MatterofMath

Solving Rational Equations · Examples

Listen up, fraction fans! In today’s lesson, you will learn and practice solving rational equations. As you will see, these are any equation involving a fraction, also known as a rational number in math talk!

By the end, you will know the difference between rational and irrational numbers and have two tricks for solving rational equations.

You could even tackle one of the tricky challenges to form a rational equation using the Pythagorean theorem , or to simplify an expression involving some radicals!

What is a Rational Equation? How to Solve Rational Equations Step 1: Eliminate the Denominators Step 2: Simplify the Equation Step 3: Solve the Equation Step 4: Check Solutions Practice & Challenges Question 1 Question 2 Challenge 1 Challenge 2 Worksheet To Sum Up (Pun Intended!)

What is a Rational Equation?

A rational equation is simply an equation involving a rational number.

A ratio -nal number can be written as a ratio of two integers – an irratio -nal number cannot.

Most of the numbers you know and love such as \(\Large\frac{2}{7}\), \(\Large\frac{1}{2}\) and \(-\Large\frac{20817}{43}\) are rational. Some common irrational numbers are π, \(\sqrt{2}\) and Euler’s number, e. These cannot be written as a fraction of integers.

Rational and Irrational Numbers

Numberphile has an interesting video about All the Numbers , which categorizes number types, including rational and irrational numbers .

Technically speaking, basic equations like x+2=5 are rational because each term is a rational number. However, the rational equations you will solve today won’t be so easy!

An example of what you will more likely see in an exam is something like this:

Each term is shown as a fraction.

Rational equations can also include radicals:

Or other operations such as division:

Luckily, the technique you learn now will work for every type of rational equation!

How to Solve Rational Equations

The method to solve these equations is pretty much the same for every type of rational equation. You’ll see questions of varying difficulty in this lesson; don’t be afraid to tackle the challenges later on!

Step 1: The Denominator Elimination Round!

First, you need to deal with the elephant in the room: what should you do with the denominators!?

Solving rational equations is just like solving any other equation once you complete this step.

If it’s a simple case, where you have one fraction being equal to one other fraction, you can cross multiply .

Multiply both sides by the values of both denominators. In this example, both sides are multiplied by 3, then 5.

The 3 cancels with the left denominator and the 5 cancels with the right denominator, leaving you with 5(x+4)=3×2.

Cross Multiplying

See why it’s called cross multiplying?

The product of the left denominator and right numerator equals the product of the right denominator and left numerator !

The more general way to deal with the denominators is to find their lowest common multiple (LCM) . This is the smallest number which all denominators divide neatly, leaving no remainder.

If you cannot find the LCM by inspection – if you cannot “just see it” – you need to factor every denominator like you would with a polynomial.

If you have more than one constant term, you may need to find their prime factors.

The LCM is the smallest combination of each denominator’s factors.

You’ll now see a worked example to illustrate!

Remember, you can only cross multiply when each side has only one fraction, so in this case, your first step is to find the LCM.

The only factors of 3x you know for certain are 3 and x. The only factor you know of x is just x, and 4 is a constant so you can use it as it is.

Write down each denominator’s polynomial factors into rows, with similar terms lined up in the same column.

Solving Rational Equations First Worked Example

You need to include both 3 and 4 because neither is a factor of the other. You don’t need both copies of x because x is a factor of itself! So the LCM is 12x.

You might find another example of finding the LCM with the same technique helpful.

You’re now ready to eliminate the denominators by multiplying both sides by the LCM.

Step 2: Simplify the Equation

Multiply each term by the LCM. Continuing from the last example, you have:

You now have a regular equation with no fractions, which should be familiar ground!

Step 3: Solve the Equation

Solving rational equations usually produces a simple polynomial equation. Hopefully, you’ve solved lots of these before!

You could complete the square, factor the terms by inspection, or use the quadratic formula.

This example can be solved by factoring the polynomial, having found that x+2 and x+4 are factors.

You could also solve the equation by completing the square:

Or by using the quadratic formula with a=1, b=6 and c=8:

Each way of solving the simplified rational equation is valid, but you will find that some are quicker than others!

Step 4: Check Every Solution

It is important to check that your solutions are complete, meaning you’ve found all of them and that they don’t give any weird numbers when substituted into the original equation.

In the worked example, you were left with a quadratic equation and found two distinct roots.

Quadratic equations either have two distinct solutions, one repeated solution, or no real solution so the solution x=-2 or x=-4 is complete.

You must be careful that none of the rational terms in the original equation have a zero in the denominator.

Do this by going back to the beginning and substituting your answers into the denominators!

The denominators in the worked example are 3x, x, and 4. Replacing x with -2 or -4 doesn’t give you zero in any of them, so you’re safe here!

A solution that gives a zero-denominator is not allowed. That’s because dividing by zero is “illegal” in math!

Any number divided by zero gives an error on a calculator. Ever wondered why that is?

This is your time to shine – try solving rational equations for yourself and, if you’re feeling confident, tackle the challenges too.

As they say, practice makes perfect! Use the worked example for guidance if you get stuck.

Find x in the following rational equation:

The equation is two equal fractions so you can cross-multiply. You could also simplify \(\Large\frac{15}{3}\normalsize\) to 5, but this does not change the final answer.

Solution 1 Cross Multiplying

Each term is divisible by 9. Simplify the equation by dividing both sides by 9:

This form is called the difference of two squares because it can be factored like this:

So the solution is x=±3.

These must be all the solutions because quadratic equations have a maximum of two distinct real roots.

Neither denominator in the original rational equation has an x term, so substituting any value for x makes no difference to their values – there is no chance of them being zero!

This means the solutions x=3 and x=-3 are valid.

Solve the following rational equation:

There are three fractions so you cannot cross-multiply.

See that the second denominator is the difference of two squares?

LCM of Polynomial Equation Denominators1

Multiply each term by the LCM and simplify.

Polynomial Equation Denominators Simplified1

So its solution is -5, right?… STOP RIGHT THERE! Don’t forget, we can’t divide by zero!

If you put x=±5 into the original equation, at least one of the denominators is always zero, so the original equation has no solutions.

Challenge 1

Can you spot the mistake in the following example? Hint: there has been some cheating with radicals!

If you need a refresher on radicals , check out our lesson on multiplying them. That will get you on the right track!

The mistake is that radicals cannot be subtracted like normal terms.

Instead, you must square both sides of the equation to remove the radical. Similar terms can then be combined as usual.

Still confused? You can find lots of interactive questions on Lumen Learning . Radicals often pop up in rational equations, so getting comfortable with radicals is super helpful for exam success!

Challenge 2

Find the value of x, by using the Pythagorean theorem on the following right-angled triangle:

Pythagorean Theorem and Solving Rational Equations

If you need a refresher on the Pythagorean theorem or are interested in the man himself, check out our lesson. Do the worksheets and you’ll be acing triangle questions in no time !

The Pythagorean theorem states that:

Where c is the length of the hypotenuse, and a and b are the other side lengths.

This gives the rational equation:

Simplifying, you find:

The LCM is 36 so the denominators are removed by dividing each term by this:

It’s always fun when different areas of math link together!

To Sum Up (Pun Intended!)

In today’s lesson on solving rational equations, you first saw the difference between rational and irrational numbers.

Rational numbers are “nice” because they can be written as a fraction of integers. Remember that all integers are rational because they can be written with a denominator of 1!

Irrational numbers are a little more abstract. They include weird but incredibly beautiful numbers like π and e, which cannot be written as a fraction of integers.

Rational equations are solved by eliminating the denominator in every term, then simplifying and solving as normal.

Denominators can be removed by cross-multiplication if there is only one fraction on either side or by finding the LCM if the equation is more complicated.

Don’t be shy, leave a comment below if you have any questions or need help!

Still curious about rational numbers, or eager for an extra challenge? Check out our lesson on the rational root theorem , which combines algebra and equation solving.

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  • 7.4 Solve Rational Equations
  • Introduction
  • 1.1 Use the Language of Algebra
  • 1.2 Integers
  • 1.3 Fractions
  • 1.4 Decimals
  • 1.5 Properties of Real Numbers
  • Key Concepts
  • Review Exercises
  • Practice Test
  • 2.1 Use a General Strategy to Solve Linear Equations
  • 2.2 Use a Problem Solving Strategy
  • 2.3 Solve a Formula for a Specific Variable
  • 2.4 Solve Mixture and Uniform Motion Applications
  • 2.5 Solve Linear Inequalities
  • 2.6 Solve Compound Inequalities
  • 2.7 Solve Absolute Value Inequalities
  • 3.1 Graph Linear Equations in Two Variables
  • 3.2 Slope of a Line
  • 3.3 Find the Equation of a Line
  • 3.4 Graph Linear Inequalities in Two Variables
  • 3.5 Relations and Functions
  • 3.6 Graphs of Functions
  • 4.1 Solve Systems of Linear Equations with Two Variables
  • 4.2 Solve Applications with Systems of Equations
  • 4.3 Solve Mixture Applications with Systems of Equations
  • 4.4 Solve Systems of Equations with Three Variables
  • 4.5 Solve Systems of Equations Using Matrices
  • 4.6 Solve Systems of Equations Using Determinants
  • 4.7 Graphing Systems of Linear Inequalities
  • 5.1 Add and Subtract Polynomials
  • 5.2 Properties of Exponents and Scientific Notation
  • 5.3 Multiply Polynomials
  • 5.4 Dividing Polynomials
  • Introduction to Factoring
  • 6.1 Greatest Common Factor and Factor by Grouping
  • 6.2 Factor Trinomials
  • 6.3 Factor Special Products
  • 6.4 General Strategy for Factoring Polynomials
  • 6.5 Polynomial Equations
  • 7.1 Multiply and Divide Rational Expressions
  • 7.2 Add and Subtract Rational Expressions
  • 7.3 Simplify Complex Rational Expressions
  • 7.5 Solve Applications with Rational Equations
  • 7.6 Solve Rational Inequalities
  • 8.1 Simplify Expressions with Roots
  • 8.2 Simplify Radical Expressions
  • 8.3 Simplify Rational Exponents
  • 8.4 Add, Subtract, and Multiply Radical Expressions
  • 8.5 Divide Radical Expressions
  • 8.6 Solve Radical Equations
  • 8.7 Use Radicals in Functions
  • 8.8 Use the Complex Number System
  • 9.1 Solve Quadratic Equations Using the Square Root Property
  • 9.2 Solve Quadratic Equations by Completing the Square
  • 9.3 Solve Quadratic Equations Using the Quadratic Formula
  • 9.4 Solve Equations in Quadratic Form
  • 9.5 Solve Applications of Quadratic Equations
  • 9.6 Graph Quadratic Functions Using Properties
  • 9.7 Graph Quadratic Functions Using Transformations
  • 9.8 Solve Quadratic Inequalities
  • 10.1 Finding Composite and Inverse Functions
  • 10.2 Evaluate and Graph Exponential Functions
  • 10.3 Evaluate and Graph Logarithmic Functions
  • 10.4 Use the Properties of Logarithms
  • 10.5 Solve Exponential and Logarithmic Equations
  • 11.1 Distance and Midpoint Formulas; Circles
  • 11.2 Parabolas
  • 11.3 Ellipses
  • 11.4 Hyperbolas
  • 11.5 Solve Systems of Nonlinear Equations
  • 12.1 Sequences
  • 12.2 Arithmetic Sequences
  • 12.3 Geometric Sequences and Series
  • 12.4 Binomial Theorem

Learning Objectives

By the end of this section, you will be able to:

  • Solve rational equations
  • Use rational functions
  • Solve a rational equation for a specific variable

Be Prepared 7.10

Before you get started, take this readiness quiz.

Solve: 1 6 x + 1 2 = 1 3 . 1 6 x + 1 2 = 1 3 . If you missed this problem, review Example 2.9 .

Be Prepared 7.11

Solve: n 2 − 5 n − 36 = 0 . n 2 − 5 n − 36 = 0 . If you missed this problem, review Example 6.45 .

Be Prepared 7.12

Solve the formula 5 x + 2 y = 10 5 x + 2 y = 10 for y . y . If you missed this problem, review Example 2.31 .

After defining the terms ‘expression’ and ‘equation’ earlier, we have used them throughout this book. We have simplified many kinds of expressions and solved many kinds of equations . We have simplified many rational expressions so far in this chapter. Now we will solve a rational equation .

Rational Equation

A rational equation is an equation that contains a rational expression.

You must make sure to know the difference between rational expressions and rational equations. The equation contains an equal sign.

Solve Rational Equations

We have already solved linear equations that contained fractions. We found the LCD of all the fractions in the equation and then multiplied both sides of the equation by the LCD to “clear” the fractions.

We will use the same strategy to solve rational equations. We will multiply both sides of the equation by the LCD. Then, we will have an equation that does not contain rational expressions and thus is much easier for us to solve. But because the original equation may have a variable in a denominator, we must be careful that we don’t end up with a solution that would make a denominator equal to zero.

So before we begin solving a rational equation, we examine it first to find the values that would make any denominators zero. That way, when we solve a rational equation we will know if there are any algebraic solutions we must discard.

An algebraic solution to a rational equation that would cause any of the rational expressions to be undefined is called an extraneous solution to a rational equation .

Extraneous Solution to a Rational Equation

An extraneous solution to a rational equation is an algebraic solution that would cause any of the expressions in the original equation to be undefined.

We note any possible extraneous solutions, c , by writing x ≠ c x ≠ c next to the equation.

Example 7.33

How to solve a rational equation.

Solve: 1 x + 1 3 = 5 6 . 1 x + 1 3 = 5 6 .

Try It 7.65

Solve: 1 y + 2 3 = 1 5 . 1 y + 2 3 = 1 5 .

Try It 7.66

Solve: 2 3 + 1 5 = 1 x . 2 3 + 1 5 = 1 x .

The steps of this method are shown.

Solve equations with rational expressions.

  • Step 1. Note any value of the variable that would make any denominator zero.
  • Step 2. Find the least common denominator of all denominators in the equation.
  • Step 3. Clear the fractions by multiplying both sides of the equation by the LCD.
  • Step 4. Solve the resulting equation.
  • If any values found in Step 1 are algebraic solutions, discard them.
  • Check any remaining solutions in the original equation.

We always start by noting the values that would cause any denominators to be zero.

Example 7.34

How to solve a rational equation using the zero product property.

Solve: 1 − 5 y = − 6 y 2 . 1 − 5 y = − 6 y 2 .

Try It 7.67

Solve: 1 − 2 x = 15 x 2 . 1 − 2 x = 15 x 2 .

Try It 7.68

Solve: 1 − 4 y = 12 y 2 . 1 − 4 y = 12 y 2 .

In the next example, the last denominators is a difference of squares. Remember to factor it first to find the LCD.

Example 7.35

Solve: 2 x + 2 + 4 x − 2 = x − 1 x 2 − 4 . 2 x + 2 + 4 x − 2 = x − 1 x 2 − 4 .

Try It 7.69

Solve: 2 x + 1 + 1 x − 1 = 1 x 2 − 1 . 2 x + 1 + 1 x − 1 = 1 x 2 − 1 .

Try It 7.70

Solve: 5 y + 3 + 2 y − 3 = 5 y 2 − 9 . 5 y + 3 + 2 y − 3 = 5 y 2 − 9 .

In the next example, the first denominator is a trinomial . Remember to factor it first to find the LCD.

Example 7.36

Solve: m + 11 m 2 − 5 m + 4 = 5 m − 4 − 3 m − 1 . m + 11 m 2 − 5 m + 4 = 5 m − 4 − 3 m − 1 .

Try It 7.71

Solve: x + 13 x 2 − 7 x + 10 = 6 x − 5 − 4 x − 2 . x + 13 x 2 − 7 x + 10 = 6 x − 5 − 4 x − 2 .

Try It 7.72

Solve: y − 6 y 2 + 3 y − 4 = 2 y + 4 + 7 y − 1 . y − 6 y 2 + 3 y − 4 = 2 y + 4 + 7 y − 1 .

The equation we solved in the previous example had only one algebraic solution, but it was an extraneous solution. That left us with no solution to the equation. In the next example we get two algebraic solutions. Here one or both could be extraneous solutions.

Example 7.37

Solve: y y + 6 = 72 y 2 − 36 + 4 . y y + 6 = 72 y 2 − 36 + 4 .

Try It 7.73

Solve: x x + 4 = 32 x 2 − 16 + 5 . x x + 4 = 32 x 2 − 16 + 5 .

Try It 7.74

Solve: y y + 8 = 128 y 2 − 64 + 9 . y y + 8 = 128 y 2 − 64 + 9 .

In some cases, all the algebraic solutions are extraneous.

Example 7.38

Solve: x 2 x − 2 − 2 3 x + 3 = 5 x 2 − 2 x + 9 12 x 2 − 12 . x 2 x − 2 − 2 3 x + 3 = 5 x 2 − 2 x + 9 12 x 2 − 12 .

Try It 7.75

Solve: y 5 y − 10 − 5 3 y + 6 = 2 y 2 − 19 y + 54 15 y 2 − 60 . y 5 y − 10 − 5 3 y + 6 = 2 y 2 − 19 y + 54 15 y 2 − 60 .

Try It 7.76

Solve: z 2 z + 8 − 3 4 z − 8 = 3 z 2 − 16 z − 16 8 z 2 + 16 z − 64 . z 2 z + 8 − 3 4 z − 8 = 3 z 2 − 16 z − 16 8 z 2 + 16 z − 64 .

Example 7.39

Solve: 4 3 x 2 − 10 x + 3 + 3 3 x 2 + 2 x − 1 = 2 x 2 − 2 x − 3 . 4 3 x 2 − 10 x + 3 + 3 3 x 2 + 2 x − 1 = 2 x 2 − 2 x − 3 .

Try It 7.77

Solve: 15 x 2 + x − 6 − 3 x − 2 = 2 x + 3 . 15 x 2 + x − 6 − 3 x − 2 = 2 x + 3 .

Try It 7.78

Solve: 5 x 2 + 2 x − 3 − 3 x 2 + x − 2 = 1 x 2 + 5 x + 6 . 5 x 2 + 2 x − 3 − 3 x 2 + x − 2 = 1 x 2 + 5 x + 6 .

Use Rational Functions

Working with functions that are defined by rational expressions often lead to rational equations. Again, we use the same techniques to solve them.

Example 7.40

For rational function, f ( x ) = 2 x − 6 x 2 − 8 x + 15 , f ( x ) = 2 x − 6 x 2 − 8 x + 15 , ⓐ find the domain of the function, ⓑ solve f ( x ) = 1 , f ( x ) = 1 , and ⓒ find the points on the graph at this function value.

ⓐ The domain of a rational function is all real numbers except those that make the rational expression undefined. So to find them, we will set the denominator equal to zero and solve.

However, x = 3 x = 3 is outside the domain of this function, so we discard that root as extraneous.

ⓒ The value of the function is 1 when x = 7 . x = 7 . So the points on the graph of this function when f ( x ) = 1 f ( x ) = 1 is ( 7 , 1 ) ) ( 7 , 1 ) )

Try It 7.79

For rational function, f ( x ) = 8 − x x 2 − 7 x + 12 , f ( x ) = 8 − x x 2 − 7 x + 12 , ⓐ find the domain of the function ⓑ solve f ( x ) = 3 f ( x ) = 3 ⓒ find the points on the graph at this function value.

Try It 7.80

For rational function, f ( x ) = x − 1 x 2 − 6 x + 5 , f ( x ) = x − 1 x 2 − 6 x + 5 , ⓐ find the domain of the function ⓑ solve f ( x ) = 4 f ( x ) = 4 ⓒ find the points on the graph at this function value.

Solve a Rational Equation for a Specific Variable

When we solved linear equations, we learned how to solve a formula for a specific variable. Many formulas used in business, science, economics, and other fields use rational equations to model the relation between two or more variables. We will now see how to solve a rational equation for a specific variable.

When we developed the point-slope formula from our slope formula, we cleared the fractions by multiplying by the LCD.

m = y − y 1 x − x 1 Multiply both sides of the equation by x − x 1 . m ( x − x 1 ) = ( y − y 1 x − x 1 ) ( x − x 1 ) Simplify. m ( x − x 1 ) = y − y 1 Rewrite the equation with the y terms on the left. y − y 1 = m ( x − x 1 ) m = y − y 1 x − x 1 Multiply both sides of the equation by x − x 1 . m ( x − x 1 ) = ( y − y 1 x − x 1 ) ( x − x 1 ) Simplify. m ( x − x 1 ) = y − y 1 Rewrite the equation with the y terms on the left. y − y 1 = m ( x − x 1 )

In the next example, we will use the same technique with the formula for slope that we used to get the point-slope form of an equation of a line through the point ( 2 , 3 ) . ( 2 , 3 ) . We will add one more step to solve for y .

Example 7.41

Solve: m = y − 2 x − 3 m = y − 2 x − 3 for y . y .

Try It 7.81

Solve: m = y − 5 x − 4 m = y − 5 x − 4 for y . y .

Try It 7.82

Solve: m = y − 1 x + 5 m = y − 1 x + 5 for y . y .

Remember to multiply both sides by the LCD in the next example.

Example 7.42

Solve: 1 c + 1 m = 1 1 c + 1 m = 1 for c .

Try It 7.83

Solve: 1 a + 1 b = c 1 a + 1 b = c for a .

Try It 7.84

Solve: 2 x + 1 3 = 1 y 2 x + 1 3 = 1 y for y .

Access this online resource for additional instruction and practice with equations with rational expressions.

  • Equations with Rational Expressions

Section 7.4 Exercises

Practice makes perfect.

In the following exercises, solve each rational equation.

1 a + 2 5 = 1 2 1 a + 2 5 = 1 2

6 3 − 2 d = 4 9 6 3 − 2 d = 4 9

4 5 + 1 4 = 2 v 4 5 + 1 4 = 2 v

3 8 + 2 y = 1 4 3 8 + 2 y = 1 4

1 − 2 m = 8 m 2 1 − 2 m = 8 m 2

1 + 4 n = 21 n 2 1 + 4 n = 21 n 2

1 + 9 p = −20 p 2 1 + 9 p = −20 p 2

1 − 7 q = −6 q 2 1 − 7 q = −6 q 2

5 3 v − 2 = 7 4 v 5 3 v − 2 = 7 4 v

8 2 w + 1 = 3 w 8 2 w + 1 = 3 w

3 x + 4 + 7 x − 4 = 8 x 2 − 16 3 x + 4 + 7 x − 4 = 8 x 2 − 16

5 y − 9 + 1 y + 9 = 18 y 2 − 81 5 y − 9 + 1 y + 9 = 18 y 2 − 81

8 z − 10 − 7 z + 10 = 5 z 2 − 100 8 z − 10 − 7 z + 10 = 5 z 2 − 100

9 a + 11 − 6 a − 11 = 6 a 2 − 121 9 a + 11 − 6 a − 11 = 6 a 2 − 121

−10 q − 2 − 7 q + 4 = 1 −10 q − 2 − 7 q + 4 = 1

2 s + 7 − 3 s − 3 = 1 2 s + 7 − 3 s − 3 = 1

v − 10 v 2 − 5 v + 4 = 3 v − 1 − 6 v − 4 v − 10 v 2 − 5 v + 4 = 3 v − 1 − 6 v − 4

w + 8 w 2 − 11 w + 28 = 5 w − 7 + 2 w − 4 w + 8 w 2 − 11 w + 28 = 5 w − 7 + 2 w − 4

x − 10 x 2 + 8 x + 12 = 3 x + 2 + 4 x + 6 x − 10 x 2 + 8 x + 12 = 3 x + 2 + 4 x + 6

y − 5 y 2 − 4 y − 5 = 1 y + 1 + 1 y − 5 y − 5 y 2 − 4 y − 5 = 1 y + 1 + 1 y − 5

b + 3 3 b + b 24 = 1 b b + 3 3 b + b 24 = 1 b

c + 3 12 c + c 36 = 1 4 c c + 3 12 c + c 36 = 1 4 c

d d + 3 = 18 d 2 − 9 + 4 d d + 3 = 18 d 2 − 9 + 4

m m + 5 = 50 m 2 − 25 + 6 m m + 5 = 50 m 2 − 25 + 6

n n + 2 − 3 = 8 n 2 − 4 n n + 2 − 3 = 8 n 2 − 4

p p + 7 − 8 = 98 p 2 − 49 p p + 7 − 8 = 98 p 2 − 49

q 3 q − 9 − 3 4 q + 12 = 7 q 2 + 6 q + 63 24 q 2 − 216 q 3 q − 9 − 3 4 q + 12 = 7 q 2 + 6 q + 63 24 q 2 − 216

r 3 r − 15 − 1 4 r + 20 = 3 r 2 + 17 r + 40 12 r 2 − 300 r 3 r − 15 − 1 4 r + 20 = 3 r 2 + 17 r + 40 12 r 2 − 300

s 2 s + 6 − 2 5 s + 5 = 5 s 2 − 3 s − 7 10 s 2 + 40 s + 30 s 2 s + 6 − 2 5 s + 5 = 5 s 2 − 3 s − 7 10 s 2 + 40 s + 30

t 6 t − 12 − 5 2 t + 10 = t 2 − 23 t + 70 12 t 2 + 36 t − 120 t 6 t − 12 − 5 2 t + 10 = t 2 − 23 t + 70 12 t 2 + 36 t − 120

2 x 2 + 2 x − 8 − 1 x 2 + 9 x + 20 = 4 x 2 + 3 x − 10 2 x 2 + 2 x − 8 − 1 x 2 + 9 x + 20 = 4 x 2 + 3 x − 10

5 x 2 + 4 x + 3 + 2 x 2 + x − 6 = 3 x 2 − x − 2 5 x 2 + 4 x + 3 + 2 x 2 + x − 6 = 3 x 2 − x − 2

3 x 2 − 5 x − 6 + 3 x 2 − 7 x + 6 = 6 x 2 − 1 3 x 2 − 5 x − 6 + 3 x 2 − 7 x + 6 = 6 x 2 − 1

2 x 2 + 2 x − 3 + 3 x 2 + 4 x + 3 = 6 x 2 − 1 2 x 2 + 2 x − 3 + 3 x 2 + 4 x + 3 = 6 x 2 − 1

Solve Rational Equations that Involve Functions

For rational function, f ( x ) = x − 2 x 2 + 6 x + 8 , f ( x ) = x − 2 x 2 + 6 x + 8 , ⓐ find the domain of the function ⓑ solve f ( x ) = 5 f ( x ) = 5 ⓒ find the points on the graph at this function value.

For rational function, f ( x ) = x + 1 x 2 − 2 x − 3 , f ( x ) = x + 1 x 2 − 2 x − 3 , ⓐ find the domain of the function ⓑ solve f ( x ) = 1 f ( x ) = 1 ⓒ find the points on the graph at this function value.

For rational function, f ( x ) = 2 − x x 2 − 7 x + 10 , f ( x ) = 2 − x x 2 − 7 x + 10 , ⓐ find the domain of the function ⓑ solve f ( x ) = 2 f ( x ) = 2 ⓒ find the points on the graph at this function value.

For rational function, f ( x ) = 5 − x x 2 + 5 x + 6 , f ( x ) = 5 − x x 2 + 5 x + 6 , ⓐ find the domain of the function ⓑ solve f ( x ) = 3 f ( x ) = 3 ⓒ the points on the graph at this function value.

In the following exercises, solve.

C r = 2 π C r = 2 π for r . r .

I r = P I r = P for r . r .

v + 3 w − 1 = 1 2 v + 3 w − 1 = 1 2 for w . w .

x + 5 2 − y = 4 3 x + 5 2 − y = 4 3 for y . y .

a = b + 3 c − 2 a = b + 3 c − 2 for c . c .

m = n 2 − n m = n 2 − n for n . n .

1 p + 2 q = 4 1 p + 2 q = 4 for p . p .

3 s + 1 t = 2 3 s + 1 t = 2 for s . s .

2 v + 1 5 = 3 w 2 v + 1 5 = 3 w for w . w .

6 x + 2 3 = 1 y 6 x + 2 3 = 1 y for y . y .

m + 3 n − 2 = 4 5 m + 3 n − 2 = 4 5 for n . n .

r = s 3 − t r = s 3 − t for t . t .

E c = m 2 E c = m 2 for c . c .

R T = W R T = W for T . T .

3 x − 5 y = 1 4 3 x − 5 y = 1 4 for y . y .

c = 2 a + b 5 c = 2 a + b 5 for a . a .

Writing Exercises

Your class mate is having trouble in this section. Write down the steps you would use to explain how to solve a rational equation.

Alek thinks the equation y y + 6 = 72 y 2 − 36 + 4 y y + 6 = 72 y 2 − 36 + 4 has two solutions, y = −6 y = −6 and y = 4 . y = 4 . Explain why Alek is wrong.

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ On a scale of 1 − 10 , 1 − 10 , how would you rate your mastery of this section in light of your responses on the checklist? How can you improve this?

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  • Authors: Lynn Marecek, Andrea Honeycutt Mathis
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  • Book title: Intermediate Algebra 2e
  • Publication date: May 6, 2020
  • Location: Houston, Texas
  • Book URL: https://openstax.org/books/intermediate-algebra-2e/pages/1-introduction
  • Section URL: https://openstax.org/books/intermediate-algebra-2e/pages/7-4-solve-rational-equations

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Rational Function Problems

These lessons are compiled to help PreCalculus students learn about rational function problems and applications.

Related Pages Simplifying Rational Expressions Graphing Rational Functions PreCalculus Lessons

The following figure shows how to solve rational equations . Scroll down the page for examples and solutions on how to solve rational function problems and applications.

Solve Rational Equations

Rational Function Problems - Work And Tank

The video explains application problems that use rational equations. Part 1 of 2.

  • Martin can pour a concrete walkway in 6 hours working alone. Victor has more experience and can pour the same walkway in 4 hours working alone. How long will it take both people to pour the concrete walkway working together?
  • An inlet pipe can fill a water tank in 12 hours. An outlet pipe can drain the tank in 20 hours. If both pipes are mistakenly left open, how long will it take to fill the tank?

Rational Function Applications - Work And Rate

The video explains application problems that use rational equations. Part 2 of 2.

  • One person can complete a task 8 hours sooner than another person. Working together, both people can perform the task in 3 hours. How many hours does it take each person to complete the task working alone?
  • The speed of a passenger train is 12 mph faster than the speed of the freight train. The passenger train travels 330 miles in the same time it takes the freight train to travel 270 miles. Find the speed of each train.

Rational Functions Word Problems - Work, Tank And Pipe

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?
  • A water tank is emptied through two drains in 50 minutes. If only the larger drain is used, the tank will empty in 85 minutes. How long would it take to empty if only the smaller drain is used?
  • One computer can run a sorting algorithm in 24 minutes. If a second computer is used together with the first, it takes 13 minutes. How long would it take the second computer alone?
  • Two pipes are filling a tank. One pipe fills three times as fast as the other. With both pipes working, the tank fills in 84 minutes. How long would each pipe take working alone?

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example of rational problem solving

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Solving Rational Equations

Intro Harder Probs Graphs

Solve the following equation:

The rational expressions in this equation have variables in the denominators. So my first step is to check for which x -values are not allowed, because they'd cause division by zero. Setting each denominator equal to zero and solving, I get:

x ≠ −4, −1

Looking at the equation, I notice that this equation is a proportion ; that is, the equation is of the form "(one fraction) equals (another fraction)". So all I need to do here is "cross-multiply"; that is, I can use Method 3:

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Rational Equations on MathHelp.com

10(4( x + 1)) = 15( x + 4)

40 x + 40 = 15 x + 60

25 x + 40 = 60

x = 20 / 25 = 4 / 5

Since this solution won't cause any division-by-zero problems, it is a valid solution to the equation, and my answer is:

There is only one fraction, so the common denominator is the only denominator; namely, x . Also, setting that denominator equal to zero, I see that the solution to this equation cannot be x  = 0 .

Method 1: To solve, I can convert everything to this common denominator, and then solve the numerators:

x 2 + x = 72

x 2 + x − 72 = 0

( x + 9)( x − 8) = 0

x = −9 or x = 8

Method 2: To solve, I can start by multiplying through on both sides by x :

Either way, the solution is the same. Since neither solution causes a division-by-zero problem in the original equation, both solution values are valid.

x = −9, 8

First, I'll solve the denominators. Setting each equal to zero to find the disallowed values, I get:

x ≠ 0, 2

The lowest common denominator of these fractions will be x ( x  − 2) .

Method 1: I can solve the equation by converting all of the rational expressions to the common denominator, and then solving the numerators:

10 + (4 x − 8) = 5 x

10 + 4 x − 8 = 5 x

4 x + 2 = 5 x

Method 2: I can also solve this equation by multiplying through on both sides of the equation by the denominator. (The equation becomes a bit of a mess. I've used color below to highlight which parts cancel off. Use caution! )

10 + 4( x − 2) = 5( x )

Using either method, I get the same answer; namely, x  = 2 . However, checking back to the beginning, where I first noted the disallowed values for the original equation, I see that x  ≠ 2 . In other words, my only solution value would actually cause division by zero. Since the only possible solution causes division by zero, then this equation really has no solution. My answer then is:

no solution

It doesn't look like I did anything wrong, mathematically, in the previous exercise. So how did I end up with an entirely invalid solution?

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When dealing with rational expressions and equations, we're not allowed to divide by zero. When there are variables in denominators, we then will have certain values which can cause division by zero. Whichever method one uses to solve a given rational equation, one will, at some point, get rid of those denominators. In other words, at some point, one will, in effect, make those division-by-zero problems magically disappear. But they aren't actually gone; they've merely been ignored at some stage. In the end, one must go back to the beginning and check one's solution against those original disallowed values. And it is perfectly possible that a given equation will have no solution at all.

Whenever you solve a rational equation, always check your (interim) solution against the denominators (and their disallowed values) from the original equation. It is entirely possible that a problem will have an invalid (that is, an "extraneous") solution. This is especially true on tests. So always check!

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Precalculus

Course: precalculus   >   unit 4.

  • Analyzing structure word problem: pet store (1 of 2)
  • Analyzing structure word problem: pet store (2 of 2)
  • Combining mixtures example
  • Rational equations word problem: combined rates

Rational equations word problem: combined rates (example 2)

  • Mixtures and combined rates word problems
  • Rational equations word problem: eliminating solutions
  • Reasoning about unknown variables
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Video transcript

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.

Rational Decision Making: The 7-Step Process for Making Logical Decisions

Clifford Chi

Published: October 17, 2023

Psychology tells us that emotions drive our behavior, while logic only justifies our actions after the fact . Marketing confirms this theory. Humans associate the same personality traits with brands as they do with people  — choosing your favorite brand is like choosing your best friend or significant other. We go with the option that makes us feel something.

Marketer working through the rational decision making process and model

But emotions can cloud your reasoning, especially when you need to do something that could cause internal pain, like giving constructive criticism, or moving on from something you’re attached to, like scrapping a favorite topic from your team's content calendar.

Download Now: How to Be More Productive at Work [Free Guide + Templates]

There’s a way to suppress this emotional bias, though. It’s a thought process that’s completely objective and data-driven. It's called the rational decision making model, and it will help you make logically sound decisions even in situations with major ramifications , like pivoting your entire blogging strategy.

But before we learn each step of this powerful process, let’s go over what exactly rational decision making is and why it’s important.

What is Rational Decision Making?

Rational decision making is a problem-solving methodology that factors in objectivity and logic instead of subjectivity and intuition to achieve a goal. The goal of rational decision making is to identify a problem, pick a solution between multiple alternatives, and find an answer.

Rational decision making is an important skill to possess, especially in the digital marketing industry. Humans are inherently emotional, so our biases and beliefs can blur our perception of reality. Fortunately, data sharpens our view. By showing us how our audience actually interacts with our brand, data liberates us from relying on our assumptions to determine what our audience likes about us.

Rational Decision Making Model: 7 Easy Steps(+ Examples)

Rational Decision Making

1. Verify and define your problem.

To prove that you actually have a problem, you need evidence for it. Most marketers think data is the silver bullet that can diagnose any issue in our strategy, but you actually need to extract insights from your data to prove anything. If you don’t, you’re just looking at a bunch of numbers packed into a spreadsheet.

To pinpoint your specific problem, collect as much data from your area of need and analyze it to find any alarming patterns or trends.

“After analyzing our blog traffic report, we now know why our traffic has plateaued for the past year — our organic traffic increases slightly month over month but our email and social traffic decrease.”

2. Research and brainstorm possible solutions for your problem.

Expanding your pool of potential solutions boosts your chances of solving your problem. To find as many potential solutions as possible, you should gather plenty of information about your problem from your own knowledge and the internet. You can also brainstorm with others to uncover more possible solutions.

Potential Solution 1: “We could focus on growing organic, email, and social traffic all at the same time."

Potential Solution 2: “We could focus on growing email and social traffic at the same time — organic traffic already increases month over month while traffic from email and social decrease.”

Potential Solution 3: "We could solely focus on growing social traffic — growing social traffic is easier than growing email and organic traffic at the same time. We also have 2 million followers on Facebook, so we could push our posts to a ton of readers."

Potential Solution 4: "We could solely focus on growing email traffic — growing email traffic is easier than growing social and organic traffic at the same time. We also have 250,000 blog subscribers, so we could push our posts to a ton of readers."

Potential Solution 5: "We could solely focus on growing organic traffic — growing organic traffic is easier than growing social and email traffic at the same time. We also just implemented a pillar-cluster model to boost our domain’s authority, so we could attract a ton of readers from Google."

3. Set standards of success and failure for your potential solutions.

Setting a threshold to measure your solutions' success and failure lets you determine which ones can actually solve your problem. Your standard of success shouldn’t be too high, though. You’d never be able to find a solution. But if your standards are realistic, quantifiable, and focused, you’ll be able to find one.

“If one of our solutions increases our total traffic by 10%, we should consider it a practical way to overcome our traffic plateau.”

4. Flesh out the potential results of each solution.

Next, you should determine each of your solutions’ consequences. To do so, create a strength and weaknesses table for each alternative and compare them to each other. You should also prioritize your solutions in a list from best chance to solve the problem to worst chance.

Potential Result 1: ‘Growing organic, email, and social traffic at the same time could pay a lot of dividends, but our team doesn’t have enough time or resources to optimize all three channels.”

Potential Result 2: “Growing email and social traffic at the same time would marginally increase overall traffic — both channels only account for 20% of our total traffic."

Potential Result 3: “Growing social traffic by posting a blog post everyday on Facebook is challenging because the platform doesn’t elevate links in the news feed and the channel only accounts for 5% of our blog traffic. Focusing solely on social would produce minimal results.”

Potential Result 4: “Growing email traffic by sending two emails per day to our blog subscribers is challenging because we already send one email to subscribers everyday and the channel only accounts for 15% of our blog traffic. Focusing on email would produce minimal results.”

Potential Result 5: “Growing organic traffic by targeting high search volume keywords for all of our new posts is the easiest way to grow our blog’s overall traffic. We have a high domain authority, Google refers 80% of our total traffic, and we just implemented a pillar-cluster model. Focusing on organic would produce the most results.”

5. Choose the best solution and test it.

Based on the evaluation of your potential solutions, choose the best one and test it. You can start monitoring your preliminary results during this stage too.

“Focusing on organic traffic seems to be the most effective and realistic play for us. Let’s test an organic-only strategy where we only create new content that has current or potential search volume and fits into our pillar cluster model.”

6. Track and analyze the results of your test.

Track and analyze your results to see if your solution actually solved your problem.

“After a month of testing, our blog traffic has increased by 14% and our organic traffic has increased by 21%.”

7. Implement the solution or test a new one.

If your potential solution passed your test and solved your problem, then it’s the most rational decision you can make. You should implement it to completely solve your current problem or any other related problems in the future. If the solution didn’t solve your problem, then test another potential solution that you came up with.

“The results from solely focusing on organic surpassed our threshold of success. From now on, we’re pivoting to an organic-only strategy, where we’ll only create new blog content that has current or future search volume and fits into our pillar cluster model.”

Avoid Bias With A Rational Decision Making Process

As humans, it’s natural for our emotions to take over your decision making process. And that’s okay. Sometimes, emotional decisions are better than logical ones. But when you really need to prioritize logic over emotion, arming your mind with the rational decision making model can help you suppress your emotion bias and be as objective as possible.

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Module 5: Polynomial and Rational Functions

Applications of rational equations, learning objectives.

  • Define and write a proportion
  • Solve proportional problems involving scale drawings
  • Solve a rational formula for a specified variable
  • Solve work problems
  • Solve motion problem
  • Define and solve an equation that represents the concentration of a mixture

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 the planet who don’t have access to a toilet. Because, why not?

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.

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.

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}{3}[/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}{7.4\text{ billion }}[/latex]

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]

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?

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.

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]

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]

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 sayinga person’s height is 4 times the length of their femur.

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.

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

 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].

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}{ccc}\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\end{array}[/latex]

We used the scale factor 1:557 to find an unknown distance between Seattle and San Jose. We also checked 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.

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.

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]

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

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

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

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.

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.

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]

Check the solutions in the original equation.

[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.

We have solved uniform motion problems using the formula [latex]D = rt[/latex] in previous chapters. We used a table like the one below to organize the information and lead us to the equation.

[latex]\begin{array}{|c|c|c|c|} \hline & \,\text{rate}\,&\,\text{time}\,&\,\text{distance}\\ \hline \,\text{First}\,&\,&\,&\\ \hline \,\text{Second}\,&\,&\,&\\ \hline \end{array} [/latex]

The formula [latex]D=rt[/latex] assumes we know [latex]r[/latex] and  [latex]t[/latex] and use them to find  [latex]D[/latex]. If we know  [latex]D[/latex] and [latex]r[/latex] and need to find  [latex]t[/latex], we would solve the equation for  [latex]t[/latex] and get the formula [latex]\displaystyle t=\frac{D}{r}[/latex].

Greg went to a conference in a city 120 miles away. On the way back, due to road construction he had to drive 10 mph slower which resulted in the return trip taking 2 hours longer. How fast did he drive on the way to the conference?

[latex]\begin{eqnarray*} \begin{array}{|c|c|c|c|} \hline & \,\text{rate}\,& \,\text{time}\,& \,\text{distance}\\ \hline \,\text{There}\,& r & t & 120\\ \hline \,\text{Back}\,& & & \\ \hline \end{array}\,& & \begin{array}{l} \,\text{We}\,\,\text{do}\,\,\text{not}\,\,\text{know}\,\,\text{rate}, r, \,\text{or} \,\text{time}, t \,\text{he}\,\,\text{traveled}\\ \,\text{on}\,\,\text{the}\,\,\text{way}\,\,\text{to}\,\,\text{the}\,\,\text{conference}\,. \,\text{But}\,\,\text{we}\,\,\text{do}\,\,\text{know}\\ \,\text{the}\,\,\text{distance}\,\,\text{was}\,120 \,\text{miles}\,. \end{array}\\ & & \\ \begin{array}{|c|c|c|c|} \hline & \,\text{rate}\,& \,\text{time}\,& \,\text{distance}\\ \hline \,\text{There}\,& r & t & 120\\ \hline \,\text{Back}\,& r - 10 & t + 2 & 120\\ \hline \end{array}& & \begin{array}{l} \,\text{Coming}\,\,\text{back}\,\,\text{he}\,\,\text{drove}\,10 \,\text{mph} \,\text{slower}\,(r - 10)\\ \,\text{and}\,\,\text{took}\,2 \,\text{hours}\,\,\text{longer}\,(t + 2) . \,\text{The} \,\text{distance}\\ \,\text{was}\,\,\text{still}\,120 \,\text{miles}\,. \end{array}\\ & & \\ r t = 120 & & \,\text{Equations}\,\,\text{are}\,\,\text{product}\,\,\text{of} \,\text{rate}\,\,\text{and}\,\,\text{time}\\ (r - 10) (t + 2) = 120 & & \,\text{We}\,\,\text{have}\,\,\text{simultaneous} \,\text{product}\,\,\text{equations}\\ & & \\ t = \frac{120}{r}\,\,\text{and}\,t + 2 = \frac{120}{r - 10} & & \,\text{Solving}\,\,\text{for}\,\,\text{rate}, \,\text{divide}\,\,\text{by}\,r \,\text{and}\,r - 10\\ & & \\ \frac{120}{r}\,+ 2 = \frac{120}{r - 10}& & \,\text{Substitute} \frac{120}{r} \,\text{for}\,t \,\text{in}\,\,\text{the}\,\,\text{second} \,\text{equation}\\ & & \\ \frac{120 r (r - 10)}{r} + 2 r (r - 10) = \frac{120 r (r - 10)}{r - 10} & & \,\text{Multiply}\,\,\text{each}\,\,\text{term}\,\,\text{by}\,\,\text{LCD}\,: r (r - 10)\\ & & \\ 120 (r - 10) + 2 r^2 - 20 r = 120 r & & \,\text{Reduce}\,\,\text{each} \,\text{fraction}\\ 120 r - 1200 + 2 r^2 - 20 r = 120 r & & \,\text{Distribute}\\ 2 r^2 + 100 r - 1200 = 120 r & & \,\text{Combine}\,\,\text{like} \,\text{terms}\\ \underline{- 120 r - 120 r} & & \,\text{Make}\,\,\text{equation}\,\,\text{equal} \,\text{to}\,\,\text{zero}\\ 2 r^2 - 20 r - 1200 = 0 & & \,\text{Divide}\,\,\text{each}\,\,\text{term} \,\text{by}\,2\\ r^2 - 10 r - 600 = 0 & & \,\text{Factor}\\ (r - 30) (r + 20) = 0 & & \,\text{Set}\,\,\text{each}\,\,\text{factor} \,\text{equal}\,\,\text{to}\,\,\text{zero}\\ r - 30 = 0 \,\text{and}\,r + 20 = 0 & & \,\text{Solve}\,\,\text{each} \,\text{equation}\\ \underline{+ 30 + 30} \underline{- 20 - 20} & & \\ r = 30 \,\text{and}\,r = - 20 & & \,\text{Can}' t \,\text{have}\,a \,\text{negative}\,\,\text{rate}\\ 30 \,\text{mph}\,& & \,\text{Our}\,\,\text{Solution} \end{eqnarray*}[/latex]

A man rows down stream for 30 miles then turns around and returns to his original location, the total trip took 8 hours. If the current flows at 2 miles per hour, how fast would the man row in still water?

Mixtures are made of ratios of different substances that may include chemicals, foods, water, or gases. There are many different situations where mixtures may occur both in nature and as a means to produce a desired product or outcome.  For example, chemical spills, manufacturing and even biochemical reactions involve mixtures.  The thing that can make mixtures interesting mathematically is when components of the mixture are added at different rates and concentrations. In our last example we will define an equation that models the concentration  – or ratio of sugar to water – in a large mixing tank over time. You are asked whether the final concentration of sugar is greater than the concentration at the beginning.

A large mixing tank currently contains 100 gallons of water into which 5 pounds of sugar have been mixed. A tap will open pouring 10 gallons per minute of water into the tank at the same time sugar is poured into the tank at a rate of 1 pound per minute. Find the concentration (pounds per gallon) of sugar in the tank after 12 minutes. Is that a greater concentration than at the beginning?

Let t  be the number of minutes since the tap opened. Since the water increases at 10 gallons per minute, and the sugar increases at 1 pound per minute, these are constant rates of change. This tells us the amount of water in the tank is a linear equation, as is the amount of sugar in the tank. We can write an equation independently for each:

The concentration, C , will be the ratio of pounds of sugar to gallons of water

The concentration after 12 minutes is given by evaluating [latex]C\left(t\right)\\[/latex] at [latex]t=\text{ }12\\[/latex].

This means the concentration is 17 pounds of sugar to 220 gallons of water.

At the beginning, the concentration is

Since [latex]\frac{17}{220}\approx 0.08>\frac{1}{20}=0.05\\[/latex], the concentration is greater after 12 minutes than at the beginning.

In the following video, we show another example of how to use rational functions to model mixing.

  • "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 . ↵
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Mathematics LibreTexts

4.3: Rational Inequalities and Applications

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  • Page ID 119162

  • Carl Stitz & Jeff Zeager
  • Lakeland Community College & Lorain County Community College

Math 370 Learning Objectives

  • Solve rational inequalities using graphs and/or sign charts
  • Build models to solve applications involving rational functions

Solving Rational Inequalities

In this section, we solve equations and inequalities involving rational functions and explore associated application problems. Our first example showcases the critical difference in procedure between solving a rational equation and a rational inequality.

Example \( \PageIndex{1} \)

  • Solve \(\frac{x^3-2x+1}{x-1} = \frac{1}{2}x-1\).
  • Solve \(\frac{x^3-2x+1}{x-1} \geq \frac{1}{2}x-1\).
  • Use graphing technology to check your answers to 1 and 2.
  • To solve the equation, we clear denominators\[\begin{array}{rclr} \dfrac{x^3-2x+1}{x-1} & = & \dfrac{1}{2}x-1 & \\  \left(\dfrac{x^3-2x+1}{x-1}\right) \cdot 2(x-1) & = & \left( \dfrac{1}{2}x-1 \right) \cdot 2(x-1) & \\ 2x^3 - 4x + 2 & = & x^2-3x+2 & \left( \text{expand} \right) \\ 2x^3 -x^2 - x & = & 0 & \\ x(2x+1)(x-1) & = & 0 & \left( \text{factor} \right) \\ x & = & -\dfrac{1}{2}, \, 0, \, 1 & \\ \end{array}\nonumber\]Since we cleared denominators, we need to check for extraneous solutions. Sure enough, we see that \(x=1\) does not satisfy the original equation and must be discarded. Our solutions are \(x=-\frac{1}{2}\) and \(x=0\).

Screen Shot 2022-04-01 at 1.16.25 AM.png

Subsection Footnotes

1  There is no asymptote at \(x = 1\) since the graph is well behaved near \(x = 1\). According to Theorem 4.1.1 , there must be a hole there.

Applications Involving Rational Equations

Next, we explore how rational equations can be used to solve some classic problems involving rates.

Example \( \PageIndex{2} \)

Carl decides to explore the Meander River, the location of several recent Sasquatch sightings. From camp, he canoes downstream five miles to check out a purported Sasquatch nest. Finding nothing, he immediately turns around, retraces his route (this time traveling upstream), and returns to camp 3 hours after he left. If Carl canoes at a rate of 6 miles per hour in still water, how fast was the Meander River flowing on that day?

We are given information about distances, rates (speeds) and times. The basic principle relating these quantities is: \[\text{distance} = \text{rate} \cdot \text{time}\nonumber\] The first observation to make, however, is that the distance, rate and time given to us aren’t "compatible": the distance given is the distance for only part of the trip, the rate given is the speed Carl can canoe in still water, not in a flowing river, and the time given is the duration of the entire trip. Ultimately, we are after the speed of the river, so let’s call that \(R\) measured in miles per hour to be consistent with the other rate given to us. To get started, let’s divide the trip into its two parts: the initial trip downstream and the return trip upstream. For the downstream trip, all we know is that the distance traveled is \(5\) miles.

\[\begin{array}{rcl} \text{distance downstream} & = & \text{rate traveling downstream} \cdot \text{time traveling downstream} \\ 5 \, \text{miles} & = & \text{rate traveling downstream} \cdot \text{time traveling downstream} \\ \end{array}\nonumber\]

Since the return trip upstream followed the same route as the trip downstream, we know that the distance traveled upstream is also 5 miles.

\[\begin{array}{rcl} \text{distance upstream} & = & \text{rate traveling upstream} \cdot \text{time traveling upstream} \\ 5 \, \text{miles} & = & \text{rate traveling upstream} \cdot \text{time traveling upstream} \\ \end{array}\nonumber\]

We are told Carl can canoe at a rate of \(6\) miles per hour in still water. How does this figure into the rates traveling upstream and downstream? The speed the canoe travels in the river is a combination of the speed at which Carl can propel the canoe in still water, 6 miles per hour, and the speed of the river, which we’re calling \(R\). When traveling downstream, the river is helping Carl along, so we add these two speeds:

\[\begin{array}{rcl} \text{rate traveling downstream} & = & \text{rate Carl propels the canoe} + \text{speed of the river} \\ & = & 6 \dfrac{\text{miles}}{\text{hour}} + R \dfrac{\text{miles}}{\text{hour}} \\ \end{array}\nonumber\]

So our downstream speed is \((6+R) \frac{\text{miles}}{\text{hour}}\). Substituting this into our "distance-rate-time" equation for the downstream part of the trip, we get:

\[\begin{array}{rcl} 5 \, \text{miles} & = & \text{rate traveling downstream} \cdot \text{time traveling downstream} \\ 5 \, \text{miles} & = & (6+R) \dfrac{\text{miles}}{\text{hour}} \cdot \text{time traveling downstream} \\ \end{array}\nonumber\]

When traveling upstream, Carl works against the current. Since the canoe manages to travel upstream, the speed Carl can canoe in still water is greater than the river’s speed, so we subtract the river’s speed from Carl’s canoeing speed to get:

\[\begin{array}{rcl} \text{rate traveling upstream} & = & \text{rate Carl propels the canoe} - \text{river speed} \\ & = & 6 \dfrac{\text{miles}}{\text{hour}} - R \dfrac{\text{miles}}{\text{hour}} \\ \end{array}\nonumber\]

Proceeding as before, we get

\[\begin{array}{rcl} 5 \, \text{miles} & = & \text{rate traveling upstream} \cdot \text{time traveling upstream} \\ 5 \, \text{miles} & = & (6 - R) \dfrac{\text{miles}}{\text{hour}} \cdot \text{time traveling upstream} \\ \end{array}\nonumber\]

The last piece of information given to us is that the total trip lasted \(3\) hours. If we let \(t_{\text{down}}\) denote the time of the downstream trip and \(t_{\text{up}}\) the time of the upstream trip, we have: \(t_{\text{down}} + t_{\text{up}} = 3 \, \text{hours}\). Substituting \(t_{\text{down}}\) and \(t_{\text{up}}\) into the "distance-rate-time" equations, we get (suppressing the units) three equations in three unknowns: 2  \[\left\{\begin{array}{lrcl} E1 & (6+R) \, t_{\text{down}} & = & 5 \\ E2 & (6-R) \, t_{\text{up}} & = & 5 \\ E3 & t_{\text{down}} + t_{\text{up}} & = & 3 \end{array} \right.\nonumber\]

Since we are ultimately after \(R\), we need to use these three equations to get at least one equation involving only \(R\). To that end, we solve \(E1\) for \(t_{\text{down}}\) by dividing both sides 3  by the quantity \((6+R)\) to get \(t_{\text{down}} = \frac{5}{6+R}\). Similarly, we solve \(E2\) for \(t_{\text{up}}\) and get \(t_{\text{up}} = \frac{5}{6-R}\). Substituting these into \(E3\), we get: 4  \[\dfrac{5}{6+R} + \dfrac{5}{6 - R} = 3.\nonumber\] Clearing denominators, we get \(5(6-R) + 5(6+R) = 3(6+R)(6-R)\) which reduces to \(R^2 = 16\). We find \(R = \pm 4\), and since \(R\) represents the speed of the river, we choose \(R = 4\). On the day in question, the Meander River is flowing at a rate of \(4\) miles per hour.

One of the important lessons to learn from Example \( \PageIndex{2} \) is that speeds, and more generally, rates, are additive. As we see in our next example, the concept of rate and its associated principles can be applied to a wide variety of problems - not just "distance-rate-time" scenarios.

Example \( \PageIndex{3} \)

Working alone, Taylor can weed the garden in 4 hours. If Carl helps, they can weed the garden in 3 hours. How long would it take for Carl to weed the garden on his own?

The key relationship between work and time which we use in this problem is: \[\text{amount of work done} = \text{rate of work} \cdot \text{time spent working}\nonumber\]

We are told that, working alone, Taylor can weed the garden in 4 hours. In Taylor’s case then: \[\begin{array}{rcl} \text{amount of work Taylor does} & = & \text{rate of Taylor working} \cdot \text{time Taylor spent working} \\ 1 \, \text{garden} & = & (\text{rate of Taylor working}) \cdot (4 \, \text{hours}) \\ \end{array}\nonumber\]

So we have that the rate Taylor works is \(\frac{1 \, \text{garden}}{ 4 \, \text{hours}} = \frac{1}{4} \frac{\text{garden}}{\text{hour}}\). We are also told that when working together, Taylor and Carl can weed the garden in just 3 hours. We have:

\[\begin{array}{rcl} \text{amount of work done together} & = & \text{rate of working together} \cdot \text{time spent working together} \\ 1 \, \text{garden} & = & (\text{rate of working together}) \cdot (3 \, \text{hours}) \\ \end{array}\nonumber\]

From this, we find that the rate of Taylor and Carl working together is \(\frac{1 \, \text{garden}}{3 \, \text{hours}} = \frac{1}{3} \frac{\text{garden}}{\text{hour}}\). We are asked to find out how long it would take for Carl to weed the garden on his own. Let us call this unknown \(t\), measured in hours to be consistent with the other times given to us in the problem. Then:

\[\begin{array}{rcl} \text{amount of work Carl does} & = & \text{rate of Carl working} \cdot \text{time Carl spent working} \\ 1 \, \text{garden} & = & (\text{rate of Carl working}) \cdot (t \, \text{hours}) \\ \end{array}\nonumber\]

In order to find \(t\), we need to find the rate of Carl working, so let’s call this quantity \(R\), with units \(\frac{\text{garden}}{\text{hour}}\). Using the fact that rates are additive, we have:

\[\begin{array}{rcl} \text{rate working together} & = & \text{rate of Taylor working} + \text{rate of Carl working} \\[4pt] \dfrac{1}{3} \dfrac{\text{garden}}{\text{hour}} & = & \dfrac{1}{4} \dfrac{\text{garden}}{\text{hour}} + R \dfrac{\text{garden}}{\text{hour}} \\ \end{array}\nonumber\]

so that \(R = \frac{1}{12} \frac{\text{garden}}{\text{hour}}\). Substituting this into our ‘work-rate-time’ equation for Carl, we get:

\[\begin{array}{rcl} 1 \, \text{garden} & = & (\text{rate of Carl working}) \cdot (t \, \text{hours}) \\[4pt] 1 \, \text{garden} & = & \left(\dfrac{1}{12} \dfrac{\text{garden}}{\text{hour}} \right) \cdot (t \, \text{hours}) \\ \end{array}\nonumber\]

Solving \(1 = \frac{1}{12} t\), we get \(t = 12\), so it takes Carl 12 hours to weed the garden on his own.

As is common with "word problems" like Examples \( \PageIndex{2} \) and \( \PageIndex{3} \), there is no short-cut to the answer. We encourage the reader to carefully think through and apply the basic principles of rate to each (potentially different!) situation. It is time well spent. We also encourage the tracking of units, especially in the early stages of the problem. Not only does this promote uniformity in the units, it also serves as a quick means to check if an equation makes sense.

2  This is called a system of equations. No doubt, you’ve had experience with these things before, and we will study systems in greater detail in Chapter 9 .

3  While we usually discourage dividing both sides of an equation by a variable expression, we know \((6+R) \neq 0\) since otherwise we couldn’t possibly multiply it by \(t_{\text {down }}\) and get 5.

4  The reader is encouraged to verify that the units in this equation are the same on both sides. To get you started, the units on the "3" is "hours."

Applications Involving Rational Inequalities

Our next example deals with the average cost function, first introduced as applied to PortaBoy Game systems from Example 2.1.5 in Section 2.1 .

Example \( \PageIndex{4} \)

Given a cost function \(C(x)\), which returns the total cost of producing \(x\) items, recall that the average cost function, \(\overline{C}(x) = \frac{C(x)}{x}\) computes the cost per item when \(x\) items are produced. Suppose the cost \(C\), in dollars, to produce \(x\) PortaBoy game systems for a local retailer is \(C(x) = 80x + 150\), \(x \geq 0\).

  • Find an expression for the average cost function \(\overline{C}(x)\).
  • Solve \(\overline{C}(x) < 100\) and interpret.
  • Determine the behavior of \(\overline{C}(x)\) as \(x \to \infty\) and interpret.
  • From \(\overline{C}(x) = \frac{C(x)}{x}\), we obtain \(\overline{C}(x) = \frac{80x+150}{x}\). The domain of \(C\) is \(x \geq 0\), but since \(x=0\) causes problems for \(\overline{C}(x)\), we get our domain to be \(x>0\), or \((0, \infty)\).

Screen Shot 2022-04-01 at 1.37.15 AM.png

  • When we apply Theorem 4.1.2 to \(\overline{C}(x)\) we find that \(y=80\) is a horizontal asymptote to the graph of \(y=\overline{C}(x)\). To more precisely determine the behavior of \(\overline{C}(x)\) as \(x \to \infty\), we first use long division 5  and rewrite \(\overline{C}(x) = 80+\frac{150}{x}\). As \(x \to \infty\), \(\frac{150}{x} \to 0^{+}\), which means \(\overline{C}(x) \approx 80+\text { very small }(+)\). Thus the average cost per system is getting closer to \(\$ 80\) per system. If we set \(\overline{C}(x) = 80\), we get \(\frac{150}{x} = 0\), which is impossible, so we conclude that \(\overline{C}(x) > 80\) for all \(x > 0\). This means that the average cost per system is always greater than \(\$ 80\) per system, but the average cost is approaching this amount as more and more systems are produced. Looking back at Example 2.1.5 , we realize \(\$ 80\) is the variable cost per system \(-\) the cost per system above and beyond the fixed initial cost of \(\$150\). Another way to interpret our answer is that "infinitely" many systems would need to be produced to effectively "zero out" the fixed cost.

Our next example is another classic "box with no top" problem.

Example \( \PageIndex{5} \)

A box with a square base and no top is to be constructed so that it has a volume of \(1000\) cubic centimeters. Let \(x\) denote the width of the box, in centimeters as seen below.

Screen Shot 2022-04-01 at 1.40.01 AM.png

  • Express the height \(h\) in centimeters as a function of the width \(x\) and state the applied domain.
  • Solve \(h(x) \geq x\) and interpret.
  • Find and interpret the behavior of \(h(x)\) as \(x \to 0^{+}\) and as \(x \to \infty\).
  • Express the surface area \(S\) of the box as a function of \(x\) and state the applied domain.
  • Use graphing technology to approximate (to two decimal places) the dimensions of the box which minimize the surface area.
  • We are told that the volume of the box is \(1000\) cubic centimeters and that \(x\) represents the width, in centimeters. From geometry, we know \(\mbox{Volume} = \mbox{width} \times \mbox{height} \times \mbox{depth}\). Since the base of the box is a square, the width and the depth are both \(x\) centimeters. Using \(h\) for the height, we have \(1000 = x^2h\), so that \(h = \frac{1000}{x^2}\). Using function notation,   \(h(x) = \frac{1000}{x^2}\) As for the applied domain, in order for there to be a box at all, \(x > 0\), and since every such choice of \(x\) will return a positive number for the height \(h\) we have no other restrictions and conclude our domain is \((0, \infty)\).

Screen Shot 2022-04-01 at 1.41.52 AM.png

  • As \(x \to 0^{+}\), \(h(x) = \frac{1000}{x^2} \to \infty\). This means that the smaller the width \(x\) (and, in this case, depth), the larger the height \(h\) has to be in order to maintain a volume of \(1000\) cubic centimeters. As \(x \to \infty\), we find \(h(x) \to 0^{+}\), which means that in order to maintain a volume of \(1000\) cubic centimeters, the width and depth must get bigger as the height becomes smaller.
  • Since the box has no top, the surface area can be found by adding the area of each of the sides to the area of the base. The base is a square of dimensions \(x\) by \(x\), and each side has dimensions \(x\) by \(h\). We get the surface area, \(S = x^2+4xh\). To get \(S\) as a function of \(x\), we substitute \(h = \frac{1000}{x^2}\) to obtain \(S = x^2+4x \left( \frac{1000}{x^2}\right)\). Hence, as a function of \(x\), \(S(x) = x^2 + \frac{4000}{x}\). The domain of \(S\) is the same as \(h\), namely \((0, \infty)\), for the same reasons as above.
  • A first attempt at the graph of \(y=S(x)\) on the calculator may lead to frustration. Chances are good that the first window chosen to view the graph will suggest \(y=S(x)\) has the \(x\)-axis as a horizontal asymptote. From the formula \(S(x) = x^2 + \frac{4000}{x}\), however, we get \(S(x) \approx x^2\) as \(x \to \infty\), so \(S(x) \to \infty\). Readjusting the window, we find \(S\) does possess a relative minimum at \(x \approx 12.60\). Without calculus, all we can say is that we belive this is the only relative extremum, so it is the absolute minimum as well. This means that the width and depth of the box should each measure approximately \(12.60\) centimeters. To determine the height, we find \(h(12.60) \approx 6.30\), so the height of the box should be approximately \(6.30\) centimeters.

Precalc_4_3_Fig2.jpg

5  In this case, long division amounts to term-by-term division.

IMAGES

  1. Rational Problem Solving Steps Infographic Vector Stock Vector

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  2. PPT

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

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  4. Solving Rational Equations #1

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  5. How to Solve Rational Equations: Step-by-Step Tutorial

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  6. 6.5 Solving Equations Containing Rational Expressions #3

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VIDEO

  1. Classify the following Numbers as Rational or Irrational (3+√23)-√23

  2. 6.7 Solving Rational Equations Example 1

  3. A Calculus Question/ Rationalization

  4. Lesson 1.8 Problem Solving with Rational Numbers

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COMMENTS

  1. Solving Rational Equations

    Example 1: Solve the rational equation below and make sure you check your answers for extraneous values. Would it be nice if the denominators are not there? Well, we can't simply vanish them without any valid algebraic step.

  2. 15.2.1: Solving Rational Equations and Applications

    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.

  3. 7.5: Solving Rational Equations

    Solve rational equations by clearing the fractions by multiplying both sides of the equation by the least common denominator (LCD). Example 7.5.1 7.5. 1. Solve: 5 x − 13 = 1 x 5 x − 1 3 = 1 x. Solution: We first make a note that x ≠ 0 x ≠ 0 and then multiply both sides by the LCD, 3x 3 x:

  4. 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

  5. Rational equations intro (video)

    Lesson 1: Rational equations Rational equations intro Rational equations intro Equations with rational expressions Equations with rational expressions (example 2) Rational equations Finding inverses of rational functions Find inverses of rational functions Math > Algebra 2 > Equations > Rational equations

  6. Rational equations (practice)

    Lesson 1: Rational equations Rational equations intro Rational equations intro Equations with rational expressions Equations with rational expressions (example 2) Rational equations Finding inverses of rational functions Math > Algebra 2 > Equations > Rational equations Rational equations Google Classroom You might need: Calculator

  7. Solve Rational Equations

    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.

  8. Solving Rational Equations · Examples · Matter of Math

    An example of what you will more likely see in an exam is something like this: x6 + 2 x = 54 Each term is shown as a fraction. Rational equations can also include radicals: x−−√ 2 + 6 = 7 Or other operations such as division: x2+33x ÷ 54 = 13 Luckily, the technique you learn now will work for every type of rational equation!

  9. 7.4 Solve Rational Equations

    Step 2. Find the least common denominator of all denominators in the equation. Step 3. Clear the fractions by multiplying both sides of the equation by the LCD. Step 4. Solve the resulting equation. Step 5. Check: If any values found in Step 1 are algebraic solutions, discard them.

  10. 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.

  11. PDF Notes, Examples, and practice (with solutions)

    Word Problems that use Rational Expressions Example: Underground pipes can fill a swimming pool in 4 hours. A regular garden hose can fill the pool in 15 hours. If both are used at the same thne, how long will it take to fill the pool? Solving Rational Equalities/Equations Step 3: Check Answer! If time is 3.158 hours, the pipes will add

  12. Rational Function Problems (video lessons, examples and solutions)

    Here are a few examples of work problems that are solved with rational equations. Examples: 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.

  13. Rational Equations

    Understand what a rational equation is and how to solve rational equations, with examples. Learn the difference between rational equations and rational expressions. Updated: 11/21/2023

  14. Solving Rational Equations: Harder Problems

    Whenever you solve a rational equation, always check your (interim) solution against the denominators (and their disallowed values) from the original equation. It is entirely possible that a problem will have an invalid (that is, an "extraneous") solution. This is especially true on tests. So always check!

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

    Combining mixtures example. Rational equations word problem: combined rates. Rational equations word problem: combined rates (example 2) Mixtures and combined rates word problems. Rational equations word problem: eliminating solutions ... (in decks per hour) would not be A + B, but 1/A + 1/B, which would equal 1/8. Try solving the problem using ...

  16. Solving Rational Equations and Applications

    For example, is a rational equation. 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.

  17. Rational Decision Making: The 7-Step Process for Making Logical Decisions

    Example: "After analyzing our blog traffic report, we now know why our traffic has plateaued for the past year — our organic traffic increases slightly month over month but our email and social traffic decrease." 2. Research and brainstorm possible solutions for your problem.

  18. 9.7: Solve Rational Inequalities

    Step 1. Write the inequality as one quotient on the left and zero on the right. Our inequality is in this form. x − 1 x + 3 ≥ 0 x − 1 x + 3 ≥ 0. Step 2. Determine the critical points—the points where the rational expression will be zero or undefined. The rational expression will be zero when the numerator is zero.

  19. Applications of Rational Equations

    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 ...

  20. 12 Approaches To Problem-Solving for Every Situation

    1. Rational One of the most common problem-solving approaches, the rational approach is a multi-step process that works well for a wide range of problems. Many other problem-solving techniques mirror or build off of its seven steps, so it may be helpful to begin with the rational approach before moving on to other techniques.

  21. 4.3: Rational Inequalities and Applications

    In this section, we solve equations and inequalities involving rational functions and explore associated application problems. Our first example showcases the critical difference in procedure between solving a rational equation and a rational inequality. Example 4.3.1. Solve x3 − 2x + 1 x − 1 = 1 2x − 1. Solve x3 − 2x + 1 x − 1 ≥ 1 ...

  22. Rational Problem Solving

    For example, whether the solution can be accepted by all organization or not and whether each solution can be operated with breaking the limitation such as time and money. Based on this condition, you need to find a solution which is best in all aspects instead of the one just can be available.

  23. Algebra

    3x+8 x −1 < −2 3 x + 8 x − 1 < − 2 Solution. u ≤ 4 u −3 u ≤ 4 u − 3 Solution. t3 −6t2 t−2 > 0 t 3 − 6 t 2 t − 2 > 0 Solution. Here is a set of practice problems to accompany the Rational Inequalities section of the Solving Equations and Inequalities chapter of the notes for Paul Dawkins Algebra course at Lamar University.