If 1,650 meal tickets were bought for a total of $14,200, how many children and how many adults bought meal tickets?

700 children, 950 adults

Sometimes, a system of equations can inform a decision.  In our next example, we help answer the question, “Which truck rental company will give the best value?”

Example: Building a System of Linear Models to Choose a Truck Rental Company

Jamal is choosing between two truck-rental companies. The first, Keep on Trucking, Inc., charges an up-front fee of $20, then 59 cents a mile. The second, Move It Your Way, charges an up-front fee of $16, then 63 cents a mile. [1] When will Keep on Trucking, Inc. be the better choice for Jamal?

The two important quantities in this problem are the cost and the number of miles driven. Because we have two companies to consider, we will define two functions.

A linear function is of the form [latex]f\left(x\right)=mx+b[/latex]. Using the rates of change and initial charges, we can write the equations

[latex]\begin{align}K\left(d\right)=0.59d+20\\ M\left(d\right)=0.63d+16\end{align}[/latex]

Using these equations, we can determine when Keep on Trucking, Inc., will be the better choice. Because all we have to make that decision from is the costs, we are looking for when Move It Your Way, will cost less, or when [latex]K\left(d\right)<M\left(d\right)[/latex]. The solution pathway will lead us to find the equations for the two functions, find the intersection, and then see where the [latex]K\left(d\right)[/latex] function is smaller.

examples of solving problems involving linear equations in two variables

These graphs are sketched above, with K ( d ) in blue.

To find the intersection, we set the equations equal and solve:

[latex]\begin{align}K\left(d\right)&=M\left(d\right) \\ 0.59d+20&=0.63d+16 \\ 4&=0.04d \\ 100&=d \\ d&=100 \end{align}[/latex]

This tells us that the cost from the two companies will be the same if 100 miles are driven. Either by looking at the graph, or noting that [latex]K\left(d\right)[/latex] is growing at a slower rate, we can conclude that Keep on Trucking, Inc. will be the cheaper price when more than 100 miles are driven, that is [latex]d>100[/latex].

The applications for systems seems almost endless, but we will just show one more. In the next example, we determine the amount 80% methane solution to add to a 50% solution to give a final solution of 60%.

Example: Solve a Chemical Mixture Problem

A chemist has 70 mL of a 50% methane solution. How much of a 80% solution must she add so the final solution is 60% methane?

We will use the following table to help us solve this mixture problem:

We start with 70 mL of solution, and the unknown amount can be x . The part is the percentages, or concentration of solution 0.5 for start, 0.8 for add.

Add amount column to get final amount. The part for this amount is 0.6 because we want the final solution to be 60% methane.

Multiply amount by part to get total. be sure to distribute on the last row:[latex](70 + x)0.6[/latex].

If we add the start and add entries in the Total column, we get the final equation that represents the total amount and it’s concentration.

[latex]\begin{align}35+0.8x& = 42+0.6x \\ 0.2x&=7 \\ \frac{0.2}{0.2}x&=\frac{7}{0.2} \\ x&=35 \end{align}[/latex]

35mL of 80% solution must be added to 70mL of 50% solution to get a 60% solution of Methane.

The same process can be used if the starting and final amount have a price attached to them, rather than a percentage.

Key Concepts

  • A system of linear equations consists of two or more equations made up of two or more variables such that all equations in the system are considered simultaneously.
  • The solution to a system of linear equations in two variables is any ordered pair that satisfies each equation independently.
  • Systems of equations are classified as independent with one solution, dependent with an infinite number of solutions, or inconsistent with no solution.
  • One method of solving a system of linear equations in two variables is by graphing. In this method, we graph the equations on the same set of axes.
  • Another method of solving a system of linear equations is by substitution. In this method, we solve for one variable in one equation and substitute the result into the second equation.
  • A third method of solving a system of linear equations is by addition, in which we can eliminate a variable by adding opposite coefficients of corresponding variables.
  • It is often necessary to multiply one or both equations by a constant to facilitate elimination of a variable when adding the two equations together.
  • Either method of solving a system of equations results in a false statement for inconsistent systems because they are made up of parallel lines that never intersect.
  • The solution to a system of dependent equations will always be true because both equations describe the same line.
  • Systems of equations can be used to solve real-world problems that involve more than one variable, such as those relating to revenue, cost, and profit.

addition method an algebraic technique used to solve systems of linear equations in which the equations are added in a way that eliminates one variable, allowing the resulting equation to be solved for the remaining variable; substitution is then used to solve for the first variable

break-even point the point at which a cost function intersects a revenue function; where profit is zero

consistent system a system for which there is a single solution to all equations in the system and it is an independent system, or if there are an infinite number of solutions and it is a dependent system

cost function the function used to calculate the costs of doing business; it usually has two parts, fixed costs and variable costs

dependent system a system of linear equations in which the two equations represent the same line; there are an infinite number of solutions to a dependent system

inconsistent system a system of linear equations with no common solution because they represent parallel lines, which have no point or line in common

independent system a system of linear equations with exactly one solution pair [latex]\left(x,y\right)[/latex]

profit function the profit function is written as [latex]P\left(x\right)=R\left(x\right)-C\left(x\right)[/latex], revenue minus cost

revenue function the function that is used to calculate revenue, simply written as [latex]R=xp[/latex], where [latex]x=[/latex] quantity and [latex]p=[/latex] price

substitution method an algebraic technique used to solve systems of linear equations in which one of the two equations is solved for one variable and then substituted into the second equation to solve for the second variable

system of linear equations a set of two or more equations in two or more variables that must be considered simultaneously.

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Chapter 4 Linear Functions

4.1 Linear Equations in Two Variables

Learning objectives.

In this section you will:

  • Write a linear equation in two variables
  • Given the equations of two lines, determine whether their graphs are parallel or perpendicular.
  • Write the equation of a line parallel or perpendicular to a given line.

Write a Linear Equation in Two Variables

Perhaps the most familiar form of a linear equation is the slope-intercept form, written as[latex]\,y=mx+b,[/latex]where[latex]\,m=\text{slope}\,[/latex]and[latex]\,b=y\text{−intercept}\text{.}\,[/latex]Let us begin with the slope.

The Slope of a Line

The slope of a line refers to the ratio of the vertical change in y over the horizontal change in x between any two points on a line. It indicates the direction in which a line slants as well as its steepness. Slope is sometimes described as rise over run.

Coordinate plane with the x and y axes ranging from negative 10 to 10. Three linear functions are plotted: y = negative 3 times x minus 2; y = 2 times x plus 1; and y = x over 3 plus 2.

The slope of a line, m , represents the change in y over the change in x. Given two points,[latex]\,\left({x}_{1},{y}_{1}\right)\,[/latex]and[latex]\,\left({x}_{2},{y}_{2}\right),[/latex]the following formula determines the slope of a line containing these points:

Finding the Slope of a Line Given Two Points

Find the slope of a line that passes through the points[latex]\,\left(2,-1\right)\,[/latex]and[latex]\,\left(-5,3\right).[/latex]

We substitute the y- values and the x- values into the formula.

The slope is[latex]\,-\frac{4}{7}.[/latex]

It does not matter which point is called[latex]\,\left({x}_{1},{y}_{1}\right)\,[/latex]or[latex]\,\left({x}_{2},{y}_{2}\right).\,[/latex]As long as we are consistent with the order of the y terms and the order of the x terms in the numerator and denominator, the calculation will yield the same result.

Find the slope of the line that passes through the points[latex]\,\left(-2,6\right)\,[/latex]and[latex]\,\left(1,4\right).[/latex]

[latex]m=-\frac{2}{3}[/latex]

Identifying the Slope and y- intercept of a Line Given an Equation

Identify the slope and y- intercept, given the equation[latex]\,y=-\frac{3}{4}x-4.[/latex]

As the line is in[latex]\,y=mx+b\,[/latex]form, the given line has a slope of[latex]\,m=-\frac{3}{4}.\,[/latex]The y- intercept is[latex]\,b=-4.[/latex]

The y -intercept is the point at which the line crosses the y- axis. On the y- axis,[latex]\,x=0.\,[/latex]We can always identify the y- intercept when the line is in slope-intercept form, as it will always equal b. Or, just substitute[latex]\,x=0\,[/latex]and solve for y.

The Point-Slope Formula

Given the slope and one point on a line, we can find the equation of the line using the point-slope formula.

This is an important formula, as it will be used in other areas of college algebra and often in calculus to find the equation of a tangent line. We need only one point and the slope of the line to use the formula. After substituting the slope and the coordinates of one point into the formula, we simplify it and write it in slope-intercept form.

Given one point and the slope, the point-slope formula will lead to the equation of a line:

Finding the Equation of a Line Given the Slope and One Point

Write the equation of the line with slope[latex]\,m=-3\,[/latex]and passing through the point[latex]\,\left(4,8\right).\,[/latex]Write the final equation in slope-intercept form.

Using the point-slope formula, substitute[latex]\,-3\,[/latex]for m and the point[latex]\,\left(4,8\right)\,[/latex]for[latex]\,\left({x}_{1},{y}_{1}\right).[/latex]

Note that any point on the line can be used to find the equation. If done correctly, the same final equation will be obtained.

Given[latex]\,m=4,[/latex]find the equation of the line in slope-intercept form passing through the point[latex]\,\left(2,5\right).[/latex]

[latex]y=4x-3[/latex]

Finding the Equation of a Line Passing Through Two Given Points

Find the equation of the line passing through the points[latex]\,\left(3,4\right)\,[/latex]and[latex]\,\left(0,-3\right).\,[/latex]Write the final equation in slope-intercept form.

First, we calculate the slope using the slope formula and two points.

Next, we use the point-slope formula with the slope of[latex]\,\frac{7}{3},[/latex]and either point. Let’s pick the point[latex]\,\left(3,4\right)\,[/latex]for[latex]\,\left({x}_{1},{y}_{1}\right).[/latex]

In slope-intercept form, the equation is written as[latex]\,y=\frac{7}{3}x-3.[/latex]

To prove that either point can be used, let us use the second point[latex]\,\left(0,-3\right)\,[/latex]and see if we get the same equation.

We see that the same line will be obtained using either point. This makes sense because we used both points to calculate the slope.

Standard Form of a Line

Another way that we can represent the equation of a line is in standard form . Standard form is given as

where[latex]\,A,[/latex][latex]B,[/latex]and[latex]\,C[/latex]are integers. The x- and y- terms are on one side of the equal sign and the constant term is on the other side.

Finding the Equation of a Line and Writing It in Standard Form

Find the equation of the line with[latex]\,m=-6\,[/latex]and passing through the point[latex]\,\left(\frac{1}{4},-2\right).\,[/latex]Write the equation in standard form.

We begin using the point-slope formula.

Find the equation of the line in standard form with slope[latex]\,m=-\frac{1}{3}\,[/latex]and passing through the point[latex]\,\left(1,\frac{1}{3}\right).[/latex]

[latex]x+3y=2[/latex]

Vertical and Horizontal Lines

The equations of vertical and horizontal lines do not require any of the preceding formulas, although we can use the formulas to prove that the equations are correct. The equation of a vertical line is given as

where c is a constant. The slope of a vertical line is undefined, and regardless of the y- value of any point on the line, the x- coordinate of the point will be c .

Suppose that we want to find the equation of a line containing the following points:[latex]\,\left(-3,-5\right),\left(-3,1\right),\left(-3,3\right),[/latex]and[latex]\,\left(-3,5\right).\,[/latex]First, we will find the slope.

Zero in the denominator means that the slope is undefined and, therefore, we cannot use the point-slope formula. However, we can plot the points. Notice that all of the x- coordinates are the same and we find a vertical line through[latex]\,x=-3.\,[/latex]See Figure 2 .

The equation of a horizontal line is given as

where c is a constant. The slope of a horizontal line is zero, and for any x- value of a point on the line, the y- coordinate will be c .

Suppose we want to find the equation of a line that contains the following set of points:[latex]\,\left(-2,-2\right),\left(0,-2\right),\left(3,-2\right),[/latex]and[latex]\,\left(5,-2\right).[/latex]We can use the point-slope formula. First, we find the slope using any two points on the line.

Use any point for[latex]\,\left({x}_{1},{y}_{1}\right)\,[/latex]in the formula, or use the y -intercept.

The graph is a horizontal line through[latex]\,y=-2.\,[/latex]Notice that all of the y- coordinates are the same. See Figure 2.

Coordinate plane with the x-axis ranging from negative 7 to 4 and the y-axis ranging from negative 4 to 4. The function y = negative 2 and the line x = negative 3 are plotted.

Finding the Equation of a Line Passing Through the Given Points

Find the equation of the line passing through the given points:[latex]\,\left(1,-3\right)\,[/latex]and[latex]\,\left(1,4\right).[/latex]

The x- coordinate of both points is 1. Therefore, we have a vertical line,[latex]\,x=1.[/latex]

Find the equation of the line passing through[latex]\,\left(-5,2\right)\,[/latex]and[latex]\,\left(2,2\right).[/latex]

Horizontal line:[latex]\,y=2[/latex]

Determining Whether Graphs of Lines are Parallel or Perpendicular

Parallel lines have the same slope and different y- intercepts. Lines that are parallel to each other will never intersect. For example, Figure 3 shows the graphs of various lines with the same slope,[latex]\,m=2.[/latex]

Coordinate plane with the x-axis ranging from negative 8 to 8 in intervals of 2 and the y-axis ranging from negative 7 to 7. Three functions are graphed on the same plot: y = 2 times x minus 3; y = 2 times x plus 1 and y = 2 times x plus 5.

All of the lines shown in the graph are parallel because they have the same slope and different y- intercepts.

Lines that are perpendicular intersect to form a[latex]\,90°[/latex]-angle. The slope of one line is the negative reciprocal of the other. We can show that two lines are perpendicular if the product of the two slopes is[latex]\,-1:{m}_{1}\cdot {m}_{2}=-1.\,[/latex]For example, Figure 4 shows the graph of two perpendicular lines. One line has a slope of 3; the other line has a slope of[latex]\,-\frac{1}{3}.[/latex]

Coordinate plane with the x-axis ranging from negative 3 to 6 and the y-axis ranging from negative 2 to 5. Two functions are graphed on the same plot: y = 3 times x minus 1 and y = negative x/3 minus 2. Their intersection is marked by a box to show that it is a right angle.

Graphing Two Equations, and Determining Whether the Lines are Parallel, Perpendicular, or Neither

Graph the equations of the given lines, and state whether they are parallel, perpendicular, or neither:[latex]\,3y=-4x+3\,[/latex]and[latex]\,3x-4y=8.[/latex]

The first thing we want to do is rewrite the equations so that both equations are in slope-intercept form.

First equation:

Second equation:

See the graph of both lines in Figure 5.

Coordinate plane with the x-axis ranging from negative 4 to 5 and the y-axis ranging from negative 4 to 4. Two functions are graphed on the same plot: y = negative 4 times x/3 plus 1 and y = 3 times x/4 minus 2. A box is placed at the intersection to note that it forms a right angle.

From the graph, we can see that the lines appear perpendicular, but we must compare the slopes.

The slopes are negative reciprocals of each other, confirming that the lines are perpendicular.

Graph the two lines and determine whether they are parallel, perpendicular, or neither:[latex]\,2y-x=10\,[/latex]and[latex]\,2y=x+4.[/latex]

Parallel lines: equations are written in slope-intercept form.

Coordinate plane with the x-axis ranging from negative 5 to 5 and the y-axis ranging from negative 1 to 6. Two functions are graphed on the same plot: y = x/2 plus 5 and y = x/2 plus 2. The lines do not cross.

Writing the Equations of Lines Parallel or Perpendicular to a Given Line

As we have learned, determining whether two lines are parallel or perpendicular is a matter of finding the slopes. To write the equation of a line parallel or perpendicular to another line, we follow the same principles as we do for finding the equation of any line. After finding the slope, use the point-slope formula to write the equation of the new line.

Given an equation for a line, write the equation of a line parallel or perpendicular to it.

  • Find the slope of the given line. The easiest way to do this is to write the equation in slope-intercept form.
  • Use the slope and the given point with the point-slope formula.
  • Simplify the line to slope-intercept form and compare the equation to the given line.

Writing the Equation of a Line Parallel to a Given Line Passing Through a Given Point

Write the equation of line parallel to a[latex]\,5x+3y=1\,[/latex]and passing through the point[latex]\,\left(3,5\right).[/latex]

First, we will write the equation in slope-intercept form to find the slope.

The slope is[latex]\,m=-\frac{5}{3}.\,[/latex]The y- intercept is[latex]\,\frac{1}{3},[/latex]but that really does not enter into our problem, as the only thing we need for two lines to be parallel is the same slope. The one exception is that if the y- intercepts are the same, then the two lines are the same line. The next step is to use this slope and the given point with the point-slope formula.

The equation of the line is[latex]\,y=-\frac{5}{3}x+10.\,[/latex]See Figure 6.

Coordinate plane with the x-axis ranging from negative 8 to 8 in intervals of 2 and the y-axis ranging from negative 2 to 12 in intervals of 2. Two functions are graphed on the same plot: y = negative 5 times x/3 plus 1/3 and y = negative 5 times x/3 plus 10. The lines do not cross.

Find the equation of the line parallel to[latex]\,5x=7+y\,[/latex]and passing through the point[latex]\,\left(-1,-2\right).[/latex]

[latex]y=5x+3[/latex]

Finding the Equation of a Line Perpendicular to a Given Line Passing Through a Given Point

Find the equation of the line perpendicular to[latex]\,5x-3y+4=0\,\,\left(-4,1\right).[/latex]

The first step is to write the equation in slope-intercept form.

We see that the slope is[latex]\,m=\frac{5}{3}.\,[/latex]This means that the slope of the line perpendicular to the given line is the negative reciprocal, or [latex]-\frac{3}{5}.\,[/latex]Next, we use the point-slope formula with this new slope and the given point.

Key Concepts

  • Given two points, we can find the slope of a line using the slope formula.
  • We can identify the slope and y -intercept of an equation in slope-intercept form.
  • We can find the equation of a line given the slope and a point.
  • We can also find the equation of a line given two points. Find the slope and use the point-slope formula.
  • The standard form of a line has no fractions.
  • Horizontal lines have a slope of zero and are defined as[latex]\,y=c,[/latex]where c is a constant.
  • Vertical lines have an undefined slope (zero in the denominator), and are defined as[latex]\,x=c,[/latex]where c is a constant.
  • Parallel lines have the same slope and different y- intercepts.
  • Perpendicular lines have slopes that are negative reciprocals of each other unless one is horizontal and the other is vertical.
  • A line parallel to another line, passing through a given point, may be found by substituting the slope value of the line and the x – and y -values of the given point into the equation,[latex]\,f\left(x\right)=mx+b,\,[/latex]and using the[latex]\,b\,[/latex]that results. Similarly, the point-slope form of an equation can also be used.
  • A line perpendicular to another line, passing through a given point, may be found in the same manner, with the exception of using the negative reciprocal slope.

Section Exercises

  • What does it mean when we say that two lines are parallel?

It means they have the same slope.

  • What is the relationship between the slopes of perpendicular lines (assuming neither is horizontal nor vertical)?
  • How do we recognize when an equation, for example[latex]\,y=4x+3,[/latex]will be a straight line (linear) when graphed?

The exponent of the[latex]\,x\,[/latex]variable is 1. It is called a first-degree equation.

For the following exercises, find the equation of the line using the point-slope formula. Write all the final equations using the slope-intercept form.

  • [latex]\left(0,3\right)\,[/latex]with a slope of[latex]\,\frac{2}{3}[/latex]
  • [latex]\left(1,2\right)\,[/latex]with a slope of[latex]\,\frac{-4}{5}[/latex]

[latex]y=\frac{-4}{5}x+\frac{14}{5}[/latex]

  • x -intercept is 1, and[latex]\,\left(-2,6\right)[/latex]
  • y -intercept is 2, and[latex]\,\left(4,-1\right)[/latex]

[latex]y=\frac{-3}{4}x+2[/latex]

  • [latex]\left(-3,10\right)\,[/latex]and[latex]\,\left(5,-6\right)[/latex]
  • [latex]\left(1,3\right)\text{ and }\left(5,5\right)[/latex]

[latex]y=\frac{1}{2}x+\frac{5}{2}[/latex]

  • parallel to[latex]\,y=2x+5\,[/latex]and passes through the point[latex]\,\left(4,3\right)[/latex]
  • perpendicular to[latex]\,\text{3}y=x-4\,[/latex]and passes through the point[latex]\,\left(-2,1\right)[/latex].

[latex]y=-3x-5[/latex]

For the following exercises, find the equation of the line using the given information.

  • [latex]\left(-2,0\right)\,[/latex]and[latex]\,\left(-2,5\right)[/latex]
  • [latex]\left(1,7\right)\,[/latex]and[latex]\,\left(3,7\right)[/latex]

[latex]y=7[/latex]

  • The slope is undefined and it passes through the point[latex]\,\left(2,3\right).[/latex]
  • The slope equals zero and it passes through the point[latex]\,\left(1,-4\right).[/latex]

[latex]y=-4[/latex]

  • The slope is[latex]\,\frac{3}{4}\,[/latex]and it passes through the point[latex]\,\text{(1,4)}\text{.}[/latex]
  • [latex]\left(-1,3\right)\,[/latex]and[latex]\,\left(4,-5\right)[/latex]

[latex]8x+5y=7[/latex]

For the following exercises, graph the pair of equations on the same axes, and state whether they are parallel, perpendicular, or neither.

  • [latex]\begin{array}{l}\\ \begin{array}{l}y=2x+7\hfill \\ y=\frac{-1}{2}x-4\hfill \end{array}\end{array}[/latex]
  • [latex]\begin{array}{l}3x-2y=5\hfill \\ 6y-9x=6\hfill \end{array}[/latex]

Coordinate plane with the x and y axes ranging from negative 10 to 10. The functions 3 times x minus 2 times y = 5 and 6 times y minus 9 times x = 6 are graphed on the same plot. The lines do not cross.

  • [latex]\begin{array}{l}y=\frac{3x+1}{4}\hfill \\ y=3x+2\hfill \end{array}[/latex]
  • [latex]\begin{array}{l}x=4\\ y=-3\end{array}[/latex]

Coordinate plane with the x and y axes ranging from negative 10 to 10. The function y = negative 3 and the line x = 4 are graphed on the same plot. These lines cross at a 90 degree angle.

Perpendicular

For the following exercises, find the slope of the line that passes through the given points.

  • [latex]\left(5,4\right)\,[/latex]and[latex]\,\left(7,9\right)[/latex]
  • [latex]\left(-3,2\right)\,[/latex]and[latex]\,\left(4,-7\right)[/latex]

[latex]m=\frac{-9}{7}[/latex]

  • [latex]\left(-5,4\right)\,[/latex]and[latex]\,\left(2,4\right)[/latex]
  • [latex]\left(-1,-2\right)\,[/latex]and[latex]\,\left(3,4\right)[/latex]

[latex]m=\frac{3}{2}[/latex]

  • [latex]\,\left(3,-2\right)[/latex]and[latex]\,\left(3,-2\right)[/latex]

For the following exercises, find the slope of the lines that pass through each pair of points and determine whether the lines are parallel or perpendicular.

  • [latex]\begin{array}{l}\left(-1,3\right)\text{ and }\left(5,1\right)\\ \left(-2,3\right)\text{ and }\left(0,9\right)\end{array}[/latex]

[latex]{m}_{1}=\frac{-1}{3},\text{ }{m}_{2}=3;\text{ }\text{Perpendicular}\text{.}[/latex]

  • [latex]\begin{array}{l}\left(2,5\right)\text{ and }\left(5,9\right)\\ \left(-1,-1\right)\text{ and }\left(2,3\right)\end{array}[/latex]

For the following exercises, express the equations in slope intercept form (rounding each number to the thousandths place). Enter this into a graphing calculator as Y1, then adjust the ymin and ymax values for your window to include where the y -intercept occurs. State your ymin and ymax values.

  • [latex]0.537x-2.19y=100[/latex]

[latex]y=0.245x-45.662.\,[/latex]Answers may vary.[latex]\,{y}_{\text{min}}=-50,\text{ }{y}_{\text{max}}=-40[/latex]

  • [latex]4,500x-200y=9,528[/latex]
  • [latex]\frac{200-30y}{x}=70[/latex]

[latex]y=-2.333x+6.667.\,[/latex]Answers may vary.[latex]\,{y}_{\mathrm{min}}=-10, {y}_{\mathrm{max}}=10[/latex]

  • Starting with the point-slope formula[latex]\,y-{y}_{1}=m\left(x-{x}_{1}\right),[/latex]solve this expression for[latex]\,x\,[/latex]in terms of[latex]\,{x}_{1},y,{y}_{1},[/latex]and[latex]\,m.[/latex]
  • Starting with the standard form of an equation[latex]\,\text{A}x\text{ + B}y\text{ = C,}[/latex]solve this expression for y in terms of[latex]\,A,B,C,\,[/latex]and[latex]\,x.\,[/latex]Then put the expression in slope-intercept form.

[latex]y=\frac{-A}{B}x+\frac{C}{B}[/latex]

  • Use the above derived formula to put the following standard equation in slope intercept form:[latex]\,7x-5y=25.[/latex]
  • Given that the following coordinates are the vertices of a rectangle, prove that this truly is a rectangle by showing the slopes of the sides that meet are perpendicular.

[latex]\left(-1,1\right),\left(2,0\right),\left(3,3\right)\text{,}[/latex]and[latex]\,\left(0,4\right)[/latex]

[latex]\begin{array}{l}\text{The slope for }\left(-1,1\right)\text{ to }\left(0,4\right)\text{ is }3.\\ \text{The slope for }\left(-1,1\right)\text{ to }\left(2,0\right)\text{ is }\frac{-1}{3}.\\ \text{The slope for }\left(2,0\right)\text{ to }\left(3,3\right)\text{ is }3.\\ \text{The slope for }\left(0,4\right)\text{ to }\left(3,3\right)\text{ is }\frac{-1}{3}.\end{array}[/latex]

Yes they are perpendicular.

  • Find the slopes of the diagonals in the previous exercise. Are they perpendicular?

Real-World Applications

  • The slope for a wheelchair ramp for a home has to be[latex]\,\frac{1}{12}.\,[/latex]If the vertical distance from the ground to the door bottom is 2.5 ft, find the distance the ramp has to extend from the home in order to comply with the needed slope.

Right triangle with legs of x feet and 2.5 feet

  • If the profit equation for a small business selling[latex]\,x\,[/latex]number of item one and[latex]\,y\,[/latex]number of item two is[latex]\,p=3x+4y,[/latex]find the[latex]\,y\,[/latex]value when[latex]\,p=\text{\$}453\text{ and }x=75.[/latex]

For the following exercises, use this scenario: The cost of renting a car is $45/wk plus $0.25/mi traveled during that week. An equation to represent the cost would be[latex]\,y=45+.25x,[/latex]where[latex]\,x\,[/latex]is the number of miles traveled.

  • What is your cost if you travel 50 mi?
  • If your cost were[latex]\,\text{\$}63.75,[/latex]how many miles were you charged for traveling?
  • Suppose you have a maximum of $100 to spend for the car rental. What would be the maximum number of miles you could travel?

Media Attributions

  • 4.1 Figure 1 © OpenStax Algebra and Trigonometry is licensed under a CC BY (Attribution) license
  • 4.1 Figure 2 © OpenStax Algebra and Trigonometry is licensed under a CC BY (Attribution) license
  • 4.1 Figure 3 Parallel Lines © OpenStax Algebra and Trigonometry is licensed under a CC BY (Attribution) license
  • 4.1 Figure 4 Perpendicular Lines © OpenStax Algebra and Trigonometry is licensed under a CC BY (Attribution) license
  • 4.1 Figure 5 © OpenStax Algebra and Trigonometry is licensed under a CC BY (Attribution) license
  • 4.1 Graph of Parallel Lines © OpenStax Algebra and Trigonometry is licensed under a CC BY (Attribution) license
  • 4.1 Figure 6 © OpenStax Algebra and Trigonometry is licensed under a CC BY (Attribution) license
  • 4.1 Exercise #19 Solution © OpenStax Algebra and Trigonometry is licensed under a CC BY (Attribution) license
  • 4.1 Exercise #21 Solution © OpenStax Algebra and Trigonometry is licensed under a CC BY (Attribution) license
  • 4.1 Exercise #37 © OpenStax Algebra and Trigonometry is licensed under a CC BY (Attribution) license

College Algebra Copyright © 2024 by LOUIS: The Louisiana Library Network is licensed under a Creative Commons Attribution 4.0 International License , except where otherwise noted.

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Linear Equations in Two Variables Questions

Linear equations in two variables questions presented here cover a variety of questions asked regarding linear equations in two variables with solutions and proper explanations. By practising these questions students will develop problem-solving skills.

Linear equations in two variables are linear polynomials with two unknowns. They are of the general form ax + by + c = 0, where x and y are the two variables, a and b are non-zero real numbers and c is a constant. The graphical representation of a linear equation in two variables is a straight line.

Linear Equations in Two Variables Questions with Solutions

Below are some practice questions on linear equations in two variables with detailed solutions.

Question 1: Solve for x and y:

\(\begin{array}{l}\frac{1}{2x}-\frac{1}{y}=-1,\:\:\frac{1}{x}+\frac{1}{2y}=8\:\:\:(x\neq0,\;y\neq0)\end{array} \)

Put 1/x = u and 1/y = v. The given equations become

u/2 – v = –1 ⇒ u – 2v = – 2 ….(i)

u + v/2 = 8 ⇒ 2u + v = 16 ….(ii)

Multiplying equation (ii) by 2 on both sides and adding (i) and (ii), we get

(u + 4u) + ( –2v + 2v) = –2 + 32

⇒ u = 6 ⇒ x = ⅙ and y = ¼

Question 2: Solve the system of linear equations 2x + 3y = 17 and 3x – 2y = 6 by the cross multiplication method.

By cross multiplication

\(\begin{array}{l}\frac{x}{\left\{ 3\times (-6)-(-2)\times(-17)\right\}}=\frac{y}{\left\{ -17\times 3-(-6)\times 2 \right\}}=\frac{1}{\left\{2\times (-2)-3 \times3\right\}}\end{array} \)

⇒ x/( –52) = y/( –39) = 1/( – 13)

⇒ x = 52/13 = 4 and y = 39/13 = 3

Hence x = 4 and y = 3 is the solution of given equations.

Question 3: Solve the following system of equations by substitution method:

2x + 3y = 0 and 3x + 4y = 5

Given equations,

2x + 3y = 0 ….(i)

3x + 4y = 5 …..(ii)

From (i) we get, y = – 2x/3, substituting value of y in (ii), we get

3x + 4(–2x/3) = 5

⇒ 9x – 8x = 15

Then y = (–2 × 15)/3 = – 10

Therefore, x = 15 and y = – 10 is solution of given system of equations.

Video Lesson on Consistent and Inconsistent Equations

examples of solving problems involving linear equations in two variables

Question 4: Find the value of k for which the given system of equations has infinitely many solutions: x + (k + 1)y = 5 and (k + 1)x + 9y + (1 – 8k) = 0.

The given equations will have infinitely many solutions if a 1 /a 2 = b 1 /b 2 = c 1 /c 2

Hence, 1/(k + 1) = (k + 1)/9 = – 5/(1 – 8k)

Solving the equations we get k = 2.

Question 5: If the lines given by 3x + 2ky = 2 and 2x + 5y + 1 = 0 are parallel. Find the value of k.

If the lines are parallel, then they are inconsistent system of equations and a 1 /a 2 = b 1 /b 2 ≠ c 1 /c 2

Now, it should be 3/2 = 2k/5

Then we have 2k/5 = 30/20 = 3/2, which satisfies the condition of inconsistency.

Question 6: Find the value of k for which the system of equations has a non-zero solution

5x + 3y = 0 and 10x + ky = 0

The given equations are homogenous equations, they will have a non-zero solution if a 1 /a 2 = b 1 /b 2

Then, 5/10 = 3/k

⇒ 1/2 = 3/k

Question7: The monthly incomes of A and B are in the ratio 8:7 and their expenditures are in the ratio 19:16. If each saves ₹ 5000 per month, find the monthly income of each.

Let the monthly incomes of A and B be 8x and 7x rupees respectively, and let their monthly expenditure be 19y and 16y rupees respectively.

Monthly savings of A = 8x – 19y = 5000 ….(i)

Monthly savings of B = 7x – 16y = 5000 ….(ii)

Multiplying (i) 16 and (ii) by 19 and subtracting (ii) from (i) we get

(16 × 8x – 19 × 7x) = 5000 (16 – 19)

⇒ 5x = 15000 ⇒ x = 3000

Monthly income of A is (8 × 3000) = ₹24,000

Monthly income of B is (7 × 3000) = ₹21,000

Question 8: The sum of a two-digit number obtained by reversing the order of its digits is 99. If the digits differ by 3, find the original number.

Let the original number be (10x + y)

According to the question,

(10x + y) + (10y + x) = 99

⇒ 11(x + y) = 99

⇒ x + y = 9 ….(i)

And x – y = 3 ….(ii)

Adding equations (i) and (ii), we get,

Hence the required number is 63.

Question 9: A man’s age is three times the sum of the ages of his two sons. After 5 years, his age will be twice the sum of his two son’s age. Find the age of the man.

Let the age of the man be x and the sum of the ages of his two sons be y.

x = 3y ⇒ x – 3y = 0 ….(i)

And (x + 5) = 2(y + 5 + 5)

⇒ x – 2y = 15 ….(ii)

Subtracting equation (i) from (ii) we get

Y = 15 and from (i) x = 45.

The present age of the man is 45 years.

Question 10: A man can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours, Find his speed of rowing in still water. Also, find the speed of the stream.

Let the speed of the man in still water be x km/hr and let the speed of the current be y kn/hr.

Speed in downstream = (x + y) km/hr

Speed in upstream = (x – y) km/hr

But speed in downstream = 20/2 km/hr = 10 km/hr

And speed in upstream = 4/2 km/hr = 2 km/hr

∴ x + y = 10 and x – y = 2

Solving both the equations we get;

x = 6 and y = 4.

Hence, the speed of the man in still water is 6 km/hr and the speed of the current is 4 km/hr.

Question 11: Find the four angles of a cyclic Quadrilateral ABCD in which ∠A = (2x – 1) o , ∠B = (y + 5) o , ∠C = (2y + 15) o , and ∠D = (4x – 7) o .

We know that sum of opposite angles of a cyclic quadrilateral is 180 o

∴ ∠A + ∠C = 180 o and ∠B + ∠C = 180 o

∠A + ∠C = 180 o ⇒ (2x – 1) + (2y + 15) = 180 o

⇒ x + y = 83 ….(i)

∠B + ∠C = 180 o ⇒ (y + 5) + (4x – 7) = 180 o

⇒ 4x + y = 182 …..(ii)

Subtracting (i) from (ii) we get

3x = 182 – 83 ⇒ x = 33

Substituting in (i), we get y = 50

∴ ∠A = 2 × 33 – 1 = 65 o

∠B = 50 + 5 = 55 o

∠C = 2 × 50 + 15 = 115 o

∠D = 4 × 33 – 7 = 125 o .

Question 12: 8 men and 12 boys can finish a piece of work in 5 days, while 6 men and 8 boys can finish it in 7 days. Find time taken by a man and a boy alone to finish the same work.

Let 1 man can finish the work in x days and let 1 boy can finish the work in y days.

1 man’s one day work = 1/x

1 boy’s one day work = 1/y

8 men’s 1 day’s work + 12 boy’s one day’s work = ⅕

⇒ 8/x + 12/y = ⅕

⇒ 8u + 12v = ⅕ ….(i) where u = 1/x and v = 1/y

Similarly, 6u + 8v = 1/7 ….(ii)

On solving (i) and (ii) we get, x = 70 and y = 140

∴ One man alone can finish the work in 70 days and one boy alone can finish the work in 140 days.

Related Articles:

Practice questions:.

1. Five years ago Anna was three times older than Mira and ten years later Anna will be two times older than Mira. What are the present ages of Anna and Mira?

2. The difference of two numbers is 4 and the difference of their reciprocals is 4/21. Find the numbers.

3. Find the value of k for which the system of equations 5x – 3y = 0, and 2x + ky = 0 has a non-zero solution.

4. Find the value of a and b for which each of the following systems of linear equations

(a – 1)x + 3y = 2 and 6x + (1 – 2b)y = 6 has infinite number of solutions.

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Linear Equations In Two Variables Worksheets

Linear equations in two variables worksheets can help encourage students to read and think about the questions, rather than simply recognizing a pattern to the solutions. linear equations in two variables worksheets come with the answer key and detailed solutions which the students can refer to anytime.

Benefits of Linear Equations in Two Variables Worksheets

Linear equations in two variables worksheets can help students to understand about the linear equation that it is an equation in which there are two variables used. the variable also can be used many times and/or can be used on either side of the equation; everything that matters is that the variable will remain the same.

Linear equations in two variables worksheets give students the opportunity to solve a wide variety of problems helping them to build a robust mathematical foundation. Linear equations in two variables worksheets help kids to improve their speed, accuracy, logical and reasoning skills in performing simple calculations related to the topic of linear equations in two variables.

These worksheets come with visual simulation for students to see the problems in action, and provides a detailed step-by-step solution for students to understand the process better, and a worksheet properly explained about the linear equations.

Download Linear Equations in Two Variables Worksheet PDFs

These math worksheets should be practiced regularly and are free to download in PDF formats.

  • 5.10 Systems of Linear Inequalities in Two Variables
  • Introduction
  • 1.1 Basic Set Concepts
  • 1.2 Subsets
  • 1.3 Understanding Venn Diagrams
  • 1.4 Set Operations with Two Sets
  • 1.5 Set Operations with Three Sets
  • Key Concepts
  • Formula Review
  • Chapter Review
  • Chapter Test
  • 2.1 Statements and Quantifiers
  • 2.2 Compound Statements
  • 2.3 Constructing Truth Tables
  • 2.4 Truth Tables for the Conditional and Biconditional
  • 2.5 Equivalent Statements
  • 2.6 De Morgan’s Laws
  • 2.7 Logical Arguments
  • 3.1 Prime and Composite Numbers
  • 3.2 The Integers
  • 3.3 Order of Operations
  • 3.4 Rational Numbers
  • 3.5 Irrational Numbers
  • 3.6 Real Numbers
  • 3.7 Clock Arithmetic
  • 3.8 Exponents
  • 3.9 Scientific Notation
  • 3.10 Arithmetic Sequences
  • 3.11 Geometric Sequences
  • 4.1 Hindu-Arabic Positional System
  • 4.2 Early Numeration Systems
  • 4.3 Converting with Base Systems
  • 4.4 Addition and Subtraction in Base Systems
  • 4.5 Multiplication and Division in Base Systems
  • 5.1 Algebraic Expressions
  • 5.2 Linear Equations in One Variable with Applications
  • 5.3 Linear Inequalities in One Variable with Applications
  • 5.4 Ratios and Proportions
  • 5.5 Graphing Linear Equations and Inequalities
  • 5.6 Quadratic Equations with Two Variables with Applications
  • 5.7 Functions
  • 5.8 Graphing Functions
  • 5.9 Systems of Linear Equations in Two Variables
  • 5.11 Linear Programming
  • 6.1 Understanding Percent
  • 6.2 Discounts, Markups, and Sales Tax
  • 6.3 Simple Interest
  • 6.4 Compound Interest
  • 6.5 Making a Personal Budget
  • 6.6 Methods of Savings
  • 6.7 Investments
  • 6.8 The Basics of Loans
  • 6.9 Understanding Student Loans
  • 6.10 Credit Cards
  • 6.11 Buying or Leasing a Car
  • 6.12 Renting and Homeownership
  • 6.13 Income Tax
  • 7.1 The Multiplication Rule for Counting
  • 7.2 Permutations
  • 7.3 Combinations
  • 7.4 Tree Diagrams, Tables, and Outcomes
  • 7.5 Basic Concepts of Probability
  • 7.6 Probability with Permutations and Combinations
  • 7.7 What Are the Odds?
  • 7.8 The Addition Rule for Probability
  • 7.9 Conditional Probability and the Multiplication Rule
  • 7.10 The Binomial Distribution
  • 7.11 Expected Value
  • 8.1 Gathering and Organizing Data
  • 8.2 Visualizing Data
  • 8.3 Mean, Median and Mode
  • 8.4 Range and Standard Deviation
  • 8.5 Percentiles
  • 8.6 The Normal Distribution
  • 8.7 Applications of the Normal Distribution
  • 8.8 Scatter Plots, Correlation, and Regression Lines
  • 9.1 The Metric System
  • 9.2 Measuring Area
  • 9.3 Measuring Volume
  • 9.4 Measuring Weight
  • 9.5 Measuring Temperature
  • 10.1 Points, Lines, and Planes
  • 10.2 Angles
  • 10.3 Triangles
  • 10.4 Polygons, Perimeter, and Circumference
  • 10.5 Tessellations
  • 10.7 Volume and Surface Area
  • 10.8 Right Triangle Trigonometry
  • 11.1 Voting Methods
  • 11.2 Fairness in Voting Methods
  • 11.3 Standard Divisors, Standard Quotas, and the Apportionment Problem
  • 11.4 Apportionment Methods
  • 11.5 Fairness in Apportionment Methods
  • 12.1 Graph Basics
  • 12.2 Graph Structures
  • 12.3 Comparing Graphs
  • 12.4 Navigating Graphs
  • 12.5 Euler Circuits
  • 12.6 Euler Trails
  • 12.7 Hamilton Cycles
  • 12.8 Hamilton Paths
  • 12.9 Traveling Salesperson Problem
  • 12.10 Trees
  • 13.1 Math and Art
  • 13.2 Math and the Environment
  • 13.3 Math and Medicine
  • 13.4 Math and Music
  • 13.5 Math and Sports
  • A | Co-Req Appendix: Integer Powers of 10

Learning Objectives

After completing this section, you should be able to:

  • Demonstrate whether an ordered pair is a solution to a system of linear inequalities.
  • Solve systems of linear inequalities using graphical methods.
  • Graph systems of linear inequalities.
  • Interpret and solve applications of linear inequalities.

In this section, we will learn how to solve systems of linear inequalities in two variables. In Systems of Linear Equations in Two Variables , we learned how to solve for systems of linear equations in two variables and found a solution that would work in both equations. We can solve systems of inequalities by graphing each inequality (as discussed in Graphing Linear Equations and Inequalities ) and putting these on the same coordinate system. The double-shaded part will be our solution to the system. There are many real-life examples for solving systems of linear inequalities.

Consider Ming who has two jobs to help her pay for college. She works at a local coffee shop for $7.50 per hour and at a research lab on campus for $12 per hour. Due to her busy class schedule, she cannot work more than 15 hours per week. If she needs to make at least $150 per week, can she work seven hours at the coffee shop and eight hours in the lab?

Determining If an Ordered Pair Is a Solution of a System of Linear Inequalities

The definition of a system of linear inequalities is similar to the definition of a system of linear equations. A system of linear inequalities looks like a system of linear equations, but it has inequalities instead of equations. A system of two linear inequalities is shown here.

To solve a system of linear inequalities, we will find values of the variables that are solutions to both inequalities. We solve the system by using the graphs of each inequality and show the solution as a graph. We will find the region on the plane that contains all ordered pairs ( x , y ) ( x , y ) that make both inequalities true. The solution of a system of linear inequalities is shown as a shaded region in the xy xy -coordinate system that includes all the points whose ordered pairs make the inequalities true.

To determine if an ordered pair is a solution to a system of two inequalities, substitute the values of the variables into each inequality. If the ordered pair makes both inequalities true, it is a solution to the system.

Example 5.89

Determining whether an ordered pair is a solution to a system.

Determine whether the ordered pair is a solution to the system:

  • ( − 2 , 4 ) ( − 2 , 4 )
  • ( 3 , 1 ) ( 3 , 1 )

Is the ordered pair ( − 2 , 4 ) ( − 2 , 4 ) a solution?

We substitute x = − 2 x = − 2 and y = 4 y = 4 into both inequalities.

The ordered pair ( − 2 , 4 ) ( − 2 , 4 ) made both inequalities true. Therefore ( − 2 , 4 ) ( − 2 , 4 ) is a solution to this system.

Is the ordered pair ( 3 , 1 ) ( 3 , 1 ) a solution?

We substitute x = 3 x = 3 and y = 1 y = 1 into both inequalities.

The ordered pair ( 3 , 1 ) ( 3 , 1 ) made one inequality true, but the other one false. Therefore ( 3 , 1 ) ( 3 , 1 ) is not a solution to this system.

Your Turn 5.89

Solving systems of linear inequalities using graphical methods.

The solution to a single linear inequality was the region on one side of the boundary line that contains all the points that make the inequality true. The solution to a system of two linear inequalities is a region that contains the solutions to both inequalities. We will review graphs of linear inequalities and solve the linear inequality from its graph.

Example 5.90

Solving a system of linear inequalities by graphing.

Use Figure 5.88 to solve the system of linear inequalities:

To solve the system of linear inequalities we look at the graph and find the region that satisfies BOTH inequalities. To do this we pick a test point and check. Let's us pick ( − 1 , − 1 ) ( − 1 , − 1 ) .

Is ( - 1 , - 1 ) ( - 1 , - 1 ) a solution to y ≥ 2 x − 1 ? y ≥ 2 x − 1 ?

Is ( - 1 , - 1 ) ( - 1 , - 1 ) a solution to y < x + 1 ? y < x + 1 ?

The region containing ( − 1 , − 1 ) ( − 1 , − 1 ) is the solution to the system of linear inequalities. Notice that the solution is all the points in the area shaded twice, which appears as the darkest shaded region.

Your Turn 5.90

Graphing systems of linear inequalities.

We learned that the solution to a system of two linear inequalities is a region that contains the solutions to both inequalities. To find this region by graphing, we will graph each inequality separately and then locate the region where they are both true. The solution is always shown as a graph.

Step 1: Graph the first inequality.

Graph the boundary line.

Shade in the side of the boundary line where the inequality is true.

Step 2: On the same grid, graph the second inequality.

Shade in the side of that boundary line where the inequality is true.

Step 3: The solution is the region where the shading overlaps.

Step 4: Check by choosing a test point.

Example 5.91

Solve the system by graphing:

Graph x − y > 3 x − y > 3 by graphing x − y = 3 x − y = 3 and testing a point ( Figure 5.89 ). The intercepts are x = 3 x = 3 and y = − 3 y = − 3 and the boundary line will be dashed. Test ( 0 , 0 ) ( 0 , 0 ) which makes the inequality false so shade the side that does not contain ( 0 , 0 ) ( 0 , 0 ) .

Graph y < − 1 5 x + 4 y < − 1 5 x + 4 by graphing y = − 1 5 x + 4 y = − 1 5 x + 4 using the slope m = − 1 5 m = − 1 5 and y y -intercept b = 4 b = 4 ( Figure 5.90 ). The boundary line will be dashed. Test ( 0 , 0 ) ( 0 , 0 ) which makes the inequality true, so shade the side that contains ( 0 , 0 ) ( 0 , 0 ) .

Choose a test point in the solution and verify that it is a solution to both inequalities. The point of intersection of the two lines is not included as both boundary lines were dashed. The solution is the area shaded twice—which appears as the darkest shaded region.

Your Turn 5.91

Example 5.92, graphing a system of linear inequalities.

Graph x − 2 y < 5 x − 2 y < 5 by graphing x − 2 y = 5 x − 2 y = 5 ( Figure 5.91 ) and testing a point. The intercepts are x = 5 x = 5 and y = − 2.5 y = − 2.5 and the boundary line will be dashed. Test ( 0 , 0 ) ( 0 , 0 ) , which makes the inequality true, so shade the side that contains ( 0 , 0 ) ( 0 , 0 ) .

Graph y > − 4 y > − 4 by graphing y = − 4 y = − 4 and recognizing that it is a horizontal line through y = − 4 y = − 4 ( Figure 5.92 ). The boundary line will be dashed. Test ( 0 , 0 ) ( 0 , 0 ) , which makes the inequality true so shade the side that contains ( 0 , 0 ) ( 0 , 0 ) .

The point ( 0 , 0 ) ( 0 , 0 ) is in the solution, and we have already found it to be a solution of each inequality. The point of intersection of the two lines is not included as both boundary lines were dashed. The solution is the area shaded twice, which appears as the darkest shaded region.

Your Turn 5.92

Systems of linear inequalities where the boundary lines are parallel might have no solution. We will see this in the next example.

Example 5.93

Graphing parallel boundary lines with no solution.

Graph 4 x + 3 y ≥ 12 4 x + 3 y ≥ 12 , by graphing 4 x + 3 y = 12 4 x + 3 y = 12 ( Figure 5.93 ) and testing a point. The intercepts are x = 3 x = 3 and y = 4 y = 4 and the boundary line will be solid. Test ( 0 , 0 ) ( 0 , 0 ) , which makes the inequality false, so shade the side that does not contain ( 0 , 0 ) ( 0 , 0 ) .

Graph y < − 4 3 x + 1 y < − 4 3 x + 1 by graphing y = − 4 3 x + 1 y = − 4 3 x + 1 using the slope m = − 4 3 m = − 4 3 and y y -intercept b = 1 b = 1 ( Figure 5.94 ). The boundary line will be dashed. Test ( 0 , 0 ) ( 0 , 0 ) , which makes the inequality true, so shade the side that contains ( 0 , 0 ) ( 0 , 0 ) .

No shared point exists in both shaded regions, so the system has no solution.

Your Turn 5.93

Some systems of linear inequalities where the boundary lines are parallel will have a solution. We will see this in the next example.

Example 5.94

Graphing parallel boundary lines with a solution.

Graph y > 1 2 x − 4 y > 1 2 x − 4 by graphing y = 1 2 x − 4 y = 1 2 x − 4 using the slope m = 1 2 m = 1 2 and the y y -intercept b = − 4 b = − 4 ( Figure 5.95 ). The boundary line will be dashed. Test ( 0 , 0 ) ( 0 , 0 ) , which makes the inequality true, so shade the side that contains ( 0 , 0 ) ( 0 , 0 ) .

Graph x − 2 y < − 4 x − 2 y < − 4 by graphing x − 2 y = − 4 x − 2 y = − 4 ( Figure 5.96 ) and testing a point. The intercepts are x = − 4 x = − 4 and y = 2 y = 2 and the boundary line will be dashed. Choose a test point in the solution and verify that it is a solution to both inequalities. Test ( 0 , 0 ) ( 0 , 0 ) , which makes the inequality false, so shade the side that does not contain ( 0 , 0 ) ( 0 , 0 ) .

No point on the boundary lines is included in the solution as both lines are dashed. The solution is the region that is shaded twice which is also the solution to x − 2 y < − 4 x − 2 y < − 4 .

Your Turn 5.94

Interpreting and solving applications of linear inequalities.

When solving applications of systems of inequalities, first translate each condition into an inequality. Then graph the system, as we did above, to see the region that contains the solutions. Many situations will be realistic only if both variables are positive, so add inequalities to the system as additional requirements.

Example 5.95

Applying linear inequalities to calculating photo costs.

A photographer sells their prints at a booth at a street fair. At the start of the day, they want to have at least 25 photos to display at their booth. Each small photo they display costs $4 and each large photo costs $10. They do not want to spend more than $200 on photos to display.

  • Write a system of inequalities to model this situation.
  • Graph the system.
  • Could they display 10 small and 20 large photos?
  • Could they display 20 large and 10 small photos?

Since x ≥ 0 x ≥ 0 and y ≥ 0 y ≥ 0 (both are greater than or equal to) all solutions will be in the first quadrant. As a result, our graph shows only Quadrant I. To graph x + y ≥ 25 x + y ≥ 25 , graph x + y = 25 x + y = 25 as a solid line. Choose ( 0 , 0 ) ( 0 , 0 ) as a test point. Since it does not make the inequality true, shade the side that does not include the point ( 0 , 0 ) ( 0 , 0 ) .

To graph 4 x + 10 y ≤ 200 4 x + 10 y ≤ 200 , graph 4 x + 10 y = 200 4 x + 10 y = 200 as a solid line. Choose ( 0 , 0 ) ( 0 , 0 ) as a test point. Since it does make the inequality true, shade (bottom left) the side that include the point ( 0 , 0 ) ( 0 , 0 ) .

The solution of the system is the region of Figure 5.97 that is shaded the darkest. The boundary line sections that border the darkly shaded section are included in the solution as are the points on the x x -axis from ( 25 , 0 ) ( 25 , 0 ) to ( 55 , 0 ) ( 55 , 0 ) .

  • To determine if 10 small and 20 large photos would work, we look at the graph to see if the point ( 10 , 20 ) ( 10 , 20 ) is in the solution region. We could also test the point to see if it is a solution of both equations. It is not, so the photographer would not display 10 small and 20 large photos.
  • To determine if 20 small and 10 large photos would work, we look at the graph to see if the point ( 20 , 10 ) ( 20 , 10 ) is in the solution region. We could also test the point to see if it is a solution of both equations. It is, so the photographer could choose to display 20 small and 10 large photos. Notice that we could also test the possible solutions by substituting the values into each inequality.

Your Turn 5.95

Solving Systems of Linear Inequalities by Graphing

Systems of Linear Inequalities

Check Your Understanding

Section 5.10 exercises.

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  • Authors: Donna Kirk
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  • Book title: Contemporary Mathematics
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  • Section URL: https://openstax.org/books/contemporary-mathematics/pages/5-10-systems-of-linear-inequalities-in-two-variables

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Solving equations

Here you will learn about solving equations, including linear and quadratic algebraic equations, and how to solve them.

Students will first learn about solving equations in grade 8 as a part of expressions and equations, and again in high school as a part of reasoning with equations and inequalities.

What is solving an equation?

Solving equations is a step-by-step process to find the value of the variable. A variable is the unknown part of an equation, either on the left or right side of the equals sign. Sometimes, you need to solve multi-step equations which contain algebraic expressions.

To do this, you must use the order of operations, which is a systematic approach to equation solving. When you use the order of operations, you first solve any part of an equation located within parentheses. An equation is a mathematical expression that contains an equals sign.

For example,

\begin{aligned}y+6&=11\\\\ 3(x-3)&=12\\\\ \cfrac{2x+2}{4}&=\cfrac{x-3}{3}\\\\ 2x^{2}+3&x-2=0\end{aligned}

There are two sides to an equation, with the left side being equal to the right side. Equations will often involve algebra and contain unknowns, or variables, which you often represent with letters such as x or y.

You can solve simple equations and more complicated equations to work out the value of these unknowns. They could involve fractions, decimals or integers.

What is solving an equation?

Common Core State Standards

How does this relate to 8 th grade and high school math?

  • Grade 8 – Expressions and Equations (8.EE.C.7) Solve linear equations in one variable.
  • High school – Reasoning with Equations and Inequalities (HSA.REI.B.3) Solve linear equations and inequalities in one variable, including equations with coefficients represented by letters.

[FREE] Math Equations Check for Understanding Quiz (Grade 6 to 8)

[FREE] Math Equations Check for Understanding Quiz (Grade 6 to 8)

Use this quiz to check your grade 6 to 8 students’ understanding of math equations. 10+ questions with answers covering a range of 6th, 7th and 8th grade math equations topics to identify areas of strength and support!

How to solve equations

In order to solve equations, you need to work out the value of the unknown variable by adding, subtracting, multiplying or dividing both sides of the equation by the same value.

  • Combine like terms .
  • Simplify the equation by using the opposite operation to both sides.
  • Isolate the variable on one side of the equation.

Solving equations examples

Example 1: solve equations involving like terms.

Solve for x.

Combine like terms.

Combine the q terms on the left side of the equation. To do this, subtract 4q from both sides.

The goal is to simplify the equation by combining like terms. Subtracting 4q from both sides helps achieve this.

After you combine like terms, you are left with q=9-4q.

2 Simplify the equation by using the opposite operation on both sides.

Add 4q to both sides to isolate q to one side of the equation.

The objective is to have all the q terms on one side. Adding 4q to both sides accomplishes this.

After you move the variable to one side of the equation, you are left with 5q=9.

3 Isolate the variable on one side of the equation.

Divide both sides of the equation by 5 to solve for q.

Dividing by 5 allows you to isolate q to one side of the equation in order to find the solution. After dividing both sides of the equation by 5, you are left with q=1 \cfrac{4}{5} \, .

Example 2: solve equations with variables on both sides

Combine the v terms on the same side of the equation. To do this, add 8v to both sides.

7v+8v=8-8v+8v

After combining like terms, you are left with the equation 15v=8.

Simplify the equation by using the opposite operation on both sides and isolate the variable to one side.

Divide both sides of the equation by 15 to solve for v. This step will isolate v to one side of the equation and allow you to solve.

15v \div 15=8 \div 15

The final solution to the equation 7v=8-8v is \cfrac{8}{15} \, .

Example 3: solve equations with the distributive property

Combine like terms by using the distributive property.

The 3 outside the parentheses needs to be multiplied by both terms inside the parentheses. This is called the distributive property.

\begin{aligned}& 3 \times c=3 c \\\\ & 3 \times(-5)=-15 \\\\ &3 c-15-4=2\end{aligned}

Once the 3 is distributed on the left side, rewrite the equation and combine like terms. In this case, the like terms are the constants on the left, –15 and –4. Subtract –4 from –15 to get –19.

Simplify the equation by using the opposite operation on both sides.

The goal is to isolate the variable, c, on one side of the equation. By adding 19 to both sides, you move the constant term to the other side.

\begin{aligned}& 3 c-19+19=2+19 \\\\ & 3 c=21\end{aligned}

Isolate the variable to one side of the equation.

To solve for c, you want to get c by itself.

Dividing both sides by 3 accomplishes this.

On the left side, \cfrac{3c}{3} simplifies to c, and on the right, \cfrac{21}{3} simplifies to 7.

The final solution is c=7.

As an additional step, you can plug 7 back into the original equation to check your work.

Example 4: solve linear equations

Combine like terms by simplifying.

Using steps to solve, you know that the goal is to isolate x to one side of the equation. In order to do this, you must begin by subtracting from both sides of the equation.

\begin{aligned} & 2x+5=15 \\\\ & 2x+5-5=15-5 \\\\ & 2x=10 \end{aligned}

Continue to simplify the equation by using the opposite operation on both sides.

Continuing with steps to solve, you must divide both sides of the equation by 2 to isolate x to one side.

\begin{aligned} & 2x \div 2=10 \div 2 \\\\ & x= 5 \end{aligned}

Isolate the variable to one side of the equation and check your work.

Plugging in 5 for x in the original equation and making sure both sides are equal is an easy way to check your work. If the equation is not equal, you must check your steps.

\begin{aligned}& 2(5)+5=15 \\\\ & 10+5=15 \\\\ & 15=15\end{aligned}

Example 5: solve equations by factoring

Solve the following equation by factoring.

Combine like terms by factoring the equation by grouping.

Multiply the coefficient of the quadratic term by the constant term.

2 x (-20) = -40

Look for two numbers that multiply to give you –40 and add up to the coefficient of 3. In this case, the numbers are 8 and –5 because 8 x -5=–40, and 8+–5=3.

Split the middle term using those two numbers, 8 and –5. Rewrite the middle term using the numbers 8 and –5.

2x^2+8x-5x-20=0

Group the terms in pairs and factor out the common factors.

2x^2+8x-5x-20=2x(x + 4)-5(x+4)=0

Now, you’ve factored the equation and are left with the following simpler equations 2x-5 and x+4.

This step relies on understanding the zero product property, which states that if two numbers multiply to give zero, then at least one of those numbers must equal zero.

Let’s relate this back to the factored equation (2x-5)(x+4)=0

Because of this property, either (2x-5)=0 or (x+4)=0

Isolate the variable for each equation and solve.

When solving these simpler equations, remember that you must apply each step to both sides of the equation to maintain balance.

\begin{aligned}& 2 x-5=0 \\\\ & 2 x-5+5=0+5 \\\\ & 2 x=5 \\\\ & 2 x \div 2=5 \div 2 \\\\ & x=\cfrac{5}{2} \end{aligned}

\begin{aligned}& x+4=0 \\\\ & x+4-4=0-4 \\\\ & x=-4\end{aligned}

The solution to this equation is x=\cfrac{5}{2} and x=-4.

Example 6: solve quadratic equations

Solve the following quadratic equation.

Combine like terms by factoring the quadratic equation when terms are isolated to one side.

To factorize a quadratic expression like this, you need to find two numbers that multiply to give -5 (the constant term) and add to give +2 (the coefficient of the x term).

The two numbers that satisfy this are -1 and +5.

So you can split the middle term 2x into -1x+5x: x^2-1x+5x-5-1x+5x

Now you can take out common factors x(x-1)+5(x-1).

And since you have a common factor of (x-1), you can simplify to (x+5)(x-1).

The numbers -1 and 5 allow you to split the middle term into two terms that give you common factors, allowing you to simplify into the form (x+5)(x-1).

Let’s relate this back to the factored equation (x+5)(x-1)=0.

Because of this property, either (x+5)=0 or (x-1)=0.

Now, you can solve the simple equations resulting from the zero product property.

\begin{aligned}& x+5=0 \\\\ & x+5-5=0-5 \\\\ & x=-5 \\\\\\ & x-1=0 \\\\ & x-1+1=0+1 \\\\ & x=1\end{aligned}

The solutions to this quadratic equation are x=1 and x=-5.

Teaching tips for solving equations

  • Use physical manipulatives like balance scales as a visual aid. Show how you need to keep both sides of the equation balanced, like a scale. Add or subtract the same thing from both sides to keep it balanced when solving. Use this method to practice various types of equations.
  • Emphasize the importance of undoing steps to isolate the variable. If you are solving for x and 3 is added to x, subtracting 3 undoes that step and isolates the variable x.
  • Relate equations to real-world, relevant examples for students. For example, word problems about tickets for sports games, cell phone plans, pizza parties, etc. can make the concepts click better.
  • Allow time for peer teaching and collaborative problem solving. Having students explain concepts to each other, work through examples on whiteboards, etc. reinforces the process and allows peers to ask clarifying questions. This type of scaffolding would be beneficial for all students, especially English-Language Learners. Provide supervision and feedback during the peer interactions.

Easy mistakes to make

  • Forgetting to distribute or combine like terms One common mistake is neglecting to distribute a number across parentheses or combine like terms before isolating the variable. This error can lead to an incorrect simplified form of the equation.
  • Misapplying the distributive property Incorrectly distributing a number across terms inside parentheses can result in errors. Students may forget to multiply each term within the parentheses by the distributing number, leading to an inaccurate equation.
  • Failing to perform the same operation on both sides It’s crucial to perform the same operation on both sides of the equation to maintain balance. Forgetting this can result in an imbalanced equation and incorrect solutions.
  • Making calculation errors Simple arithmetic mistakes, such as addition, subtraction, multiplication, or division errors, can occur during the solution process. Checking calculations is essential to avoid errors that may propagate through the steps.
  • Ignoring fractions or misapplying operations When fractions are involved, students may forget to multiply or divide by the common denominator to eliminate them. Misapplying operations on fractions can lead to incorrect solutions or complications in the final answer.

Related math equations lessons

  • Math equations
  • Rearranging equations
  • How to find the equation of a line
  • Solve equations with fractions
  • Linear equations
  • Writing linear equations
  • Substitution
  • Identity math
  • One step equation

Practice solving equations questions

1. Solve 4x-2=14.

GCSE Quiz False

Add 2 to both sides.

Divide both sides by 4.

2. Solve 3x-8=x+6.

Add 8 to both sides.

Subtract x from both sides.

Divide both sides by 2.

3. Solve 3(x+3)=2(x-2).

Expanding the parentheses.

Subtract 9 from both sides.

Subtract 2x from both sides.

4. Solve \cfrac{2 x+2}{3}=\cfrac{x-3}{2}.

Multiply by 6 (the lowest common denominator) and simplify.

Expand the parentheses.

Subtract 4 from both sides.

Subtract 3x from both sides.

5. Solve \cfrac{3 x^{2}}{2}=24.

Multiply both sides by 2.

Divide both sides by 3.

Square root both sides.

6. Solve by factoring:

Use factoring to find simpler equations.

Set each set of parentheses equal to zero and solve.

x=3 or x=10

Solving equations FAQs

The first step in solving a simple linear equation is to simplify both sides by combining like terms. This involves adding or subtracting terms to isolate the variable on one side of the equation.

Performing the same operation on both sides of the equation maintains the equality. This ensures that any change made to one side is also made to the other, keeping the equation balanced and preserving the solutions.

To handle variables on both sides of the equation, start by combining like terms on each side. Then, move all terms involving the variable to one side by adding or subtracting, and simplify to isolate the variable. Finally, perform any necessary operations to solve for the variable.

To deal with fractions in an equation, aim to eliminate them by multiplying both sides of the equation by the least common denominator. This helps simplify the equation and make it easier to isolate the variable. Afterward, proceed with the regular steps of solving the equation.

The next lessons are

  • Inequalities
  • Types of graph
  • Coordinate plane

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Mathematics LibreTexts

4.5: Solving Systems of Linear Inequalities (Two Variables)

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  • Save as PDF
  • Page ID 18351

Learning Objectives

  • Check solutions to systems of linear inequalities with two variables.
  • Solve systems of linear inequalities.

Solutions to Systems of Linear Inequalities

A system of linear inequalities consists of a set of two or more linear inequalities with the same variables. The inequalities define the conditions that are to be considered simultaneously. For example,

We know that each inequality in the set contains infinitely many ordered pair solutions defined by a region in a rectangular coordinate plane. When considering two of these inequalities together, the intersection of these sets defines the set of simultaneous ordered pair solutions. When we graph each of the above inequalities separately, we have

IMAGES

  1. Introduction to Solving Linear Equations with 2 Variables

    examples of solving problems involving linear equations in two variables

  2. SSC Maths solutions for Pair of Linear Equations in Two Variables class

    examples of solving problems involving linear equations in two variables

  3. Linear Equations in Two Variables

    examples of solving problems involving linear equations in two variables

  4. how do you solve a problem with two variables

    examples of solving problems involving linear equations in two variables

  5. Math 8 Module 13: Solving Problems Involving Linear Equations in Two

    examples of solving problems involving linear equations in two variables

  6. How to solve a linear equation in two variables? (1 of 2)

    examples of solving problems involving linear equations in two variables

VIDEO

  1. Solving Real Life Problems involving Linear Equations || Solution of Word Problems || Pages by Aapi

  2. LINEAR EQUATIONS IN ONE VARIABLE

  3. Solving Problems Involving Linear Equations

  4. Lesson 12: Solving Problems Involving Linear Equations in Two Variables

  5. Solving Linear Equation in one Variable

  6. Linear Equations with two variables

COMMENTS

  1. 4.1: Solve Systems of Linear Equations with Two Variables

    An example of a system of two linear equations is shown below. We use a brace to show the two equations are grouped together to form a system of equations. {2x + y x − 2y = 7 = 6 { 2 x + y = 7 x − 2 y = 6. A linear equation in two variables, such as 2x + y = 7 2 x + y = 7, has an infinite number of solutions.

  2. Linear Equations in Two Variables

    Step 1: To solve a system of two equations in two variables graphically, we graph each equation. To know how, click here or follow steps 2 and 3 below. Step 2: To graph an equation manually, first convert it to the form y = mx+b by solving the equation for y.

  3. Algebra

    2x+3y = 20 7x+2y = 53 2 x + 3 y = 20 7 x + 2 y = 53 Solution Here is a set of practice problems to accompany the Linear Systems with Two Variables section of the Systems of Equations chapter of the notes for Paul Dawkins Algebra course at Lamar University.

  4. 3.3: Applications of Linear Systems with Two Variables

    Problems Involving Relationships between Two Variables. If we translate an application to a mathematical setup using two variables, then we need to form a linear system with two equations. Setting up word problems with two variables often simplifies the entire process, particularly when the relationships between the variables are not so clear.

  5. Linear Equations with Two Variables

    1. Substitute the value of a variable into the equation. If we know the value of a variable, we plug it into the given equation. If we don't know the value of a variable, the equation automatically has an infinite number of solutions. 2. Simplify both sides of the equation. Remove the parentheses (using the distributive property) if there are any.

  6. 7.1 Systems of Linear Equations: Two Variables

    For example, consider the following system of linear equations in two variables. 2x + y = 15 3x- y = 5. The solution to a system of linear equations in two variables is any ordered pair that satisfies each equation independently. In this example, the ordered pair (4, 7) is the solution to the system of linear equations.

  7. 5.8: Linear Equations in Two Variables

    Solution to an Equation in Two Variables. We have discovered that an equation is a mathematical way of expressing the relationship of equality between quantities. If the relationship is between two quantities, the equation will contain two variables. We say that an equation in two variables has a solution if an ordered pair of values can be ...

  8. Solutions to 2-variable equations (practice)

    Choose 1 answer: Only ( 3, 2) A Only ( 3, 2) Only ( − 3, 3) B Only ( − 3, 3) Both ( 3, 2) and ( − 3, 3) C Both ( 3, 2) and ( − 3, 3) Neither D Neither Stuck? Review related articles/videos or use a hint. Report a problem Do 4 problems

  9. Solutions to 2-variable equations (video)

    Solving linear equations is a foundation step for solving systems of linear equations, which is a foundation step for linear programming (which, surprisingly, is not computer programming) or linear optimization. Linear programming is a method for calculating an optimal result given a set of constraints.

  10. PDF Linear Equations in Two Variables

    A solution of a linear equation in two variables ax+by = r is a specific point in R2 such that when when the x-coordinate of the point is multiplied by a, and the y-coordinate of the point is multiplied by b, and those two numbers are added together, the answer equals r.

  11. Linear Equations in Two Variables: Solving Methods & Examples

    Example: x + y- 3 = 0 x + y - 3 = 0 is a linear equation in two variables x and y. Solution of Linear Equations in Two Variables x and y are a solution of the linear equation ax + by + c = 0 a x + b y + c = 0 if and only if a + b + c = 0 a + b + c = 0, where and are real numbers.

  12. 5.2

    Learning Objectives. (5.2.1) - Solve cost and revenue problems. Specify what the variables in a cost/ revenue system of linear equations represent. Determine and apply an appropriate method for solving the system. (5.2.2) - Solve value problems with a system of linear equations. (5.2.3) - Solve mixture problems with a system of linear ...

  13. Linear Equations in Two Variables (Definition and Solutions)

    Definition An equation is said to be linear equation in two variables if it is written in the form of ax + by + c=0, where a, b & c are real numbers and the coefficients of x and y, i.e a and b respectively, are not equal to zero. For example, 10x+4y = 3 and -x+5y = 2 are linear equations in two variables.

  14. 4.1 Solve Systems of Linear Equations with Two Variables

    { 2 x + y = 7 x − 2 y = 6 A linear equation in two variables, such as 2 x + y = 7, has an infinite number of solutions. Its graph is a line. Remember, every point on the line is a solution to the equation and every solution to the equation is a point on the line.

  15. PDF Section 12.1 Solving Systems of Linear Equations in Two Variables

    Solving a System of Equations by the Method of Elimination. Step 1: Choose a variable to eliminate. Step 2: Multiply one or both equations by an appropriate nonzero constant so that the sum of the coefficients of one of the variables is zero. Step 3: Add the two equations together to obtain an equation in one variable.

  16. Algebra

    Section 7.1 : Linear Systems with Two Variables. A linear system of two equations with two variables is any system that can be written in the form. ax+by = p cx+dy = q a x + b y = p c x + d y = q. where any of the constants can be zero with the exception that each equation must have at least one variable in it.

  17. Systems of Linear Equations: Two Variables

    2x + y = 15 3x − y = 5 The solution to a system of linear equations in two variables is any ordered pair that satisfies each equation independently. In this example, the ordered pair (4, 7) is the solution to the system of linear equations.

  18. 4.1 Linear Equations in Two Variables

    First, we find the slope using any two points on the line. m = − 2 − ( − 2) 0 − ( − 2) = 0 2 = 0. Use any point for (x1, y1) in the formula, or use the y -intercept. y − ( − 2) = 0(x − 3) y + 2 = 0 y = − 2. The graph is a horizontal line through y = − 2. Notice that all of the y- coordinates are the same. See Figure 2.

  19. Linear Equations in Two Variables

    Solution of Linear Equation in Two Variables: If x = 𝛼 and y = 𝛽 is the solution of the expression ax + by + c then a𝛼 + b𝛽 + c = 0. Simultaneous Equations: Pair of linear equations in two variables are said to be simultaneous equations if both the equations have the same solution.

  20. Linear Equations In Two Variables Worksheets

    Linear equations in two variables worksheets can help students to understand about the linear equation that it is an equation in which there are two variables used. the variable also can be used many times and/or can be used on either side of the equation; everything that matters is that the variable will remain the same.

  21. 5.10 Systems of Linear Inequalities in Two Variables

    To find this region by graphing, we will graph each inequality separately and then locate the region where they are both true. The solution is always shown as a graph. Step 1: Graph the first inequality. Graph the boundary line. Shade in the side of the boundary line where the inequality is true.

  22. Solving Equations

    Solve linear equations in one variable. High school - Reasoning with Equations and Inequalities (HSA.REI.B.3) ... Example 1: solve equations involving like terms. Solve for x. 5q-4q=9 . ... Allow time for peer teaching and collaborative problem solving. Having students explain concepts to each other, work through examples on whiteboards, etc ...

  23. 4.5: Solving Systems of Linear Inequalities (Two Variables)

    Therefore, to solve these systems, graph the solution sets of the inequalities on the same set of axes and determine where they intersect. This intersection, or overlap, defines the region of common ordered pair solutions. Example 4.5.1. Graph the solution set: {− 2x + y > − 4 3x − 6y ≥ 6. Solution:

  24. PDF Quarter 1 Module 13: Solving Problems Involving Linear Equations in Two

    on how to solve problems involving linear equations in two variables. The scope of this module permits it to be used in many different learning situations. The language and numeric used recognizes the diverse vocabulary and numeracy level of students. The lessons are arranged to follow the standard sequence of the course. But the order