Rational Equations Word Problems Lesson

  • Demonstrate an understanding of how to solve a word problem
  • Demonstrate an understanding of how to solve an equation with rational expressions
  • Learn how to solve word problems that involve rational equations

How to Solve a Word Problem with Rational Equations

Six-step method for solving word problems with rational expressions.

  • Read the problem carefully and determine what you are asked to find
  • Assign a variable to represent the unknown
  • Write out an equation that describes the given situation
  • Solve the equation
  • State the answer using a nice clear sentence
  • Check the result by reading back through the problem

Solving a Proportion Problem

Motion word problems with rational expressions, rate of work word problems, skills check:.

Solve each word problem.

Working alone, it takes Steve 11 hours to complete a restoration project on a truck. Jacob can perform the same task in 110 hours. How long would it take if they worked together?

Please choose the best answer.

Jamie’s hot tub has an outlet pipe that can empty the hot tub in 6 minutes. Additionally, her hot tub has an inlet pipe that can fill the hot tub in 3 minutes. If both pipes were turned on, how long would it take to fill a completely empty hot tub?

On her drive from Port Smith to Maryland, Stephanie averaged 51 miles per hour. If she had been able to average 60 miles per hour, she would have reached Maryland 3 hours earlier. What is the driving distance between Port Smith and Maryland?

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Applications of Rational Expressions and Word Problems - Expii

Applications of rational expressions and word problems, explanations (3).

steps in solving word problems involving rational algebraic expressions

Word Problems Involving Rational Equations

Seven divided by the sum of a number and two is equal to half the difference of the number and three. Find all such numbers.

Let's begin by defining our variable. Let x=the number Now we can set up our equation. Seven divided by the sum of a number and two 7x+2 is equal to 7x+2= half the difference of the number and three. 7x+2=x−32 Before we solve, take note that there's a variable in the denominator which means we will have a restricted value. So, x≠−2 because that would make the denominator zero which would make the expression undefined. Now we can cross multiply and solve! 7x+2=x−327⋅2=(x+2)(x−3)14=x2−x−6−14=          −140=x2−x−200=(x−5)(x+4)x=5 or x=−4 None of the solutions are our restricted value so we know both these will work. Finally we state our answer. The numbers are 5 and−4.

Related Lessons

Distance, rate, and time word problems.

Let's talk about how to solve word problems involving distance, rate, and time. These problems can be tricky, but here is a good guide on how to solve them.

A boat that can travel fifteen miles per hour in still water can travel thirty-six miles downstream in the same amount of time that it can travel twenty-four miles upstream. Find the speed of the current in the river.

The general formula we use for these problems is:

Overall Rate = speed of the vehicle (with no current/wind) ± speed of the water or wind

Since these are distance, rate, and time problems, we will be using the distance formula as well: Distance=Rate×Timed=rt

A 2x3 grid. First row labeled "Upstream," second row labeled "Downstream," first column labeled "Distance," second column labeled "rate," third column labeled "time."

Here's the skeleton of the table we make to help solve these problems.

Once we make the outline of our table we begin by filling in the information that was given to us. The problem directly tells us that you travel thirty-six miles downstream and twenty-four miles upstream.

A 2x3 grid. First row labeled "Upstream," second row labeled "Downstream," first column labeled "Distance," second column labeled "rate," third column labeled "time." First row/first column contains the number 24, and second row/first column contains the number 36.

Next we're given that the boat can travel fifteen miles per hour in still water so we have a rate of 15 mph. But this is in still water, not going upstream or downstream. So we need to use our formula given above for overall rate.

So we know the speed of the vehicle with no current or wind. That's the 15 mph. So let's write that in under rate.

A 2x3 grid. First row labeled "Upstream," second row labeled "Downstream," first column labeled "Distance," second column labeled "rate," third column labeled "time." First row/first column contains the number 24, first row/second column contains the number 15, second row/first column contains the number 36, second row second column contains the number 15.

Now we need to figure out what gets added or subtracted. When going upstream, are you adding speed or losing speed? You're losing speed because you have to fight the current. So it would be a minus. When going downstream you're gaining speed because you're moving with the current. So it would be a plus.

A 2x3 grid. First row labeled "Upstream," second row labeled "Downstream," first column labeled "Distance," second column labeled "rate," third column labeled "time." First row/first column contains the number 24, first row/second column contains the number 15-, second row/first column contains the number 36, second row second column contains the number 15+.

So, we're adding or subtracting the speed of the current. Well, what's the speed of the current? That's what the problem is asking us to find. So, we can define our variable here: Let x=speed of current

A 2x3 grid. First row labeled "Upstream," second row labeled "Downstream," first column labeled "Distance," second column labeled "rate," third column labeled "time." First row/first column contains the number 24, first row/second column contains the number 15-x, second row/first column contains the number 36, second row second column contains the number 15+x.

Now what about the time? Well, we don't know the time. But we do know that our distance equals our rate times our time: d=r⋅t We also have values for distance and rate. So we can solve this equation for time. d=r⋅tdr=r⋅trt=dr So we can just put in the distance divided by the rate into our slot for time.

A 2x3 grid. First row labeled "Upstream," second row labeled "Downstream," first column labeled "Distance," second column labeled "rate," third column labeled "time." First row/first column contains the number 24, first row/second column contains the number 15-x, first row/third column contains the number 24/(15-x). Second row/first column contains the number 36, second row second column contains the number 15+x, second row/third column contains the number 36/(15+x).

Now what does it say about the time in the problem? It can travel thirty-six miles downstream in the same amount of time that it can travel twenty-four miles upstream. The time it takes to travel upstream is equal to the time it takes to travel downstream. 2415−x=3615+x And now we have a rational equation which we can solve with cross multiplication! 2415−x=3615+x24(15+x)=36(15−x)24(15+x)12=36(15−x)122(15+x)=3(15−x)30+2x=45−3x−30         =−302x=15−3x+3x=    +3x5x=155x5=155x=3

Finally, we state our answer. The speed of the current is 3 miles per hour.

(Video) Word Problems with Rational Equations

by larryschmidt

steps in solving word problems involving rational algebraic expressions

Here is a really good video by larryschmidt showing you how to work through some word problems involving rational expressions.

One type of problem is when one person can work at a certain rate, and another person can work at a different rate, and you have to calculate how long it would take if these two people worked together.

The first problem he looks at is:

Sandra can paint a kitchen in 6 hours and Roger can paint the same kitchen in 7 hours. How long would it take for both working together to point the kitchen?

We start by setting up some rates per hour. It says that Sandra can paint 1 kitchen in 6 hours, or: 16 kitchen/hour

Roger can paint 1 kitchen in 7 hours, or: 17 kitchen/hour

Finally, let's let t be the time it takes when they work together. So they can paint 1 kitchen in t hours, or: 1t kitchen/hour

If they work together, this is the same as their rates being added, so we get the equation: 16+17=1t Let's solve for t. 16+17=1t77⋅16+17⋅66=1t742+642=1t1342=1t4213=t3.23≈t

So it takes approximately 3.23 hours for them to paint the kitchen together.

Test Yourself: Sam can beat a video game in 4 hours. Sara can beat the game in 3.5 hours. If they team up, how long will it take to beat the game.

2.48 hours

1.87 hours

1.13 hours

0.54 hours

Chapter 7: Rational Expressions and Equations

Solve rational equations, learning objectives.

  • Solve rational equations by clearing denominators
  • Identify extraneous solutions in a rational equation
  • Solve for a variable in a rational formula
  • Identify the components of a work equation
  • Solve a work equation
  • Define and write a proportion
  • Solve proportional problems involving scale drawings
  • Define direct variation, and solve problems involving direct variation
  • Define inverse variation and solve problems involving inverse variation
  • Define joint variation and solve problems involving joint variation

Equations that contain rational expressions are called rational equations . For example, [latex] \frac{2x+1}{4}=\frac{x}{3}[/latex] is a rational equation. Rational equations can be useful for representing real-life situations and for finding answers to real problems. In particular, they are quite good for describing a variety of proportional relationships.

One of the most straightforward ways to solve a rational equation is to eliminate denominators with the common denominator, then use properties of equality to isolate the variable. This method is often used to solve linear equations that involve fractions as in the following example:

Solve  [latex]\frac{1}{2}x-3=2-\frac{3}{4}x[/latex] by clearing the fractions in the equation first.

Multiply both sides of the equation by 4, the common denominator of the fractional coefficients.

[latex]\begin{array}{c}\frac{1}{2}x-3=2-\frac{3}{4}x\\ 4\left(\frac{1}{2}x-3\right)=4\left(2-\frac{3}{4}x\right)\\\text{}\\\,\,\,\,4\left(\frac{1}{2}x\right)-4\left(3\right)=4\left(2\right)+4\left(-\frac{3}{4}x\right)\\2x-12=8-3x\\\underline{+3x}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{+3x}\\5x-12=8\,\,\,\,\,\,\,\,\\\,\,\,\,\,\,\,\underline{+12}\,\,\,\,\underline{+12} \\5x=20\\x=4\end{array}[/latex]

We could have found a common denominator and worked with fractions, but that often leads to more mistakes. We can apply the same idea to solving rational equations.  The difference between a linear equation and a rational equation is that rational equations can have polynomials in the numerator and denominator of the fractions. This means that clearing the denominator may sometimes mean multiplying the whole rational equation by a polynomial. In the next example, we will clear the denominators of a rational equation with a terms that has a polynomial in the numerator.

Solve the equation [latex] \frac{x+5}{8}=\frac{7}{4}[/latex].

[latex]\begin{array}{l}4=2\cdot2\\8=2\cdot2\cdot2\cdot2\\\text{LCM}=2\cdot2\cdot2\\\text{LCM}=8\end{array}[/latex]

The LCM of 4 and 8 is also the lowest common denominator for the two fractions.

Multiply both sides of the equation by the common denominator, 8, to keep the equation balanced and to eliminate the denominators.

[latex]\begin{array}{r}8\cdot \frac{x+5}{8}=\frac{7}{4}\cdot 8\,\,\,\,\,\,\,\\\\\frac{8(x+5)}{8}=\frac{7(8)}{4}\,\,\,\,\,\,\\\\\frac{8}{8}\cdot (x+5)=\frac{7(4\cdot 2)}{4}\\\\\frac{8}{8}\cdot (x+5)=7\cdot 2\cdot \frac{4}{4}\\\\1\cdot (x+5)=14\cdot 1\,\,\,\end{array}[/latex]

Simplify and solve for x .

[latex]\begin{array}{r}x+5=14\\x=9\,\,\,\end{array}[/latex]

Check the solution by substituting 9 for x in the original equation.

[latex]\begin{array}{r}\frac{x+5}{8}=\frac{7}{4}\\\\\frac{9+5}{8}=\frac{7}{4}\\\\\frac{14}{8}=\frac{7}{4}\\\\\frac{7}{4}=\frac{7}{4}\end{array}[/latex]

[latex]x=9[/latex]

In the next example, we show how to solve a rational equation with a binomial in the denominator of one term. We will use the common denominator to eliminate the denominators from both fractions. Note that the LCD is the product of both denominators because they don’t share any common factors.

Solve the equation [latex] \frac{8}{x+1}=\frac{4}{3}[/latex].

[latex]\begin{array}{c}3\left(x+1\right)\left(\frac{8}{x+1}\right)=3\left(x+1\right)\left(\frac{4}{3}\right)\end{array}[/latex]

Simplify common factors.

[latex]\begin{array}{c}3\cancel{\left(x+1\right)}\left(\frac{8}{\cancel{x+1}}\right)=\cancel{3}\left(x+1\right)\left(\frac{4}{\cancel{3}}\right)\\24=4\left(x+1\right)\\24=4x+4\end{array}[/latex]

Now this looks like a linear equation, and we can use the addition and multiplication properties of equality to solve it.

[latex]\begin{array}{c}24=4x+4\\\underline{-4}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{-4}\\20=4x\,\,\,\,\,\,\,\,\\\\x=5\,\,\,\,\,\,\,\,\,\end{array}[/latex]

Check the solution in the original equation.

[latex]\begin{array}{r}\,\,\,\,\,\frac{8}{\left(x+1\right)}=\frac{4}{3}\\\\\frac{8}{\left(5+1\right)}=\frac{4}{3}\\\\\frac{8}{6}=\frac{4}{3}\end{array}[/latex]

Reduce the fraction [latex]\frac{8}{6}[/latex] by simplifying the common factor of 2:

[latex]\large\frac{\cancel{2}\cdot4}{\cancel{2}\cdot3}\normalsize=\large\frac{4}{3}[/latex]

[latex]x=1[/latex]

You could also solve this problem by multiplying each term in the equation by 3 to eliminate the fractions altogether. Here is how it would look.

Solve the equation [latex]\frac{x}{3}+1=\frac{4}{3}[/latex].

[latex] 3\left( \frac{x}{3}+1 \right)=3\left( \frac{4}{3} \right)[/latex]

Apply the distributive property and multiply 3 by each term within the parentheses. Then simplify and solve for x .

[latex]\begin{array}{r}3\left( \frac{x}{3} \right)+3\left( 1 \right)=3\left( \frac{4}{3} \right)\\\\\cancel{3}\left( \frac{x}{\cancel{3}} \right)+3\left( 1 \right)=\cancel{3}\left( \frac{4}{\cancel{3}} \right)\\\\ x+3=4\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{-3}\,\,\,\,\,\underline{-3}\\\\x=1\end{array}[/latex]

In the video that follows we present two ways to solve rational equations with both integer and variable denominators.

Excluded Values and Extraneous Solutions

Some rational expressions have a variable in the denominator. When this is the case, there is an extra step in solving them. Since division by 0 is undefined, you must exclude values of the variable that would result in a denominator of 0. These values are called excluded values . Let’s look at an example.

Solve the equation [latex] \frac{2x-5}{x-5}=\frac{15}{x-5}[/latex].

[latex] \frac{2x-5}{x-5}=\frac{15}{x-5}[/latex]

5 is an excluded value because it makes the denominator [latex]x-5[/latex] equal to 0.

Since the denominator of each expression in the equation is the same, the numerators must be equal. Set the numerators equal to one another and solve for x.

[latex]\begin{array}{r}2x-5=15\\2x=20\\x=10\end{array}[/latex]

[latex]\begin{array}{r}\frac{2x-5}{x-5}=\frac{15}{x-5}\,\,\\\\\frac{2(10)-5}{10-5}=\frac{15}{10-5}\\\\\frac{20-5}{10-5}=\frac{15}{10-5}\\\\\frac{15}{5}=\frac{15}{5}\,\,\,\,\,\,\,\,\,\end{array}[/latex]

[latex]x=10[/latex]

In the following video we present an example of solving a rational equation with variables in the denominator.

You’ve seen that there is more than one way to solve rational equations. Because both of these techniques manipulate and rewrite terms, sometimes they can produce solutions that don’t work in the original form of the equation. These types of answers are called extraneous solutions . That’s why it is always important to check all solutions in the original equations—you may find that they yield untrue statements or produce undefined expressions.

Solve the equation [latex] \frac{16}{m+4}=\frac{{{m}^{2}}}{m+4}[/latex].

Since the denominator of each expression in the equation is the same, the numerators must be equal. Set the numerators equal to one another and solve for m.

[latex]\begin{array}{l}16=m^{2}\\\,\,\,0={{m}^{2}}-16\\\,\,\,0=\left( m+4 \right)\left( m-4 \right)\end{array}[/latex]

[latex]\begin{array}{c}0=m+4\,\,\,\,\,\,\text{or}\,\,\,\,\,\,0=m-4\\m=-4\,\,\,\,\,\,\text{or}\,\,\,\,\,\,m=4\\m=4,-4\end{array}[/latex]

Check the solutions in the original equation.

Since [latex]m=−4[/latex] leads to division by 0, it is an extraneous solution.

[latex]\begin{array}{c}\frac{16}{m+4}=\frac{{{m}^{2}}}{m+4}\\\\\frac{16}{-4+4}=\frac{{{(-4)}^{2}}}{-4+4}\\\\\frac{16}{0}=\frac{16}{0}\end{array}[/latex]

[latex]-4[/latex] is excluded because it leads to division by 0.

[latex]\begin{array}{c}\frac{16}{4+4}=\frac{{{(4)}^{2}}}{4+4}\\\\\frac{16}{8}=\frac{16}{8}\end{array}[/latex]

[latex]m=4[/latex]

Rational formulas

Rational formulas can be useful tools for representing real-life situations and for finding answers to real problems. Equations representing direct, inverse, and joint variation are examples of rational formulas that can model many real-life situations. As you will see, if you can find a formula, you can usually make sense of a situation.

When solving problems using rational formulas, it is often helpful to first solve the formula for the specified variable. For example, work problems ask you to calculate how long it will take different people working at different speeds to finish a task. The algebraic models of such situations often involve rational equations derived from the work formula, [latex]W=rt[/latex]. The amount of work done ( W ) is the product of the rate of work ( r ) and the time spent working ( t ). Using algebra, you can write the work formula 3 ways:

[latex]W=rt[/latex]

Find the time (t): [latex] t=\frac{W}{r}[/latex] (divide both sides by r)

Find the rate (r): [latex] r=\frac{W}{t}[/latex] (divide both sides by t)

The formula for finding the density of an object is [latex] D=\frac{m}{v}[/latex], where D is the density, m is the mass of the object and v is the volume of the object. Rearrange the formula to solve for the mass ( m ) and then for the volume ( v ).

[latex] D=\frac{m}{v}[/latex]

Multiply both side of the equation by v to isolate m.

[latex] v\cdot D=\frac{m}{v}\cdot v[/latex]

Simplify and rewrite the equation, solving for m .

[latex]\begin{array}{l}v\cdot D=m\cdot \frac{v}{v}\\v\cdot D=m\cdot 1\\v\cdot D=m\end{array}[/latex]

To solve the equation [latex] D=\frac{m}{v}[/latex] in terms of v , you will need do the same steps to this point, and then divide both sides by D .

[latex]\begin{array}{r}\frac{v\cdot D}{D}=\frac{m}{D}\\\\\frac{D}{D}\cdot v=\frac{m}{D}\\\\1\cdot v=\frac{m}{D}\\\\v=\frac{m}{D}\end{array}[/latex]

[latex] m=D\cdot v[/latex] and [latex] v=\frac{m}{D}[/latex]

Now let’s look at an example using the formula for the volume of a cylinder.

The formula for finding the volume of a cylinder is [latex]V=\pi{r^{2}}h[/latex], where V is the volume, r is the radius and h is the height of the cylinder. Rearrange the formula to solve for the height ( h ).

[latex] V=\pi{{r}^{2}}h[/latex]

Divide both sides by [latex] \pi {{r}^{2}}[/latex] to isolate h.

[latex] \frac{V}{\pi {{r}^{2}}}=\frac{\pi {{r}^{2}}h}{\pi {{r}^{2}}}[/latex]

Simplify. You find the height, h , is equal to [latex] \frac{V}{\pi {{r}^{2}}}[/latex].

[latex] \frac{V}{\pi {{r}^{2}}}=h[/latex]

[latex] h=\frac{V}{\pi {{r}^{2}}}[/latex]

In the following video we give another example of solving for a variable in a formula, or as they are also called, a literal equation.

Applications of Rational Equations

Rational equations can be used to solve a variety of problems that involve rates, times and work. Using rational expressions and equations can help you answer questions about how to combine workers or machines to complete a job on schedule.

Man with a lunch box walking. THere is a caption above him that says "Boy! I sure did a good day's work today"

A Good Day’s Work

A “work problem” is an example of a real life situation that can be modeled and solved using a rational equation. Work problems often ask you to calculate how long it will take different people working at different speeds to finish a task. The algebraic models of such situations often involve rational equations derived from the work formula, [latex]W=rt[/latex]. (Notice that the work formula is very similar to the relationship between distance, rate, and time, or [latex]d=rt[/latex].) The amount of work done ( W ) is the product of the rate of work ( r ) and the time spent working ( t ). The work formula has 3 versions.

[latex]\begin{array}{l}W=rt\\\\\,\,\,\,\,t=\frac{W}{r}\\\\\,\,\,\,\,r=\frac{W}{t}\end{array}[/latex]

Some work problems include multiple machines or people working on a project together for the same amount of time but at different rates. In that case, you can add their individual work rates together to get a total work rate. Let’s look at an example.

Myra takes 2 hours to plant 50 flower bulbs. Francis takes 3 hours to plant 45 flower bulbs. Working together, how long should it take them to plant 150 bulbs?

Myra: [latex] \frac{50\,\,\text{bulbs}}{2\,\,\text{hours}}[/latex], or [latex] \frac{25\,\,\text{bulbs}}{1\,\,\text{hour}}[/latex]

Francis: [latex] \frac{45\,\,\text{bulbs}}{3\,\,\text{hours}}[/latex], or [latex] \frac{15\,\,\text{bulbs}}{1\,\,\text{hour}}[/latex]

Combine their hourly rates to determine the rate they work together.

Myra and Francis together:

[latex] \frac{25\,\,\text{bulbs}}{1\,\,\text{hour}}+\frac{15\,\,\text{bulbs}}{1\,\,\text{hour}}=\frac{40\,\,\text{bulbs}}{1\,\,\text{hour}}[/latex]

Use one of the work formulas to write a rational equation, for example [latex] r=\frac{W}{t}[/latex]. You know r , the combined work rate, and you know W , the amount of work that must be done. What you don’t know is how much time it will take to do the required work at the designated rate.

[latex] \frac{40}{1}=\frac{150}{t}[/latex]

Solve the equation by multiplying both sides by the common denominator, then isolating t .

[latex]\begin{array}{c}\frac{40}{1}\cdot 1t=\frac{150}{t}\cdot 1t\\\\40t=150\\\\t=\frac{150}{40}=\frac{15}{4}\\\\t=3\frac{3}{4}\text{hours}\end{array}[/latex]

It should take 3 hours 45 minutes for Myra and Francis to plant 150 bulbs together.

Other work problems go the other way. You can calculate how long it will take one person to do a job alone when you know how long it takes people working together to complete the job.

Joe and John are planning to paint a house together. John thinks that if he worked alone, it would take him 3 times as long as it would take Joe to paint the entire house. Working together, they can complete the job in 24 hours. How long would it take each of them, working alone, to complete the job?

Let x = time it takes Joe to complete the job

3 x = time it takes John to complete the job

The work is painting 1 house or 1. Write an expression to represent each person’s rate using the formula [latex] r=\frac{W}{t}[/latex] .

Joe’s rate: [latex] \frac{1}{x}[/latex]

John’s rate: [latex] \frac{1}{3x}[/latex]

Their combined rate is the sum of their individual rates. Use this rate to write a new equation using the formula [latex]W=rt[/latex].

combined rate: [latex] \frac{1}{x}+\frac{1}{3x}[/latex]

The problem states that it takes them 24 hours together to paint a house, so if you multiply their combined hourly rate [latex] \left( \frac{1}{x}+\frac{1}{3x} \right)[/latex] by 24, you will get 1, which is the number of houses they can paint in 24 hours.

[latex] \begin{array}{l}1=\left( \frac{1}{x}+\frac{1}{3x} \right)24\\\\1=\frac{24}{x}+\frac{24}{3x}\end{array}[/latex]

Now solve the equation for x . (Remember that x represents the number of hours it will take Joe to finish the job.)

[latex]\begin{array}{l}\,\,\,1=\frac{3}{3}\cdot \frac{24}{x}+\frac{24}{3x}\\\\\,\,\,1=\frac{3\cdot 24}{3x}+\frac{24}{3x}\\\\\,\,\,1=\frac{72}{3x}+\frac{24}{3x}\\\\\,\,\,1=\frac{72+24}{3x}\\\\\,\,\,1=\frac{96}{3x}\\\\3x=96\\\\\,\,\,x=32\end{array}[/latex]

[latex]\begin{array}{l}1=\left( \frac{1}{x}+\frac{1}{3x} \right)24\\\\1=\left[ \frac{\text{1}}{\text{32}}+\frac{1}{3\text{(32})} \right]24\\\\1=\frac{24}{\text{32}}+\frac{24}{3\text{(32})}\\\\1=\frac{24}{\text{32}}+\frac{24}{96}\\\\1=\frac{3}{3}\cdot \frac{24}{\text{32}}+\frac{24}{96}\\\\1=\frac{72}{96}+\frac{24}{96}[\end{array}[/latex]

The solution checks. Since [latex]x=32[/latex], it takes Joe 32 hours to paint the house by himself. John’s time is 3 x , so it would take him 96 hours to do the same amount of work.

It takes 32 hours for Joe to paint the house by himself and 96 hours for John the paint the house himself.

In the video that follows, we show another example of finding one person’s work rate given a combined work rate.

As shown above, many work problems can be represented by the equation [latex] \frac{t}{a}+\frac{t}{b}=1[/latex], where t is the time to do the job together, a is the time it takes person A to do the job, and b is the time it takes person B to do the job. The 1 refers to the total work done—in this case, the work was to paint 1 house.

The key idea here is to figure out each worker’s individual rate of work. Then, once those rates are identified, add them together, multiply by the time t , set it equal to the amount of work done, and solve the rational equation.

We present another example of two people painting at different rates in the following video.

Proportions

Matryoshka, or nesting dolls.

Matryoshka, or nesting dolls.

A proportion is a statement that two ratios are equal to each other.  There are many things that can be represented with ratios, including the actual distance on the earth that is represented on a map.  In fact, you probably use proportional reasoning on a regular basis and don’t realize it.  For example, say you have volunteered to provide drinks for a community event.  You are asked to bring enough drinks for 35-40 people.  At the store  you see that drinks come in packages of 12. You multiply 12 by 3 and get 36 – this may not be enough if 40 people show up, so you decide to buy 4 packages of drinks just to be sure.

This process can also be expressed as a proportional equation and solved using mathematical principles. First, we can express the number of drinks in a package as a ratio:

[latex]\frac{12\text{ drinks }}{1\text{ package }}[/latex]

Then we express the number of people who we are buying drinks for as a ratio with the unknown number of packages we need. We will use the maximum so we have enough.

[latex]\frac{40\text{ people }}{x\text{ packages }}[/latex]

We can find out how many packages to purchase by setting the expressions equal to each other:

[latex]\frac{12\text{ drinks }}{1\text{ package }}=\frac{40\text{ people }}{x\text{ packages }}[/latex]

To solve for x, we can use techniques for solving linear equations, or we can cross multiply as a shortcut.

[latex]\begin{array}{l}\,\,\,\,\,\,\,\frac{12\text{ drinks }}{1\text{ package }}=\frac{40\text{ people }}{x\text{ packages }}\\\text{}\\x\cdot\frac{12\text{ drinks }}{1\text{ package }}=\frac{40\text{ people }}{x\text{ packages }}\cdot{x}\\\text{}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,12x=40\\\text{}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x=\frac{40}{12}=\frac{10}{3}=3.33\end{array}[/latex]

We can round up  to 4 since it doesn’t make sense to by 0.33 of a package of drinks.  Of course, you don’t write out your thinking this way when you are in the grocery store, but doing so helps you to be able to apply the concepts to less obvious problems.  In the following example we will show how to use a proportion to find the number of people on teh planet who don’t have access to a toilet.

As of March, 2016 the world’s population was estimated at 7.4 billion.  [1] .  According to water.org , 1 out of every 3 people on the planet lives without access to a toilet.  Find the number of people on the planet that do not have access to a toilet.

Read and Understand:  We can use a proportion to find the unknown number of people who live without a toilet since we are given that 1 in 3 don’t have access, and we are given the population of the planet.

Define and Translate:  We know that 1 out of every 3 people don’t have access, so we can write that as a ratio (fraction)

[latex]\frac{1\text{ doesn't }}{3\text{ do }}[/latex].

Let the number of people without access to a toilet be x. The ratio of people with and without toilets is then

[latex]\frac{x\text{ don't }}{7.4\text{ billion do}}[/latex]

Notice how it helps to use descriptions or units to know where to place the given numbers in the proportion.

Write and Solve:  Equate the two ratios since they are representing the same fractional amount of the population.

[latex]\frac{1}{3}=\frac{x}{7.4\text{ billion }}[/latex]

[latex]\begin{array}{l}\frac{1}{3}=\frac{x}{7.4}\\\text{}\\7.4\cdot\frac{1}{3}=\frac{x}{7.4}\cdot{7.4}\\\text{}\\2.46=x\end{array}[/latex]

Interpret:  The original units were billions of people, so our answer is [latex]2.46[/latex] billion people don’t have access to a toilet.  Wow, that’s a lot of people.

2.46 billion people don’t have access to a toilet.

In the next example, we will use the length of a person’t femur to estimate their height.  This process is used in forensic science and anthropology, and has been found in many scientific studies to be a very good estimate.

It has been shown that a person’s height is proportional to the length of their femur  [2] . Given that a person who is 71 inches tall has a femur length of 17.75 inches, how tall is someone with a femur length of 16 inches?

Read and Understand:  Height and femur length are proportional for everyone, so we can define a ratio with the given height and femur length.  We can then use this to write a proportion to find the unknown height.

Define and Translate:  Let x be the unknown height.  Define the ratio of femur length and height for both people using the given measurements.

Person 1:  [latex]\frac{\text{femur length}}{\text{height}}=\frac{17.75\text{inches}}{71\text{inches}}[/latex]

Person 2:  [latex]\frac{\text{femur length}}{\text{height}}=\frac{16\text{inches}}{x\text{inches}}[/latex]

Write and Solve:  Equate the ratios, since we are assuming height and femur length are proportional for everyone.

[latex]\frac{17.75\text{inches}}{71\text{inches}}=\frac{16\text{inches}}{x\text{inches}}[/latex]

 Solve by using the common denominator to clear fractions.  The common denominator is [latex]71x[/latex]

[latex]\begin{array}{c}\frac{17.75}{71}=\frac{16}{x}\\\\71x\cdot\frac{17.75}{71}=\frac{16}{x}\cdot{71x}\\\\17.75\cdot{x}=16\cdot{71}\\\\x=\frac{16\cdot{71}}{17.75}=64\end{array}[/latex]

Interpret:  The unknown height of person 2 is 64 inches. In general, we can reduce the fraction [latex]\frac{17.75}{71}=0.25=\frac{1}{4}[/latex] to find a general rule for everyone.  This would translate to saying for every one femur length, a person’s height is 4 times that length.

Another way to describe the ratio of femur length to height that we found in the last example is to say there’s a 1:4 ratio between femur length and height, or 1 to 4.

Ratios are also used in scale drawings. Scale drawings are enlarged or reduced drawings of objects, buildings, roads, and maps. Maps are smaller than what they represent and a drawing of dendritic cells in your brain is most likely larger than what it represents. The scale of the drawing is a ratio that represents a comparison of the length of the actual object and it’s representation in the drawing. The image below shows a map of the us with a scale of 1 inch representing 557 miles. We could write the scale factor as a fraction [latex]\frac{1}{557}[/latex] or as we did with the femur-height relationship, 1:557.

map of the lower 48 states with a scale factor of 1 inch equals 557 miles.

Map with scale factor

In the next example we will use the scale factor given in the image above to find the distance between Seattle Washington and San Jose California.

Given a scale factor of 1:557 on a map of the US, if the distance from Seattle, WA to San Jose, CA is 1.5 inches on the map,  define a proportion to find the actual distance between them.

Read and Understand:  We need to define a proportion to solve for the unknown distance between Seattle and San Jose.

Define and Translate: T he scale factor is 1:557, and we will call the unknown distance x. The ratio of inches to miles is [latex]\frac{1}{557}[/latex].

We know inches between the two cities, but we don’t know miles, so the ratio that describes the distance between them is [latex]\frac{1.5}{x}[/latex].

Write and Solve:  The proportion that will help us solve this problem is [latex]\frac{1}{557}=\frac{1.5}{x}[/latex].

Solve using the common denominator [latex]557x[/latex] to clear fractions.

[latex]\begin{array}{c}\frac{1}{557}=\frac{1.5}{x}\\\\557x\cdot\frac{1}{557}=\frac{1.5}{x}\cdot{557x}\\\\x=1.5\cdot{557}=835.5[/latex]

Interpret:  We used the scale factor 1:557 to find an unknown distance between Seattle and San Jose. We also check our answer of 835.5 miles with Google maps, and found that the distance is 839.9 miles, so we did pretty well!

In the next example, we will find a scale factor given the length between two cities on a map, and their actual distance from each other.

Two cities are 2.5 inches apart on a map.  Their actual distance from each other is 325 miles.  Write a proportion to represent and solve for the scale factor for one inch of the map.

Read and Understand: We know that for each 2.5 inches on the map, it represents 325 actual miles. We are looking for the scale factor for one inch of the map.

Define and Translate:  The ratio we want is [latex]\frac{1}{x}[/latex] where x is the actual distance represented by one inch on the map.  We know that for every 2.5 inches, there are 325 actual miles, so we can define that relationship as [latex]\frac{2.5}{325}[/latex]

Write and Solve:  We can use a proportion to equate the two ratios and solve for the unknown distance.

[latex]\begin{array}{c}\frac{1}{x}=\frac{2.5}{325}\\\\325x\cdot\frac{1}{x}=\frac{2.5}{325}\cdot{325x}\\\\325=2.5x\\\\x=130[/latex]

Interpret:  The scale factor for one inch on the map is 1:130, or for every inch of map there are 130 actual miles.

The video that follows show another example of finding an actual distance using the scale factor from a map.

In the video that follows, we present an example of using proportions to obtain the correct amount of medication for a patient, as well as finding a desired mixture of coffees.

Huge parking lot full of cars.

So many cars, so many tires.

Direct Variation

Variation equations are examples of rational formulas and are used to describe the relationship between variables. For example, imagine a parking lot filled with cars. The total number of tires in the parking lot is dependent on the total number of cars. Algebraically, you can represent this relationship with an equation.

[latex]\text{number of tires}=4\cdot\text{number of cars}[/latex]

The number 4 tells you the rate at which cars and tires are related. You call the rate the constant of variation . It’s a constant because this number does not change. Because the number of cars and the number of tires are linked by a constant, changes in the number of cars cause the number of tires to change in a proportional, steady way. This is an example of direct variation , where the number of tires varies directly with the number of cars.

You can use the car and tire equation as the basis for writing a general algebraic equation that will work for all examples of direct variation. In the example, the number of tires is the output, 4 is the constant, and the number of cars is the input. Let’s enter those generic terms into the equation. You get [latex]y=kx[/latex]. That’s the formula for all direct variation equations.

[latex]\text{number of tires}=4\cdot\text{number of cars}\\\text{output}=\text{constant}\cdot\text{input}[/latex]

Solve for k , the constant of variation, in a direct variation problem where [latex]y=300[/latex] and [latex]x=10[/latex].

[latex]y=kx[/latex]

Substitute known values into the equation.

[latex]300=k\left(10\right)[/latex]

Solve for k by dividing both sides of the equation by 10.

[latex]\begin{array}{l}\frac{300}{10}=\frac{10k}{10}\\\\\,\,\,\,30=k\end{array}[/latex]

The constant of variation, k , is 30.

In the video that follows, we present an example of solving a direct variation equation.

Inverse Variation

Another kind of variation is called inverse variation . In these equations, the output equals a constant divided by the input variable that is changing. In symbolic form, this is the equation [latex] y=\frac{k}{x}[/latex].

One example of an inverse variation is the speed required to travel between two cities in a given amount of time.

Let’s say you need to drive from Boston to Chicago, which is about 1,000 miles. The more time you have, the slower you can go. If you want to get there in 20 hours, you need to go 50 miles per hour (assuming you don’t stop driving!), because [latex] \frac{1,000}{20}=50[/latex]. But if you can take 40 hours to get there, you only have to average 25 miles per hour, since [latex] \frac{1,000}{40}=25[/latex].

The equation for figuring out how fast to travel from the amount of time you have is [latex] speed=\frac{miles}{time}[/latex]. This equation should remind you of the distance formula [latex] d=rt[/latex]. If you solve [latex] d=rt[/latex] for r , you get [latex] r=\frac{d}{t}[/latex], or [latex] speed=\frac{miles}{time}[/latex].

In the case of the Boston to Chicago trip, you can write [latex] s=\frac{1,000}{t}[/latex]. Notice that this is the same form as the inverse variation function formula, [latex] y=\frac{k}{x}[/latex].

Solve for k , the constant of variation, in an inverse variation problem where [latex]x=5[/latex] and [latex]y=25[/latex].

[latex] y=\frac{k}{x}[/latex]

[latex] 25=\frac{k}{5}[/latex]

Solve for k by multiplying both sides of the equation by 5.

[latex] \begin{array}{c}5\cdot 25=\frac{k}{5}\cdot 5\\\\125=\frac{5k}{5}\\\\125=k\,\,\,\end{array}[/latex]

The constant of variation, k , is 125.

In the next example, we will find the water temperature in the ocean at a depth of 500 meters.  Water temperature is inversely proportional to depth in the ocean.

Scuba divers in the ocean.

Water temperature in the ocean varies inversely with depth.

The water temperature in the ocean varies inversely with the depth of the water. The deeper a person dives, the colder the water becomes. At a depth of 1,000 meters, the water temperature is 5º Celsius. What is the water temperature at a depth of 500 meters?

[latex]\begin{array}{l}\,\,\,\,\,\,\,\,\,\,y=\frac{k}{x}\\\\temp=\frac{k}{depth}\end{array}[/latex]

[latex] 5=\frac{k}{1,000}[/latex]

Solve for k .

[latex]\begin{array}{l}1,000\cdot5=\frac{k}{1,000}\cdot 1,000\\\\\,\,\,\,\,\,\,\,5,000=\frac{1,000k}{1,000}\\\\\,\,\,\,\,\,\,\,5,000=k\end{array}[/latex]

Now that k , the constant of variation is known, use that information to solve the problem: find the water temperature at 500 meters.

[latex]\begin{array}{l}temp=\frac{k}{depth}\\\\temp=\frac{5,000}{500}\\\\temp=10\end{array}[/latex]

At 500 meters, the water temperature is 10º C.

In the video that follows, we present an example of inverse variation.

Joint Variation

A third type of variation is called joint variation . Joint variation is the same as direct variation except there are two or more quantities. For example, the area of a rectangle can be found using the formula [latex]A=lw[/latex], where l is the length of the rectangle and w is the width of the rectangle. If you change the width of the rectangle, then the area changes and similarly if you change the length of the rectangle then the area will also change. You can say that the area of the rectangle “varies jointly with the length and the width of the rectangle.”

The formula for the volume of a cylinder, [latex]V=\pi {{r}^{2}}h[/latex] is another example of joint variation. The volume of the cylinder varies jointly with the square of the radius and the height of the cylinder. The constant of variation is [latex] \pi [/latex].

The area of a triangle varies jointly with the lengths of its base and height. If the area of a triangle is 30 inches[latex]^{2}[/latex] when the base is 10 inches and the height is 6 inches, find the variation constant and the area of a triangle whose base is 15 inches and height is 20 inches.

[latex]\begin{array}{l}\,\,\,\,\,\,\,\,\,\,y=kxz\\Area=k(base)(height)\end{array}[/latex]

Substitute known values into the equation, and solve for k .

[latex]30=k\left(10\right)\left(6\right)\\30=60k\\\\\frac{30}{60}=\frac{60k}{60}\\\\\frac{1}{2}=k[/latex]

Now that k is known, solve for the area of a triangle whose base is 15 inches and height is 20 inches.

[latex]\begin{array}{l}Area=k(base)(height)\\\\Area=(15)(20)(\frac{1}{2})\\\\Area=\frac{300}{2}\\\\Area=150\,\,\text{square inches}\end{array}[/latex]

The constant of variation, k , is [latex] \frac{1}{2}[/latex], and the area of the triangle is 150 square inches.

Finding k to be [latex] \frac{1}{2}[/latex] shouldn’t be surprising. You know that the area of a triangle is one-half base times height, [latex] A=\frac{1}{2}bh[/latex]. The [latex] \frac{1}{2}[/latex] in this formula is exactly the same [latex] \frac{1}{2}[/latex] that you calculated in this example!

In the following video, we show an example of finding the constant of variation for a jointly varying relation.

Direct, Joint, and Inverse Variation

k is the constant of variation. In all cases, [latex]k\neq0[/latex].

  • Direct variation: [latex]y=kx[/latex]
  • Inverse variation: [latex] y=\frac{k}{x}[/latex]
  • Joint variation: [latex]y=kxz[/latex]

Rational formulas can be used to solve a variety of problems that involve rates, times, and work. Direct, inverse, and joint variation equations are examples of rational formulas. In direct variation, the variables have a direct relationship—as one quantity increases, the other quantity will also increase. As one quantity decreases, the other quantity decreases. In inverse variation, the variables have an inverse relationship—as one variable increases, the other variable decreases, and vice versa. Joint variation is the same as direct variation except there are two or more variables.

You can solve rational equations by finding a common denominator. By rewriting the equation so that all terms have the common denominator, you can solve for the variable using just the numerators. Or, you can multiply both sides of the equation by the least common multiple of the denominators so that all terms become polynomials instead of rational expressions.

An important step in solving rational equations is to reject any extraneous solutions from the final answer. Extraneous solutions are solutions that don’t satisfy the original form of the equation because they produce untrue statements or are excluded values that make a denominator equal to 0.

  • "Current World Population." World Population Clock: 7.4 Billion People (2016). Accessed June 21, 2016. http://www.worldometers.info/world-population/ . "Current World Population." World Population Clock: 7.4 Billion People (2016). Accessed June 21, 2016. http://www.worldometers.info/world-population/ . "Current World Population." World Population Clock: 7.4 Billion People (2016). Accessed June 21, 2016. http://www.worldometers.info/world-population/ . ↵
  • Obialor, Ambrose, Churchill Ihentuge, and Frank Akapuaka. "Determination of Height Using Femur Length in Adult Population of Oguta Local Government Area of Imo State Nigeria." Federation of American Societies for Experimental Biology, April 2015. Accessed June 22, 2016. http://www.fasebj.org/content/29/1_Supplement/LB19.short . ↵
  • Revision and Adaptation. Provided by : Lumen Learning. License : CC BY: Attribution
  • Solve Basic Rational Equations. Authored by : James Sousa (Mathispower4u.com) for Lumen Learning. Located at : https://youtu.be/R9y2D9VFw0I . License : CC BY: Attribution
  • Solve Rational Equations with Like Denominators. Authored by : James Sousa (Mathispower4u.com) for Lumen Learning. Located at : https://youtu.be/gGA-dF_aQQQ . License : CC BY: Attribution
  • Screenshot: A Good Day's Work. Provided by : Lumen Learning. License : CC BY: Attribution
  • Screenshot: Matroyshka, or nesting dolls.. Provided by : Lumen Learning. License : CC BY: Attribution
  • Screenshot: map with scale factor. Provided by : Lumen Learning. License : CC BY: Attribution
  • Proportion Applications: Map Scale Factor (Clear Fractions, No Cross Products). Authored by : James Sousa (Mathispower4u.com) for Lumen Learning. Located at : https://youtu.be/id3sp4wvmVg . License : CC BY: Attribution
  • Screenshot: so many cars, so many tires. Provided by : Lumen Learning. License : CC BY: Attribution
  • Screenshot: Water temperature in the ocean varies inversely with depth. Provided by : Lumen Learning. License : CC BY: Attribution
  • Joint Variation: Determine the Variation Constant (Volume of a Cone). Authored by : James Sousa (Mathispower4u.com) for Lumen Learning. Located at : https://youtu.be/JREPATMScbM . License : CC BY: Attribution
  • Unit 15: Rational Expressions, from Developmental Math: An Open Program. Provided by : Monterey Institute of Technology and Education. License : CC BY: Attribution
  • Ex 2: Solve a Literal Equation for a Variable. Authored by : James Sousa (Mathispower4u.com). Located at : https://www.youtube.com/watch?v=ecEUUbRLDQs&feature=youtu.be . License : CC BY: Attribution
  • Ex 1: Rational Equation Application - Painting Together. Authored by : James Sousa (Mathispower4u.com) . Located at : https://youtu.be/SzSasnDF7Ms . License : CC BY: Attribution
  • Ex: Rational Equation App - Find Individual Working Time Given Time Working Together. Authored by : James Sousa (Mathispower4u.com) . Located at : https://www.youtube.com/watch?v=kbRSYb8UYqU&feature=youtu.be . License : CC BY: Attribution
  • Ex: Proportion Applications - Mixtures . Authored by : James Sousa (Mathispower4u.com) . Located at : https://youtu.be/yGid1a_x38g . License : CC BY: Attribution
  • Ex: Direct Variation Application - Aluminum Can Usage. Authored by : James Sousa (Mathispower4u.com) . Located at : https://youtu.be/DLPKiMD_ZZw . License : CC BY: Attribution
  • Ex: Inverse Variation Application - Number of Workers and Job Time. Authored by : James Sousa (Mathispower4u.com). Located at : https://youtu.be/y9wqI6Uo6_M . License : CC BY: Attribution

Solving Rational Equations and Applications

Learning Objective(s)

·          Solve rational equations.

·          Check for extraneous solutions.

·          Solve application problems involving rational equations.

Introduction

You can solve these equations using the techniques for performing operations with rational expressions and the procedures for solving algebraic equations. Rational equations can be useful for representing real-life situations and for finding answers to real problems. In particular, they are quite good for describing distance-speed-time relationships and for modeling work problems that involve more than one person.

Solving Rational Equations

One method for solving rational equations is to rewrite the rational expressions in terms of a common denominator. Then, since you know the numerators are equal, you can solve for the variable. To illustrate this, let’s look at a very simple equation.

Since the denominator of each expression is the same, the numerators must be equivalent. This means that x = 2.

This is true for rational equations with polynomials too.

Since the denominators of each rational expression are the same, x + 4, the numerators must be equivalent for the equation to be true. So, x – 5 = 11 and x = 16.

Just as with other algebraic equations, you can check your solution in the original rational equation by substituting the value for the variable back into the equation and simplifying.

When the terms in a rational equation have unlike denominators, solving the equation will involve some extra steps. One way of solving rational equations with unlike denominators is to multiply both sides of the equation by the least common multiple of the denominators of all the fractions contained in the equation. This eliminates the denominators and turns the rational equation into a polynomial equation. Here’s an example.

Another way to solve a rational equation with unlike denominators is to rewrite each term with a common denominator and then just create an equation from the numerators. This works because if the denominators are the same, the numerators must be equal. The next example shows this approach with the same equation you just solved:

In some instances, you’ll need to take some additional steps in finding a common denominator. Consider the example below, which illustrates using what you know about denominators to rewrite one of the expressions in the equation.

You could also solve this problem by multiplying each term in the equation by 3 to eliminate the fractions altogether. Here is how it would look.

Excluded Values and Extraneous Solutions

Some rational expressions have a variable in the denominator. When this is the case, there is an extra step in solving them. Since division by 0 is undefined, you must exclude values of the variable that would result in a denominator of 0. These values are called excluded values . Let’s look at an example.

Let’s look at an example with a more complicated denominator.

You’ve seen that there is more than one way to solve rational equations. Because both of these techniques manipulate and rewrite terms, sometimes they can produce solutions that don’t work in the original form of the equation. These types of answers are called extraneous solutions . That's why it is always important to check all solutions in the original equations—you may find that they yield untrue statements or produce undefined expressions.

Since the denominator of each expression in the equation is the same, the numerators must be equal. Set the numerators equal to one another and solve for m.

-4 is excluded because

it leads to division by 0.

Check the solutions in the original equation.

Since m = −4 leads to division by 0, it is an extraneous solution.

Solving Application Problems

A “work problem” is an example of a real life situation that can be modeled and solved using a rational equation. Work problems often ask you to calculate how long it will take different people working at different speeds to finish a task. The algebraic models of such situations often involve rational equations derived from the work formula, W = rt . (Notice that the work formula is very similar to the relationship between distance, rate, and time, or d = rt .) The amount of work done ( W ) is the product of the rate of work ( r ) and the time spent working ( t ). The work formula has 3 versions.

Some work problems include multiple machines or people working on a project together for the same amount of time but at different rates. In that case, you can add their individual work rates together to get a total work rate. Let’s look at an example.

Other work problems go the other way. You can calculate how long it will take one person to do a job alone when you know how long it takes people working together to complete the job.

The key idea here is to figure out each worker’s individual rate of work. Then, once those rates are identified, add them together, multiply by the time t , set it equal to the amount of work done, and solve the rational equation.

You can solve rational equations by finding a common denominator. By rewriting the equation so that all terms have the common denominator, you can solve for the variable using just the numerators. Or, you can multiply both sides of the equation by the least common multiple of the denominators so that all terms become polynomials instead of rational expressions. Rational equations can be used to solve a variety of problems that involve rates, times and work. Using rational expressions and equations can help you answer questions about how to combine workers or machines to complete a job on schedule.

An important step in solving rational equations is to reject any extraneous solutions from the final answer. Extraneous solutions are solutions that don't satisfy the original form of the equation because they produce untrue statements or are excluded values that make a denominator equal to 0.

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Mathematics LibreTexts

7.3: Adding and Subtracting Rational Expressions

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  • Page ID 18370

Learning Objectives

  • Add and subtract rational expressions with common denominators.
  • Add and subtract rational expressions with unlike denominators.
  • Add and subtract rational functions.

Adding and Subtracting with Common Denominators

Adding and subtracting rational expressions is similar to adding and subtracting fractions. Recall that if the denominators are the same, we can add or subtract the numerators and write the result over the common denominator.

\(\begin{aligned} \frac{3}{13}+\frac{7}{13} &=\frac{3+7}{13} \\ &=\frac{10}{13} \end{aligned}\)

When working with rational expressions, the common denominator will be a polynomial. In general, given polynomials P , Q , and R , where \(Q≠0\), we have the following:

\[\frac{P}{Q} \pm \frac{R}{Q}=\frac{P \pm R}{Q}\]

In this section, assume that all variable factors in the denominator are nonzero.

Example \(\PageIndex{1}\)

\(\frac{3}{y}+\frac{7}{y}\)

Add the numerators \(3\) and \(7\), and write the result over the common denominator, y .

\(\begin{aligned} \frac{3}{y}+\frac{7}{y} &=\frac{3+7}{y} \\ &=\frac{10}{y} \end{aligned}\)

\(\frac{10}{y}\)

Example \(\PageIndex{2}\)

\(\frac{x-5}{2 x-1}-\frac{1}{2 x-1}\)

Subtract the numerators \(x−5\) and \(1\), and write the result over the common denominator, \(2x−1\).

\(\begin{aligned} \frac{x-5}{2 x-1}-\frac{1}{2 x-1} &=\frac{x-5-1}{2 x-1}\qquad\color{Cerulean}{Simplify\:the\:numerator.} \\ &=\frac{x-6}{2 x-1} \end{aligned}\)

\(\frac{x-6}{2 x-1}\)

Example \(\PageIndex{3}\)

\(\frac{2 x+7}{(x+5)(x-3)}-\frac{x+10}{(x+5)(x-3)}\)

We use parentheses to remind us to subtract the entire numerator of the second rational expression.

\(\begin{aligned} \frac{2 x+7}{(x+5)(x-3)}-\frac{x+10}{(x+5)(x-3)} &=\frac{(2 x+7)-(x+10)}{(x+5)(x-3)}\qquad\color{Cerulean}{Simplify\:the\:numerator.} \\ &=\frac{2 x+7-x-10}{(x+5)(x-3)}\qquad\quad\:\:\color{Cerulean}{Leave\:the\:denominator\:factored.} \\ &=\frac{\color{Cerulean}{\stackrel{1}{\cancel{\color{black}{x-3}}}}}{(x+5)\color{Cerulean}{\cancel{\color{black}{(x-3)}}}}\qquad\:\:\:\quad\color{Cerulean}{Cancel\:common\:factors.} \\ &=\frac{1}{x+5} \end{aligned}\)

\(\frac{1}{x+5}\)

Example \(\PageIndex{4}\)

\(\frac{2 x^{2}+10 x+3}{x^{2}-36}-\frac{x^{2}+6 x+5}{x^{2}-36}+\frac{x-4}{x^{2}-36}\)

Subtract and add the numerators. Make use of parentheses and write the result over the common denominator, \(x^{2}−36\).

\(\frac{x-1}{x-6}\)

Exercise \(\PageIndex{1}\)

\(\frac{1}{x-4}\)

Adding and Subtracting with Unlike Denominators

To add rational expressions with unlike denominators, first find equivalent expressions with common denominators. Do this just as you have with fractions. If the denominators of fractions are relatively prime, then the least common denominator (LCD) is their product. For example,

\(\frac{1}{3}+\frac{1}{5} \color{Cerulean}{\Rightarrow}\color{black}{ \mathrm{LCD}=3 \cdot 5=15}\)

Multiply each fraction by the appropriate form of 1 to obtain equivalent fractions with a common denominator.

\(\begin{aligned} \frac{1}{3}+\frac{1}{5} &=\frac{1\color{Cerulean}{ \cdot 5}}{3 \color{Cerulean}{\cdot 5}}+\frac{1 \color{Cerulean}{\cdot 3}}{5 \color{Cerulean}{\cdot 3}} \\ &=\frac{5}{15}+\frac{3}{15}\qquad\qquad\color{Cerulean}{Equivalent\:fractions\:with\:a\:common\:denominator} \\ &=\frac{5+3}{15} \\ &=\frac{8}{15} \end{aligned}\)

The process of adding and subtracting rational expressions is similar. In general, given polynomials P , Q , R , and S , where \(Q≠0\) and \(S≠0\), we have the following:

\[\frac{P}{Q} \pm \frac{R}{S}=\frac{P S \pm Q R}{Q S}\]

Example \(\PageIndex{5}\)

\(\frac{1}{x}+\frac{1}{y}\)

In this example, the \(LCD=xy\). To obtain equivalent terms with this common denominator, multiply the first term by \(\frac{y}{y}\) and the second term by \(\frac{x}{x}\).

\(\begin{aligned} \frac{1}{x}+\frac{1}{y} &=\frac{1}{x} \cdot\color{Cerulean}{ \frac{y}{y}}\color{black}{+}\frac{1}{y} \cdot \color{Cerulean}{\frac{x}{x}} \\ &=\frac{y}{x y}+\frac{x}{x y}\qquad\qquad\color{Cerulean}{Equivalent\:terms\:with\:a\:common\:denominator.} \\ &=\frac{y+x}{x y}\qquad\qquad\:\:\quad\color{Cerulean}{Add\:the\:numerators\:and\:place\:the\:result\:over\:the\:common\:denominator,\:xy.} \end{aligned}\)

\(\frac{y+x}{x y}\)

Example \(\PageIndex{6}\)

\(\frac{1}{y}-\frac{1}{y-3}\)

Since the \(LCD=y(y−3)\), multiply the first term by 1 in the form of \(\frac{(y−3)}{(y−3)}\) and the second term by \(\frac{y}{y}\).

\(\begin{aligned} \frac{1}{y}-\frac{1}{y-3} &=\frac{1}{y} \cdot\color{Cerulean}{ \frac{(y-3)}{(y-3)}}\color{black}{-}\frac{1}{y-3} \cdot\color{Cerulean}{ \frac{y}{y}} \\ &=\frac{(y-3)}{y(y-3)}-\frac{y}{y(y-3)} \\ &=\frac{y-3-y}{y(y-3)} \\ &=\frac{-3}{y(y-3)} \quad \text { or }=-\frac{3}{y(y-3)} \end{aligned}\)

\(\frac{-3}{y(y-3)}\)

It is not always the case that the LCD is the product of the given denominators. Typically, the denominators are not relatively prime; thus determining the LCD requires some thought. Begin by factoring all denominators. The LCD is the product of all factors with the highest power. For example, given

\(\frac{1}{\color{Cerulean}{x^{3}}\color{black}{(x+2)}\color{Cerulean}{(x-3)}} \quad \text { and } \quad \frac{1}{x\color{Cerulean}{(x+2)^{2}}}\)

there are three base factors in the denominator: \(x, (x+2)\), and \((x−3)\). The highest powers of these factors are \(x^{3}, (x+2)^{2}\), and \((x−3)^{1}\). Therefore,

\(\mathrm{LCD}=\color{Cerulean}{x^{3}(x+2)^{2}(x-3)}\)

The general steps for adding or subtracting rational expressions are illustrated in the following example.

Example \(\PageIndex{7}\)

Step 1 : Factor all denominators to determine the LCD.

The LCD is \((x+1)(x+3)(x−5)\).

Step 2: Multiply by the appropriate factors to obtain equivalent terms with a common denominator. To do this, multiply the first term by \(\frac{(x−5)}{(x−5)}\) and the second term by \(\frac{(x+3)}{(x+3)}\).

\(\begin{array}{l}{=\frac{x}{(x+1)(x+3)} \cdot \color{Cerulean}{\frac{(x-5)}{(x-5)}}\color{black}{-}\frac{3}{(x+1)(x-5)} \cdot\color{Cerulean}{ \frac{(x+3)}{(x+3)}}} \\ {=\frac{x(x-5)}{(x+1)(x+3)(x-5)}-\frac{3(x+3)}{(x+1)(x+3)(x-5)}}\end{array}\)

Step 3 : Add or subtract the numerators and place the result over the common denominator.

\(=\frac{x(x-5)-3(x+3)}{(x+1)(x+3)(x-5)}\)

Step 4 : Simplify the resulting algebraic fraction.

\(\frac{(x-9)}{(x+3)(x-5)}\)

Example \(\PageIndex{8}\)

It is best not to factor the numerator, \(x^{2}−9x+18\), because we will most likely need to simplify after we subtract.

\(\frac{18}{(x-4)(x-9)}\)

Example \(\PageIndex{9}\)

\(\frac{1}{x^{2}-4}-\frac{1}{2-x}\)

First, factor the denominators and determine the LCD. Notice how the opposite binomial property is applied to obtain a more workable denominator.

\(\begin{aligned} \frac{1}{x^{2}-4}-\frac{1}{2-x} &=\frac{1}{(x+2)(x-2)}-\frac{1}{-1(x-2)} \\ &=\frac{1}{(x+2)(x-2)}+\frac{1}{(x-2)} \end{aligned}\)

The LCD is \((x+2)(x−2)\). Multiply the second term by 1 in the form of \(\frac{(x+2)}{(x+2)}\).

\(\begin{array}{l}{=\frac{1}{(x+2)(x-2)}+\frac{1}{(x-2)} \cdot \color{Cerulean}{\frac{(x+2)}{(x+2)}}} \\ {=\frac{1}{(x+2)(x-2)}+\frac{x+2}{(x-2)(x+2)}}\end{array}\)

Now that we have equivalent terms with a common denominator, add the numerators and write the result over the common denominator.

\(\begin{array}{l}{=\frac{1}{(x+2)(x-2)}+\frac{x+2}{(x-2)(x+2)}} \\ {=\frac{1+x+2}{(x+2)(x-2)}} \\ {=\frac{x+3}{(x+2)(x-2)}}\end{array}\)

\(\frac{x+3}{(x+2)(x-2)}\)

Example \(\PageIndex{10}\)

\(\frac{y-1}{y+1}-\frac{y+1}{y-1}+\frac{y^{2}-5}{y^{2}-1}\)

Begin by factoring the denominator.

\(\begin{array}{c}{\frac{y-1}{y+1}-\frac{y+1}{y-1}+\frac{y^{2}-5}{y^{2}-1}} \\ {=\frac{y-1}{y+1}-\frac{y+1}{y-1}+\frac{y^{2}-5}{(y+1)(y-1)}}\end{array}\)

We can see that the LCD is \((y+1)(y−1)\). Find equivalent fractions with this denominator.

Next, subtract and add the numerators and place the result over the common denominator.

Finish by simplifying the resulting rational expression.

\(\frac{y-5}{y-1}\)

Exercise \(\PageIndex{2}\)

\(-\frac{2 x^{2}-1+x}{1+x}-\frac{5}{1-x}\)

\(\frac{x+3}{x-1}\)

Rational expressions are sometimes expressed using negative exponents. In this case, apply the rules for negative exponents before simplifying the expression.

Example \(\PageIndex{11}\)

\(y^{-2}+(y-1)^{-1}\)

Recall that \(x^{−n}=\frac{1}{x^{n}}\). We begin by rewriting the negative exponents as rational expressions.

\(\begin{aligned} & y^{-2}+(y-1)^{-1} \\=& \frac{1}{y^{2}}+\frac{1}{(y-1)^{1}}\qquad\qquad\qquad\qquad\color{Cerulean}{Replace\:negative\:exponents.} \\=& \frac{1}{y^{2}} \cdot \color{Cerulean}{\frac{(y-1)}{(y-1)}}\color{black}{+\frac{1}{(y-1)}} \cdot\color{Cerulean}{ \frac{y^{2}}{y^{2}}}\qquad\color{Cerulean}{Multiply\:by\:factors\:to\:obtain\:equivalent}\\&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\color{Cerulean}{expressions\:with\:a\:common\:denominator.} \\=&\frac{(y-1)}{y^{2}(y-1)}+\frac{y^{2}}{y^{2}(y-1)} \\=&\frac{(y-1)+y^{2}}{y^{2}(y-1)}\qquad\qquad\qquad\qquad\:\:\:\color{Cerulean}{Add\:and\:simplify.} \\ =& \frac{y^{2}+y-1}{y^{2}(y-1)}\qquad\qquad\qquad\qquad\quad\:\:\:\color{Cerulean}{The\:trinomial\:does\:not\:factor.} \end{aligned}\)

\(\frac{y^{2}+y-1}{y^{2}(y-1)}\)

Adding and Subtracting Rational Functions

We can simplify sums or differences of rational functions using the techniques learned in this section. The restrictions of the result consist of the restrictions to the domains of each function.

Example \(\PageIndex{12}\)

Calculate \((f+g)(x)\), given \(f(x)=\frac{1}{x+3}\) and \(g(x)=\frac{1}{x−2}\), and state the restrictions.

\(\begin{aligned}(f+g)(x) &=f(x)+g(x) \\ &=\frac{1}{x+3}+\frac{1}{x-2} \\ &=\frac{1}{x+3} \cdot \color{Cerulean}{\frac{(x-2)}{(x-2)}}\color{black}{+\frac{1}{x-2} \cdot}\color{Cerulean}{ \frac{(x+3)}{(x+3)}} \\ &=\frac{x-2}{(x+3)(x-2)}+\frac{x+3}{(x-2)(x+3)} \\ &=\frac{x-2+x+3}{(x+3)(x-2)} \\ &=\frac{2 x+1}{(x+3)(x-2)} \end{aligned}\)

Here the domain of f consists of all real numbers except \(−3\), and the domain of g consists of all real numbers except \(2\). Therefore, the domain of f + g consists of all real numbers except \(−3\) and \(2\).

\(\frac{2 x+1}{(x+3)(x-2)}\), where \(x\neq -3,2\)

Example \(\PageIndex{13}\)

Calculate \((f−g)(x)\), given \(f(x)=\frac{x(x−1)}{x^{2}−25}\) and \(g(x)=\frac{x−3}{x−5}\), and state the restrictions to the domain.

\(\begin{aligned}(f-g)(x) &=f(x)-g(x) \\ &=\frac{x(x-1)}{x^{2}-25}-\frac{x-3}{x-5} \\ &=\frac{x(x-1)}{(x+5)(x-5)}-\frac{(x-3)}{(x-5)} \cdot \color{Cerulean}{\frac{(x+5)}{(x+5)}} \\ &=\frac{x(x-1)-(x-3)(x+5)}{(x+5)(x-5)} \\&=\frac{x^{2}-x-(x^{2}+5x-3x-15)}{(x+5)(x-5)}\\&=\frac{x^{2}-x-(x^{2}+2x-15)}{(x+5)(x-5)}\\&=\frac{x^{2}-x-x^{2}-2x+15}{(x+5)(x-5)}\\&=\frac{-3x+15}{(x+5)(x-5)}\\&=\frac{-3\color{Cerulean}{\cancel{\color{black}{(x-5)}}}}{(x+5)\color{Cerulean}{\cancel{\color{black}{(x-5)}}}}\\&=\frac{-3}{x+5} \end{aligned}\)

The domain of f consists of all real numbers except \(5\) and \(−5\), and the domain of g consists of all real numbers except \(5\). Therefore, the domain of f − g consists of all real numbers except \(−5\) and \(5\).

\(\frac{-3}{x+5}\), where \(x\neq\pm 5\)

Key Takeaways

  • When adding or subtracting rational expressions with a common denominator, add or subtract the expressions in the numerator and write the result over the common denominator.
  • To find equivalent rational expressions with a common denominator, first factor all denominators and determine the least common multiple. Then multiply numerator and denominator of each term by the appropriate factor to obtain a common denominator. Finally, add or subtract the expressions in the numerator and write the result over the common denominator.
  • The restrictions to the domain of a sum or difference of rational functions consist of the restrictions to the domains of each function.

Exercise \(\PageIndex{3}\) Adding and Subtracting with Common Denominators

Simplify. (Assume all denominators are nonzero.)

  • \(\frac{3}{x}+\frac{7}{x}\)
  • \(\frac{9}{x}-\frac{10}{x}\)
  • \(\frac{x}{y}−\frac{3}{y}\)
  • \(\frac{4}{x−3}+\frac{6}{x−3}\)
  • \(\frac{7}{2x−1}−\frac{x}{2x−1}\)
  • \(\frac{8}{3x−8}−\frac{3x}{3x−8}\)
  • \(\frac{2}{x−9}+\frac{x−11}{x−9}\)
  • \(\frac{y+2}{2y+3}−\frac{y+3}{2y+3}\)
  • \(\frac{2x−3}{4x−1}−\frac{x−4}{4x−1}\)
  • \(\frac{2x}{x−1}−\frac{3x+4}{x−1}+\frac{x−2}{x−1}\)
  • \(\frac{1}{3y}−\frac{2y−9}{3y}−\frac{13−5y}{3y}\)
  • \(\frac{−3y+2}{5y−10}+\frac{y+7}{5y−10}−\frac{3y+4}{5y−10}\)
  • \(\frac{x}{(x+1)(x-3)}-\frac{3}{(x+1)(x-3)}\)
  • \(\frac{3x+5}{(2x−1)(x−6)}−\frac{x+6}{(2x−1)(x−6)}\)
  • \(\frac{x}{x^{2}-36}+\frac{6}{x^{2}-36}\)
  • \(\frac{x}{x^{2}−81}−\frac{9}{x^{2}−81}\)
  • \(\frac{x^{2}+2}{x^{2}+3 x-28}+\frac{x-2}{2 x^{2}+3 x-28}\)
  • \(\frac{x^{2}}{x^{2}-x-3}-\frac{3-x^{2}}{x^{2}-x-3}\)

1. \(\frac{10}{x}\)

3. \(\frac{x−3}{y}\)

5. \(\frac{7−x}{2x−1}\)

9. \(\frac{x+1}{4x−1}\)

11. \(\frac{y−1}{y}\)

13. \(\frac{1}{x+1}\)

15. \(\frac{1}{x−6}\)

17. \(\frac{x+5}{x+7}\)

Exercise \(\PageIndex{4}\) Adding and Subtracting with Unlike Denominators

  • \(\frac{1}{2}+\frac{1}{3x}\)
  • \(\frac{1}{5 x^{2}}-\frac{1}{x}\)
  • \(\frac{1}{12 y^{2}}+\frac{3}{10 y^{3}}\)
  • \(\frac{1}{x}−\frac{1}{2y}\)
  • \(\frac{1}{y}−2\)
  • \(\frac{3}{y+2}−4\)
  • \(\frac{2}{x+4}+2\)
  • \(\frac{2}{y}−\frac{1}{y^{2}}\)
  • \(\frac{3}{x+1}+\frac{1}{x}\)
  • \(\frac{1}{x−1}−\frac{2}{x}\)
  • \(\frac{1}{x−3}+\frac{1}{x+5}\)
  • \(\frac{1}{x+2}−\frac{1}{x−3}\)
  • \(\frac{x}{x+1}−\frac{2}{x−2}\)
  • \(\frac{2x−3}{x+5}−\frac{x}{x−3}\)
  • \(\frac{y+1}{y−1}+\frac{y−1}{y+1}\)
  • \(\frac{3y−1}{3y}−\frac{y+4}{y−2}\)
  • \(\frac{2x−5}{2x+5}−\frac{2x+5}{2x−5}\)
  • \(\frac{2}{2x−1}−\frac{2x+1}{1−2x}\)
  • \(\frac{3x+4}{x−8}−\frac{2}{8−x}\)
  • \(\frac{1}{y−1}+\frac{1}{1−y}\)
  • \(\frac{2x^{2}}{x^{2}−9}+\frac{x+15}{9−x^{2}}\)
  • \(\frac{x}{x+3}+\frac{1}{x−3}−\frac{1}{5}−\frac{x}{(x+3)(x−3) }\)
  • \(\frac{2 x}{3 x-1}-\frac{1}{3 x+1}+\frac{2(x-1)}{(3 x-1)(3 x+1)}\)
  • \(\frac{4 x}{2 x+1}-\frac{x}{x-5}+\frac{16 x-3}{(2 x+1)(x-5)}\)
  • \(\frac{x}{3 x}+\frac{2}{x-2}+\frac{4}{3 x(x-2)}\)
  • \(-\frac{2 x}{x+6}-\frac{3 x}{6-x}-\frac{18(x-2)}{(x+6)(x-6)}\)
  • \(\frac{x}{x+5}-\frac{1}{x-7}-\frac{25-7 x}{(x+5)(x-7)}\)
  • \(\frac{x}{x^{2}}-\frac{2}{x-3}+\frac{2}{x-3}\)
  • \(\frac{1}{x+5}-\frac{x^{2}}{x^{2}-25}\)
  • \(\frac{5x−2}{x^{2}−4}−\frac{2}{x−2}\)
  • \(\frac{1}{x+1}−\frac{6x−3}{x^{2}−7x−8}\)
  • \(\frac{3x}{9x^{2}−16}−\frac{1}{3x+4}\)
  • \(\frac{2x}{x^{2}−1}+\frac{1}{x^{2}+x}\)
  • \(\frac{x(4x−1)}{2x^{2}}+\frac{7}{x−4}−\frac{x}{4+x}\)
  • \(\frac{3x^{2}}{3x^{2}+5x−2}−\frac{2x}{3x−1}\)
  • \(\frac{2x}{x−4}−\frac{11x+4}{x^{2}−2x−8}\)
  • \(\frac{x}{2x+1}+\frac{6x−24}{2x^{2}−7x−4}\)
  • \(\frac{1}{x^{2}−x−6}+\frac{1}{x^{2}−3x−10}\)
  • \(\frac{x}{x^{2}+4x+3}−\frac{3}{x^{2}−4x−5}\)
  • \(\frac{y+1}{2y^{2}+5y−3}−\frac{y}{4y^{2}−1}\)
  • \(\frac{y−1}{y^{2}−25}−\frac{2}{y^{2}−10y+25 }\)
  • \(\frac{3x^{2}+2}{4x^{2}−2x−8}−\frac{1}{2x−4}\)
  • \(\frac{4x^{2}+2}{8x^{2}−6x−7}−\frac{2}{8x−7}\)
  • \(\frac{a}{4−a+a^{2}}−\frac{9a+18}{a^{2}−13a+36}\)
  • \(\frac{3a−12}{a^{2}−8a+16}−\frac{a+2}{4−a}\)
  • \(\frac{a^{2}−14}{2a^{2}−7a−4}−\frac{5}{1+2a}\)
  • \(\frac{1}{x+3}−\frac{x}{x^{2}−6x+9}+\frac{3}{x^{2}−9}\)
  • \(\frac{3x}{x+7}−\frac{2x}{x−2}+\frac{23x−10}{x^{2}+5x−14}\)
  • \(\frac{x+3}{x−1}+\frac{x−1}{x+2}−\frac{x(x+11)}{x^{2}+x−2}\)
  • \(−\frac{2x}{3x+1}−\frac{4}{x−2}+\frac{4(x+5)}{3x^{2}−5x−2}\)
  • \(\frac{x−1}{4x−1}−\frac{x+3}{2x+3}−\frac{3(x+5)}{8x^{2}+10x−3}\)
  • \(\frac{3x}{2x−3}−\frac{2}{2x+3}−\frac{6x^{2}−5x−9}{4x^{2}−9}\)
  • \(\frac{1}{y+1}+\frac{1}{y}+\frac{2}{y^{2}−1}\)
  • \(\frac{1}{y}−\frac{1}{y+1}+\frac{1}{y−1}\)
  • \(5^{−2}+2^{−1}\)
  • \(6^{−1}+4^{−2}\)
  • \(x^{−1}+y^{−1}\)
  • \(x^{−2}−y^{−1}\)
  • \((2x−1)^{−1}−x^{−2}\)
  • \((x−4)^{−1}−(x+1)^{−1}\)
  • \(3 x^{2}(x-1)^{-1}-2 x\)
  • \(2(y−1)^{−2}−(y−1)^{−1}\)

1. \(\frac{3x+2}{6x}\)

3. \(\frac{5y+18}{60y^{3}}\)

5. \(\frac{1−2y}{y}\)

7. \(\frac{2(x+5)}{x+4}\)

9. \(\frac{4x+1}{x(x+1)}\)

11. \(\frac{2(x+1)}{(x−3)(x+5)}\)

13. \(\frac{x^{2}−4x−2}{(x−2)(x+1)}\)

15. \(\frac{2(y2+1)}{(y+1)(y−1)}\)

17. \(−\frac{40x}{(2x+5)(2x−5)}\)

19. \(\frac{3(x+2)}{x−8}\)

21. \(\frac{2x+5}{x+3}\)

23. \(\frac{2x+1}{3x+1}\)

25. \(\frac{x^{2}+4x+4}{3x(x−2)}\)

27. \(\frac{x−6}{x−7}\)

29. \(\frac{x-5-x^{2}}{(x+5)(x-5)}\)

31. \(−\frac{5}{x−8}\)

33. \(\frac{2x−1}{x(x−1)}\)

35. \(\frac{x(x−4)}{(x+2)(3x−1)}\)

37. \(\frac{x+6}{2x+1}\)

39. \(\frac{x−9}{(x−5)(x+3)}\)

41. \(\frac{y^{2}−8y−5}{(y+5)(y−5)^{2}}\)

43. \(\frac{4x}{x+1}\)

45. \(\frac{a+5}{a−4}\)

47. \(−\frac{6x}{(x+3)(x−3)^{2}}\)

49. \(\frac{x−7}{x+2}\)

51. \(−\frac{x−5}{4x−1}\)

53. \(\frac{2y−1}{y(y−1) }\)

55. \(\frac{27}{50}\)

57. \(\frac{x+y}{xy}\)

59. \(\frac{(x−1)^{2}}{x^{2}(2x−1)}\)

61. \(\frac{x(x+2)}{x−1}\)

Exercise \(\PageIndex{5}\) Adding and Subtracting Rational Functions

Calculate \((f+g)(x)\) and \((f−g)(x)\) and state the restrictions to the domain.

  • \(f(x)=\frac{1}{3x}\) and \(g(x)=\frac{1}{x−2}\)
  • \(f(x)=\frac{1}{x−1}\) and \(g(x)=\frac{1}{x+5}\)
  • \(f(x)=\frac{x}{x−4}\) and \(g(x)=\frac{1}{4−x}\)
  • \(f(x)=\frac{x}{x−5}\) and \(g(x)=\frac{1}{2x−3}\)
  • \(f(x)=\frac{x−1}{x^{2}−4}\) and \(g(x)=\frac{4}{x^{2}-6 x-16}\)
  • \(f(x)=\frac{5}{x+2}\) and \(g(x)=\frac{3}{x+4}\)

1. \((f+g)(x)=\frac{2(2x−1)}{3x(x−2)}; (f−g)(x)=−\frac{2(x+1)}{3x(x−2)}; x≠0, 2 \)

3. \((f+g)(x)=\frac{x−1}{x−4}; (f−g)(x)=\frac{x+1}{x−4}; x≠4\)

5. \((f+g)(x)=\frac{x(x−5)}{(x+2)(x−2)(x−8)}; (f−g)(x)=\frac{x^{2}−13x+16}{(x+2)(x−2)(x−8)}; x≠−2, 2, 8\)

Exercise \(\PageIndex{6}\) Adding and Subtracting Rational Functions

Calculate \((f+f)(x)\) and state the restrictions to the domain.

  • \(f(x)=\frac{1}{x}\)
  • \(f(x)=\frac{1}{2x}\)
  • \(f(x)=\frac{x}{2x−1}\)
  • \(f(x)=\frac{1}{x+2}\)

1. \((f+f)(x)=\frac{2}{x}; x≠0\)

3. \((f+f)(x)=\frac{2x}{2x−1}; x≠\frac{1}{2}\)

Exercise \(\PageIndex{7}\) Discussion Board

  • Explain to a classmate why this is incorrect: \(\frac{1}{x^{2}}+\frac{2}{x^{2}}=\frac{3}{2x^{2}}\).
  • Explain to a classmate how to find the common denominator when adding algebraic expressions. Give an example.

1. Answer may vary

Mathematics 8 Quarter 1 – Module 5C: “Solving Problems Involving Rational Algebraic Expressions”

This module was designed and written with you in mind. It is here to help you master the skills of solving problems involving rational algebraic expressions. You are provided with varied activities to process the knowledge and skills learned and to deepen and transfer your understanding of the lesson. The scope of this module enables you to use it in many different learning situations. The lessons are arranged to follow the standard sequence of the course. But the order in which you read them can be changed to correspond with the textbook you are now using.

This module contains:

  • Lesson 1- Solving Problems Involving Rational Algebraic Expressions

After going through this module, you are expected to solve problems involving rational algebraic expressions.

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IMAGES

  1. Solving Word Problem Involving Rational Algebraic Expression

    steps in solving word problems involving rational algebraic expressions

  2. Solving Rational Equations (video lessons, examples, solutions)

    steps in solving word problems involving rational algebraic expressions

  3. PPT

    steps in solving word problems involving rational algebraic expressions

  4. Word Problems Involving Rational Algebraic Expressions

    steps in solving word problems involving rational algebraic expressions

  5. Rational Expressions Word Problems by tutorcircle team

    steps in solving word problems involving rational algebraic expressions

  6. problem solving rational algebraic expression examples

    steps in solving word problems involving rational algebraic expressions

VIDEO

  1. Algebra 2H 5 2a Simplify Rational Expressions

  2. Algebra 2 S2: Skill 6.1: Simplifying Rational Expressions

  3. Algebra I Chapter 8 Review Part 2

  4. Multiplication of Rational Algebraic Expressions Part 3

  5. Simplifying Rational Algebraic Expressions #maths #shorts #youtube

  6. Real Life Problem Involving Rational Algebraic Expressions- Work Problem

COMMENTS

  1. 7.6: Applications of Rational Equations

    In this section, the applications will often involve the key word "reciprocal.". When this is the case, we will see that the algebraic setup results in a rational equation. Example 7.6.1 7.6. 1. A positive integer is 4 4 less than another. The sum of the reciprocals of the two positive integers is 10 21 10 21.

  2. Rational Equation Word Problem Lesson

    Six-Step Method for Applications of Rational Expressions Read the problem carefully and determine what you are asked to find Assign a variable to represent the unknown Write out an equation that describes the given situation Solve the equation State the answer using a nice clear sentence Check the result by reading back through the problem

  3. Rational Equations Word Problems Lesson

    Step 1) Read the problem carefully and determine what you are asked to find We are asked to find how many additional gallons of gasoline are needed to travel 200 miles. Step 2) Assign a variable to represent the unknown Let x = the number of additional gallons of gasoline needed to travel 200 miles.

  4. Applications of Rational Expressions and Word Problems

    Text 12 Word Problems Involving Rational Equations Example 1: Seven divided by the sum of a number and two is equal to half the difference of the number and three. Find all such numbers. Let's begin by defining our variable. Let x=the number Now we can set up our equation.

  5. How to Solve a Word Problem Using a Rational Equation

    Step 1: Write the equation indicated by the word problem. Pay attention to words indicating mathematical operations or equalities. Step 2: Clear the equation of fractions. Step 3: Solve...

  6. PDF 10 Math 51 Application problems with rational expressions

    1 + = 1 4 x ⋅ 12 x ⇒ 4 x + 3 x = 12 ⇒ 7x = 12 ⇒ 12 x = ≈ 1.7 7 Answer: Together it takes about 1.7 hours to finish the job together. Variation problems Variation problems should be done in steps. The starting equations will be direct or inverse variations. • Direct variation y = kx means x and y are directly across from each other.

  7. 7.6: Solve Applications with Rational Equations

    The rational expression will be undefined when the denominator is zero. Since x + 3 = 0 when x = − 3, then -3 is a critical point. Step 3. Use the critical points to divide the number line into intervals. Step 4. Above the number line show the sign of each factor of the rational expression in each interval.

  8. Rational equations word problem: combined rates (example 2)

    Structure in rational expression. Math > Precalculus > Rational functions ... Sal solves a word problem about the combined deck-staining rates of Anya and Bill, by creating a rational equation that models the situation. ... (in decks per hour) would not be A + B, but 1/A + 1/B, which would equal 1/8. Try solving the problem using 1/A + 1/B ...

  9. 7.5: Solving Rational Equations

    Solving Rational Equations. A rational equation is an equation containing at least one rational expression. Rational expressions typically contain a variable in the denominator. For this reason, we will take care to ensure that the denominator is not 0 by making note of restrictions and checking our solutions.

  10. Rational equations word problem: eliminating solutions

    You can't choose which one, only given ( x - 4 ) ( x - 30 ) = 0. But if you choose x = 4 and find another hose speed, you will get: another hose speed = x - 10 = -6. Since negative value is not suitable for this problem, you can deduce to the following solution: x isn't 4, hence, x = 30. another hose speed = x - 10 = 20.

  11. Algebraic word problems

    To solve an algebraic word problem: Define a variable. Write an equation using the variable. Solve the equation. If the variable is not the answer to the word problem, use the variable to calculate the answer. It's important for us to keep in mind how we define our variables.

  12. Applications of Rational Expressions

    Here are a few examples of work problems that are solved with rational equations. Sam can paint a house in 5 hours. Gary can do it in 4 hours. How long will it take the two working together? Joy can file 100 claims in 5 hours. Stephen can file 100 claims in 8 hours. If they work together, how long will it take to file 100 claims?

  13. Solve Rational Equations

    Find the time (t): [latex] t=\frac {W} {r} [/latex] (divide both sides by r) Find the rate (r): [latex] r=\frac {W} {t} [/latex] (divide both sides by t) ) is the product of the rate of work () and the time spent working ( ). The work formula has 3 versions.

  14. Solving Rational Equations and Applications

    One way of solving rational equations with unlike denominators is to multiply both sides of the equation by the least common multiple of the denominators of all the fractions contained in the equation. This eliminates the denominators and turns the rational equation into a polynomial equation. Here's an example.

  15. Algebra

    Here is a set of practice problems to accompany the Rational Expressions section of the Preliminaries chapter of the notes for Paul Dawkins Algebra course at Lamar University.

  16. Solving Rational Equations

    Divide both sides by the coefficient of [latex]x [/latex]. That is it! Check the value [latex]x = - \,39 [/latex] back into the main rational equation and it should convince you that it works. Example 3: Solve the rational equation below and make sure you check your answers for extraneous values.

  17. Equations with rational expressions (video)

    Equations with rational expressions (example 2) ... The technique shown in the video eliminates some factors in the denominators making the problem more manageable and simple to solve. ... And so if we actually try to test x minus four in the original equation, not one of these intermediary steps, the original equation, I would end up dividing ...

  18. Word Problems Calculator

    Full pad Examples Frequently Asked Questions (FAQ) How do you solve word problems? To solve word problems start by reading the problem carefully and understanding what it's asking. Try underlining or highlighting key information, such as numbers and key words that indicate what operation is needed to perform.

  19. PDF Quarter 1 Module 5C

    8 Mathematics Quarter 1 - Module 5C: "Solving Problems Involving Rational Algebraic Expressions" Introductory Message This Self-Learning Module (SLM) is prepared so that you, our dear learners, can continue your studies and learn while at home.

  20. Rational expressions, equations, & functions

    This topic covers: - Simplifying rational expressions - Multiplying, dividing, adding, & subtracting rational expressions - Rational equations - Graphing rational functions (including horizontal & vertical asymptotes) - Modeling with rational functions - Rational inequalities - Partial fraction expansion

  21. Algebraic Expressions and Word Problems

    Examples, solutions, videos, worksheets, games and activities to help Algebra 1 or grade 7 students learn how to write algebraic expressions from word problems. Beginning Algebra & Word Problem Steps. Name what x is. Define everything in the problem in terms of x. Write the equation.

  22. 7.3: Adding and Subtracting Rational Expressions

    Solution: Step 1: Factor all denominators to determine the LCD. The LCD is (x + 1)(x + 3)(x − 5). Step 2: Multiply by the appropriate factors to obtain equivalent terms with a common denominator. To do this, multiply the first term by ( x − 5) ( x − 5) and the second term by ( x + 3) ( x + 3).

  23. Mathematics 8 Quarter 1

    Lesson 1- Solving Problems Involving Rational Algebraic Expressions; After going through this module, you are expected to solve problems involving rational algebraic expressions. math8_q1_mod5c_solving-problems-involving-rational-algebraic-expressions_v2