Rational Equations
Solving rational equations.
A rational equation is a type of equation where it involves at least one rational expression, a fancy name for a fraction . The best approach to address this type of equation is to eliminate all the denominators using the idea of LCD (least common denominator). By doing so, the leftover equation to deal with is usually either linear or quadratic.
In this lesson, I want to go over ten (10) worked examples with various levels of difficulty. I believe that most of us learn math by looking at many examples. Here we go!
Examples of How to Solve 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. The approach is to find the Least Common Denominator (also known Least Common Multiple) and use that to multiply both sides of the rational equation. It results in the removal of the denominators, leaving us with regular equations that we already know how to solve such as linear and quadratic. That is the essence of solving rational equations.
 The LCD is [latex]6x[/latex]. I will multiply both sides of the rational equation by [latex]6x[/latex] to eliminate the denominators. That’s our goal anyway – to make our life much easier.
 You should have something like this after distributing the LCD.
 I decided to keep the variable [latex]x[/latex] on the right side. So remove the [latex]5x[/latex] on the left by adding both sides by [latex]5x[/latex].
 Simplify. It’s obvious now how to solve this onestep equation. Divide both sides by the coefficient of [latex]5x[/latex].
 Yep! The final answer is [latex]x = 2[/latex] after checking it back into the original rational equation. It yields a true statement.
Always check your “solved answers” back into the original equation to exclude extraneous solutions. This is a critical aspect of the overall approach when dealing with problems like Rational Equations and Radical Equations .
Example 2: Solve the rational equation below and make sure you check your answers for extraneous values.
The first step in solving a rational equation is always to find the “silver bullet” known as LCD. So for this problem, finding the LCD is simple.
Here we go.
Try to express each denominator as unique powers of prime numbers, variables and/or terms.
Multiply together the ones with the highest exponents for each unique prime number, variable and/or terms to get the required LCD.
 The LCD is [latex]9x[/latex]. Distribute it to both sides of the equation to eliminate the denominators.
 To keep the variables on the left side, subtract both sides by [latex]63[/latex].
 The resulting equation is just a onestep equation. Divide both sides by the coefficient of [latex]x[/latex].
 That is it! Check the value [latex]x = – \,39[/latex] back into the main rational equation and it should convince you that it works.
Example 3: Solve the rational equation below and make sure you check your answers for extraneous values.
It looks like the LCD is already given. We have a unique and common term [latex]\left( {x – 3} \right)[/latex] for both of the denominators. The number [latex]9[/latex] has the trivial denominator of [latex]1[/latex] so I will disregard it. Therefore the LCD must be [latex]\left( {x – 3} \right)[/latex].
 The LCD here is [latex]\left( {x – 3} \right)[/latex]. Use it as a multiplier to both sides of the rational equation.
 I hope you get this linear equation after performing some cancellations.
Distribute the constant [latex]9[/latex] into [latex]\left( {x – 3} \right)[/latex].
 Combine the constants on the left side of the equation.
 Move all the numbers to the right side by adding [latex]21[/latex] to both sides.
 Not too bad. Again make it a habit to check the solved “answer” from the original equation.
It should work so yes, [latex]x = 2[/latex] is the final answer.
Example 4: Solve the rational equation below and make sure you check your answers for extraneous values.
I hope that you can tell now what’s the LCD for this problem by inspection. If not, you’ll be fine. Just keep going over a few examples and it will make more sense as you go along.
 The LCD is [latex]4\left( {x + 2} \right)[/latex]. Multiply each side of the equations by it.
 After careful distribution of the LCD into the rational equation, I hope you have this linear equation as well.
Quick note : If ever you’re faced with leftovers in the denominator after multiplication, that means you have an incorrect LCD.
Now, distribute the constants into the parenthesis on both sides.
 Combine the constants on the left side to simplify it.
 At this point, make the decision where to keep the variable.
 Keeping the [latex]x[/latex] to the left means we subtract both sides by [latex]4[/latex].
 Add both sides by [latex]3x[/latex].
 That’s it. Check your answer to verify its validity.
Example 5: Solve the rational equation below and make sure you check your answers for extraneous values.
Focusing on the denominators, the LCD should be [latex]6x[/latex]. Why?
Remember, multiply together “each copy” of the prime numbers or variables with the highest powers.
 The LCD is [latex]6x[/latex]. Distribute to both sides of the given rational equation.
 It should look like after careful cancellation of similar terms.
 Distribute the constant into the parenthesis.
 The variable [latex]x[/latex] can be combined on the left side of the equation.
 Since there’s only one constant on the left, I will keep the variable [latex]x[/latex] to the opposite side.
 So I subtract both sides by [latex]5x[/latex].
 Divide both sides by [latex]2[/latex] to isolate [latex]x[/latex].
 Yep! We got the final answer.
Example 6: Solve the rational equation below and make sure you check your answers for extraneous values.
Whenever you see a trinomial in the denominator, always factor it out to identify the unique terms. By simple factorization, I found that [latex]{x^2} + 4x – 5 = \left( {x + 5} \right)\left( {x – 1} \right)[/latex]. Not too bad?
Finding the LCD just like in previous problems.
Try to express each denominator as unique powers of prime numbers, variables and/or terms. In this case, we have terms in the form of binomials.
Multiply together the ones with the highest exponents for each unique copy of a prime number, variable and/or terms to get the required LCD.
 Before I distribute the LCD into the rational equations, factor out the denominators completely.
This aids in the cancellations of the commons terms later.
 Multiply each side by the LCD.
 Wow! It’s amazing how quickly the “clutter” of the original problem has been cleaned up.
 Get rid of the parenthesis by the distributive property.
You should end up with a very simple equation to solve.
Example 7: Solve the rational equation below and make sure you check your answers for extraneous values.
Since the denominators are two unique binomials, it makes sense that the LCD is just their product.
 The LCD is [latex]\left( {x + 5} \right)\left( {x – 5} \right)[/latex]. Distribute this into the rational equation.
 It results in a product of two binomials on both sides of the equation.
It makes a lot of sense to perform the FOIL method. Does that ring a bell?
 I expanded both sides of the equation using FOIL. You should have a similar setup up to this point. Now combine like terms (the [latex]x[/latex]) in both sides of the equation.
 What’s wonderful about this is that the squared terms are exactly the same! They should cancel each other out. We could have bumped into a problem if their signs are opposite.
 Subtract both sides by [latex]{x^2}[/latex].
 The problem is reduced to a regular linear equation from a quadratic.
 To isolate the variable [latex]x[/latex] on the left side implies adding both sides by [latex]6x[/latex].
 Move all constant to the right.
 Add both sides by [latex]30[/latex].
 Finally, divide both sides by [latex]5[/latex] and we are done.
Example 8: Solve the rational equation below and make sure you check your answers for extraneous values.
This one looks a bit intimidating. But if we stick to the basics, like finding the LCD correctly, and multiplying it across the equation carefully, we should realize that we can control this “beast” quite easily.
Expressing each denominator as unique powers of terms
Multiply each unique terms with the highest power to obtain the LCD
 Factor out the denominators.
 Multiply both sides by the LCD obtained above.
Be careful now with your cancellations.
 You should end up with something like this when done right.
 Next step, distribute the constants into the parenthesis.
This is getting simpler in each step!
I would combine like terms on both sides also to simplify further.
 This is just a multistep equation with variables on both sides. Easy!
 To keep [latex]x[/latex] on the left side, subtract both sides by [latex]10x[/latex].
 Move all the pure numbers to the right side.
 Subtract both sides by [latex]15[/latex].
 A simple onestep equation.
 Divide both sides by [latex]5[/latex] to get the final answer. Again, don’t forget to check the value back into the original equation to verify.
Example 9: Solve the rational equation below and make sure you check your answers for extraneous values.
Let’s find the LCD for this problem, and use it to get rid of all the denominators.
Express each denominator as unique powers of terms.
Multiply each unique term with the highest power to determine the LCD.
 Factor out the denominators completely
 Distribute the LCD found above into the given rational equation to eliminate all the denominators.
 We reduced the problem into a very easy linear equation. That’s the “magic” of using LCD.
Multiply the constants into the parenthesis.
 Combine similar terms
 Keep the variable to the left side by subtracting [latex]x[/latex] on both sides.
 Keep constants to the right.
 Add both sides by [latex]8[/latex] to solve for [latex]x[/latex]. Done!
Example 10: Solve the rational equation below and make sure you check your answers for extraneous values.
Start by determining the LCD. Express each denominator as powers of unique terms. Then multiply together the expressions with the highest exponents for each unique term to get the required LCD.
So then we have,
 Factor out the denominators completely.
 Distribute the LCD found above into the rational equation to eliminate all the denominators.
 Critical Step : We are dealing with a quadratic equation here. Therefore keep everything (both variables and constants) on one side forcing the opposite side to equal zero.
 I can make the left side equal to zero by subtracting both sides by [latex]3x[/latex].
 At this point, it is clear that we have a quadratic equation to solve.
Always start with the simplest method before trying anything else. I will utilize the factoring method of the form [latex]x^2+bx+c=0[/latex] since the trinomial is easily factorable by inspection .
 The factors of [latex]{x^2} – 5x + 4 = \left( {x – 1} \right)\left( {x – 4} \right)[/latex]. You can check it by the FOIL method .
 Use the Zero Product Property to solve for [latex]x[/latex].
Set each factor equal to zero, then solve each simple onestep equation.
Again, always check the solved answers back into the original equations to make sure they are valid.
You may also be interested in these related math lessons or tutorials:
Adding and Subtracting Rational Expressions
Multiplying Rational Expressions
Solving Rational Inequalities
8.6 Solve Rational Equations
Learning objectives.
By the end of this section, you will be able to:
 Solve rational equations
 Solve a rational equation for a specific variable
Be Prepared 8.19
Before you get started, take this readiness quiz.
If you miss a problem, go back to the section listed and review the material.
Solve: 1 6 x + 1 2 = 1 3 . 1 6 x + 1 2 = 1 3 . If you missed this problem, review Example 2.48 .
Be Prepared 8.20
Solve: n 2 − 5 n − 36 = 0 . n 2 − 5 n − 36 = 0 . If you missed this problem, review Example 7.73 .
Be Prepared 8.21
Solve for y y in terms of x x : 5 x + 2 y = 10 5 x + 2 y = 10 for y . y . If you missed this problem, review Example 2.65 .
After defining the terms expression and equation early in Foundations , 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 rational equations.
The definition of a rational equation is similar to the definition of equation we used in Foundations .
Rational Equation
A rational equation is two rational expressions connected by an equal sign.
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.
Here is an example we did when we worked with linear equations:
We multiplied both sides by the LCD.  
Then we distributed.  
We simplified—and then we had an equation with no fractions.  
Finally, we solved that equation.  
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 .
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 8.59
How to solve equations with rational expressions.
Solve: 1 x + 1 3 = 5 6 . 1 x + 1 3 = 5 6 .
Try It 8.117
Solve: 1 y + 2 3 = 1 5 . 1 y + 2 3 = 1 5 .
Try It 8.118
Solve: 2 3 + 1 5 = 1 x . 2 3 + 1 5 = 1 x .
The steps of this method are shown below.
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 8.60
Solve: 1 − 5 y = − 6 y 2 . 1 − 5 y = − 6 y 2 .
Note any value of the variable that would make any denominator zero.  
Find the least common denominator of all denominators in the equation. The LCD is .  
Clear the fractions by multiplying both sides of the equation by the LCD.  
Distribute.  
Multiply.  
Solve the resulting equation. First write the quadratic equation in standard form.  
Factor.  
Use the Zero Product Property.  
Solve.  
Check.  
We did not get 0 as an algebraic solution.  
Try It 8.119
Solve: 1 − 2 a = 15 a 2 . 1 − 2 a = 15 a 2 .
Try It 8.120
Solve: 1 − 4 b = 12 b 2 . 1 − 4 b = 12 b 2 .
Example 8.61
Solve: 5 3 u − 2 = 3 2 u . 5 3 u − 2 = 3 2 u .
Note any value of the variable that would make any denominator zero.  
Find the least common denominator of all denominators in the equation. The LCD is .  
Clear the fractions by multiplying both sides of the equation by the LCD.  
Remove common factors.  
Simplify.  
Multiply.  
Solve the resulting equation.  
We did not get 0 or as algebraic solutions.  
Try It 8.121
Solve: 1 x − 1 = 2 3 x . 1 x − 1 = 2 3 x .
Try It 8.122
Solve: 3 5 n + 1 = 2 3 n . 3 5 n + 1 = 2 3 n .
When one of the denominators is a quadratic, remember to factor it first to find the LCD.
Example 8.62
Solve: 2 p + 2 + 4 p − 2 = p − 1 p 2 − 4 . 2 p + 2 + 4 p − 2 = p − 1 p 2 − 4 .
Note any value of the variable that would make any denominator zero.  
Find the least common denominator of all denominators in the equation. The LCD is .  
Clear the fractions by multiplying both sides of the equation by the LCD.  
Distribute.  
Remove common factors.  
Simplify.  
Distribute.  
Solve.  
We did not get as algebraic solutions.  
Try It 8.123
Solve: 2 x + 1 + 1 x − 1 = 1 x 2 − 1 . 2 x + 1 + 1 x − 1 = 1 x 2 − 1 .
Try It 8.124
Solve: 5 y + 3 + 2 y − 3 = 5 y 2 − 9 . 5 y + 3 + 2 y − 3 = 5 y 2 − 9 .
Example 8.63
Solve: 4 q − 4 − 3 q − 3 = 1 . 4 q − 4 − 3 q − 3 = 1 .
Note any value of the variable that would make any denominator zero.  
Find the least common denominator of all denominators in the equation. The LCD is .  
Clear the fractions by multiplying both sides of the equation by the LCD.  
Distribute.  
Remove common factors.  
Simplify.  
Simplify.  
Combine like terms.  
Solve. First write in standard form.  
Factor.  
Use the Zero Product Property.  
We did not get 4 or 3 as algebraic solutions.  
Try It 8.125
Solve: 2 x + 5 − 1 x − 1 = 1 . 2 x + 5 − 1 x − 1 = 1 .
Try It 8.126
Solve: 3 x + 8 − 2 x − 2 = 1 . 3 x + 8 − 2 x − 2 = 1 .
Example 8.64
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 .
Factor all the denominators, so we can note any value of the variable the would make any denominator zero.  
Find the least common denominator of all denominators in the equation. The LCD is .  
Clear the fractions.  
Distribute.  
Remove common factors.  
Simplify.  
Solve the resulting equation.  
Check. The only algebraic solution was 4, but we said that 4 would make a denominator equal to zero. The algebraic solution is an extraneous solution. There is no solution to this equation. 
Try It 8.127
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 8.128
Solve: y − 14 y 2 + 3 y − 4 = 2 y + 4 + 7 y − 1 . y − 14 y 2 + 3 y − 4 = 2 y + 4 + 7 y − 1 .
The equation we solved in Example 8.64 had only one algebraic solution, but it was an extraneous solution. That left us with no solution to the equation. Some equations have no solution.
Example 8.65
Solve: n 12 + n + 3 3 n = 1 n . n 12 + n + 3 3 n = 1 n .
Note any value of the variable that would make any denominator zero.  
Find the least common denominator of all denominators in the equation. The LCD is .  
Clear the fractions by multiplying both sides of the equation by the LCD.  
Distribute.  
Remove common factors.  
Simplify.  
Solve the resulting equation.  
Check.  
is an extraneous solution.  
Try It 8.129
Solve: x 18 + x + 6 9 x = 2 3 x . x 18 + x + 6 9 x = 2 3 x .
Try It 8.130
Solve: y + 5 5 y + y 15 = 1 y . y + 5 5 y + y 15 = 1 y .
Example 8.66
Solve: y y + 6 = 72 y 2 − 36 + 4 . y y + 6 = 72 y 2 − 36 + 4 .
Factor all the denominators, so we can note any value of the variable that would make any denominator zero.  
Find the least common denominator. The LCD is .  
Clear the fractions.  
Simplify.  
Simplify.  
Solve the resulting equation.  
Check.  
is an extraneous solution.  
Try It 8.131
Solve: x x + 4 = 32 x 2 − 16 + 5 . x x + 4 = 32 x 2 − 16 + 5 .
Try It 8.132
Solve: y y + 8 = 128 y 2 − 64 + 9 . y y + 8 = 128 y 2 − 64 + 9 .
Example 8.67
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 .
We will start by factoring all denominators, to make it easier to identify extraneous solutions and the LCD.  
Note any value of the variable that would make any denominator zero.  
Find the least common denominator.The LCD is  
Clear the fractions.  
Simplify.  
Simplify.  
Solve the resulting equation.  
Check.  
and are extraneous solutions. The equation has no solution. 
Try It 8.133
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 8.134
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 .
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.
We’ll start with a formula relating distance, rate, and time. We have used it many times before, but not usually in this form.
Example 8.68
Solve: D T = R for T . D T = R for T .
Note any value of the variable that would make any denominator zero.  
Clear the fractions by multiplying both sides of the equations by the LCD, .  
Simplify.  
Divide both sides by to isolate .  
Simplify. 
Try It 8.135
Solve: A L = W A L = W for L . L .
Try It 8.136
Solve: F A = M F A = M for A . A .
Example 8.69 uses the formula for slope that we used to get the pointslope form of an equation of a line.
Example 8.69
Solve: m = x − 2 y − 3 for y . m = x − 2 y − 3 for y .
Note any value of the variable that would make any denominator zero.  
Clear the fractions by multiplying both sides of the equations by the LCD, .  
Simplify.  
Isolate the term with .  
Divide both sides by to isolate .  
Simplify. 
Try It 8.137
Solve: y − 2 x + 1 = 2 3 y − 2 x + 1 = 2 3 for x . x .
Try It 8.138
Solve: x = y 1 − y x = y 1 − y for y . y .
Be sure to follow all the steps in Example 8.70 . It may look like a very simple formula, but we cannot solve it instantly for either denominator.
Example 8.70
Solve 1 c + 1 m = 1 for c . 1 c + 1 m = 1 for c .
Note any value of the variable that would make any denominator zero.  
Clear the fractions by multiplying both sides of the equations by the LCD, .  
Distribute.  
Simplify.  
Collect the terms with to the right.  
Factor the expression on the right.  
To isolate , divide both sides by .  
Simplify by removing common factors. 
Notice that even though we excluded c = 0 and m = 0 c = 0 and m = 0 from the original equation, we must also now state that m ≠ 1 m ≠ 1 .
Try It 8.139
Solve: 1 a + 1 b = c 1 a + 1 b = c for a . a .
Try It 8.140
Solve: 2 x + 1 3 = 1 y 2 x + 1 3 = 1 y for y . y .
Section 8.6 Exercises
Practice makes perfect.
In the following exercises, solve.
1 a + 2 5 = 1 2 1 a + 2 5 = 1 2
5 6 + 3 b = 1 3 5 6 + 3 b = 1 3
5 2 − 1 c = 3 4 5 2 − 1 c = 3 4
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 7 + 2 3 = 1 w 3 7 + 2 3 = 1 w
7 9 + 1 x = 2 3 7 9 + 1 x = 2 3
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
1 r + 3 = 4 2 r 1 r + 3 = 4 2 r
3 t − 6 = 1 t 3 t − 6 = 1 t
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 = 7 a 2 − 121 9 a + 11 + 6 a − 11 = 7 a 2 − 121
1 q + 4 − 2 q − 2 = 1 1 q + 4 − 2 q − 2 = 1
3 r + 10 − 4 r − 4 = 1 3 r + 10 − 4 r − 4 = 1
1 t + 7 − 5 t − 5 = 1 1 t + 7 − 5 t − 5 = 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 − 3 y 2 − 4 y − 5 = 1 y + 1 + 8 y − 5 y − 3 y 2 − 4 y − 5 = 1 y + 1 + 8 y − 5
z 16 + z + 2 4 z = 1 2 z z 16 + z + 2 4 z = 1 2 z
a 9 + a + 3 3 a = 1 a a 9 + a + 3 3 a = 1 a
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 = 8 n 2 − 4 + 3 n n + 2 = 8 n 2 − 4 + 3
p p + 7 = 98 p 2 − 49 + 8 p p + 7 = 98 p 2 − 49 + 8
q 3 q − 9 − 3 4 q + 12 q 3 q − 9 − 3 4 q + 12 = 7 q 2 + 6 q + 63 24 q 2 − 216 = 7 q 2 + 6 q + 63 24 q 2 − 216
r 3 r − 15 − 1 4 r + 20 r 3 r − 15 − 1 4 r + 20 = 3 r 2 + 17 r + 40 12 r 2 − 300 = 3 r 2 + 17 r + 40 12 r 2 − 300
s 2 s + 6 − 2 5 s + 5 s 2 s + 6 − 2 5 s + 5 = 5 s 2 − s − 18 10 s 2 + 40 s + 30 = 5 s 2 − s − 18 10 s 2 + 40 s + 30
t 6 t − 12 − 5 2 t + 10 t 6 t − 12 − 5 2 t + 10 = t 2 − 23 t + 70 12 t 2 + 36 t − 120 = t 2 − 23 t + 70 12 t 2 + 36 t − 120
C r = 2 π for r C r = 2 π for r
I r = P for r I r = P for r
V h = l w for h V h = l w for h
2 A b = h for b 2 A b = h for b
v + 3 w − 1 = 1 2 for w v + 3 w − 1 = 1 2 for w
x + 5 2 − y = 4 3 for y x + 5 2 − y = 4 3 for y
a = b + 3 c − 2 for c a = b + 3 c − 2 for c
m = n 2 − n for n m = n 2 − n for n
1 p + 2 q = 4 for p 1 p + 2 q = 4 for p
3 s + 1 t = 2 for s 3 s + 1 t = 2 for s
2 v + 1 5 = 3 w for w 2 v + 1 5 = 3 w for w
6 x + 2 3 = 1 y for y 6 x + 2 3 = 1 y for y
m + 3 n − 2 = 4 5 for n m + 3 n − 2 = 4 5 for n
E c = m 2 for c E c = m 2 for c
3 x − 5 y = 1 4 for y 3 x − 5 y = 1 4 for y
R T = W for T R T = W for T
r = s 3 − t for t r = s 3 − t for t
c = 2 a + b 5 for a c = 2 a + b 5 for a
Everyday Math
House Painting Alain can paint a house in 4 days. Spiro would take 7 days to paint the same house. Solve the equation 1 4 + 1 7 = 1 t 1 4 + 1 7 = 1 t for t to find the number of days it would take them to paint the house if they worked together.
Boating Ari can drive his boat 18 miles with the current in the same amount of time it takes to drive 10 miles against the current. If the speed of the boat is 7 knots, solve the equation 18 7 + c = 10 7 − c 18 7 + c = 10 7 − c for c to find the speed of the current.
Rational Equations
Correct Answer
Your answer.
x^{\msquare}  \log_{\msquare}  \sqrt{\square}  \nthroot[\msquare]{\square}  \le  \ge  \frac{\msquare}{\msquare}  \cdot  \div  x^{\circ}  \pi  
\left(\square\right)^{'}  \frac{d}{dx}  \frac{\partial}{\partial x}  \int  \int_{\msquare}^{\msquare}  \lim  \sum  \infty  \theta  (f\:\circ\:g)  H_{2}O 
▭\:\longdivision{▭}  \times \twostack{▭}{▭}  + \twostack{▭}{▭}   \twostack{▭}{▭}  \left(  \right)  \times  \square\frac{\square}{\square} 
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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 reallife situations and for finding answers to real problems. In particular, they are quite good for describing distancespeedtime 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.
 
Problem 

 
 4 = 2 • 2 8 =
LCM = 2 • 2 • 2 LCM = 8
 Find the least common multiple (LCM) of 4 and 8. Remember, to find the LCM, identify the greatest number of times each factor appears in each factorization. Here, 2 appears 3 times, so 2 • 2 • 2, or 8, will be the LCM.  

 The LCM of 4 and 8 is also the lowest common denominator for the two fractions.
Multiply both sides of the equation by the common denominator, 8, to keep the equation balanced and to eliminate the denominators.  

 Simplify and solve for .  

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


 
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:
 
Problem 

 

 Multiply the right side of the equation by to get a common denominator of 8. (Multiplying by  

 Since the denominators are the same, the numerators must be equal for the equation to be true. Solve for  


 
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.
 
Problem 

 

 Rewrite the expression using a common denominator.  

 Since the denominator for each expression is 3, the numerators must be equal.  

 Check the solution in the original 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.
 
Problem 

 

 Both fractions in the equation have a denominator of 3. Multiply both of the equation (not just the fractions!) by 3 to eliminate the denominators.  

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


 
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.
 
Problem 

 
5 is an excluded value because it makes the denominator  5 equal to 0.  Determine any values for that would make the denominator 0.  

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

 Check the solution in the original equation.  


 
Give the excluded values for . Do not solve.
A) B) 2 C) −2, 2 D) −2, 2, 4
Incorrect. Excluded values are those values of the variable which result in a 0 in denominator, not in the numerator. The correct answer is −2, 2.
B) 2 Incorrect. 2 is an excluded value, but −2 also results in a 0 in the denominator. The correct answer is −2, 2.
C) −2, 2 Correct. −2 and 2, when substituted into the equation, result in a 0 in the denominator. Since division by 0 is undefined, both of these values are excluded from the solution.
D) −2, 2, 4 Incorrect. While −2 and 2 are excluded, 4 is not excluded because it does not cause the denominator to be 0. The correct answer is −2, 2.

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

3 is an excluded value because it maxes – 3 and equal to 0.
−3 is an excluded value because it makes + 3 and equal to 0.  Determine any values for that would make the denominator 0. 

 Since – 9 or ( ‒ 3)( 3) is a common multiple of ‒ 3 and 3, you can multiply both sides of the equation by ( ‒ 3)( 3) to clear the denominator from the equation. Solve for 

 Check the solution in the original equation. 



Solve the equation 0 or 2
A) 2 B) no solution C) = 8 2 Incorrect. You probably found the common denominator correctly, but forgot to distribute when you were simplifying. You also forgot to check your solution or note the excluded values; ≠ 2 because it makes the expression on the right side undefined. Multiplying both sides by the common denominator gives . The correct answer is = 8.
B) no solution Incorrect. . The solution, 8, is not an excluded value. The correct answer is = 8.
C) = 8 Correct. Multiplying both sides of the equation by the common denominator gives . The correct answer is = 8.

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.
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3a + 2 − 6aa2 − 4 = 1a − 2. The students is incorrect. There are no solutions to this equation because first, you would find the LCD which is (a2) (a2). Next, you would simplify making 3 (a2)6aa2. Then, you would expand making 3a6a2. The next step is adding 6 to both sides. Soon, you get 4a/4 which equals 8/4.
After multiplying each side of the equation by the LCD and simplifying, the resulting equation is. C. What are the solutions to the equation? A. From least to greatest, the solutions are. x = 8. x = 4. Which of the following statements are true about the given rational equation? Check all of the boxes that apply.
Study with Quizlet and memorize flashcards containing terms like Find the value of the variable using these steps. 0.4x + 3.9 = 5.78 1. Subtraction property of equality 2. Division property of equality, Solve the linear equation for x. 4.8(6.3x  4.18) = 58.56 x = 2.6, A group of friends are going to see the newest action movie. The price of a ticket is $6.25. As a group, they spent $12 on ...
19 −. {4, 1} {. − 4} 1. Create your own worksheets like this one with Infinite Algebra 2. Free trial available at KutaSoftware.com.
Solving Rational Equations. A rational equation is a type of equation where it involves at least one rational expression, a fancy name for a fraction. The best approach to address this type of equation is to eliminate all the denominators using the idea of LCD (least common denominator). By doing so, the leftover equation to deal with is ...
Solve rational equations. Solve a rational equation for a specific variable. Solve: 16x + 12 = 13 1 6 x + 1 2 = 1 3. If you missed this problem, review Exercise 2.5.1. Solve: n2 − 5n − 36 = 0 n 2 − 5 n − 36 = 0. If you missed this problem, review Exercise 7.6.13. Solve for y in terms of x: 5x+2y=10 for y. If you missed this problem ...
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 ...
Step 2: Identify the restrictions. In this case, they are and . Step 3: Multiply both sides of the equation by the LCD. Distribute carefully and then simplify. Step 4: Solve the resulting equation. Here the result is a quadratic equation. Rewrite it in standard form, factor, and then set each factor equal to .
Rational equations intro. When we have an equation where the variable is in the denominator of a quotient, that's a rational equation. We can solve it by multiplying both sides by the denominator, but we have to look out for extraneous solutions in the process. Created by Sal Khan.
Rational equations. Find all solutions to the equation. Learn for free about math, art, computer programming, economics, physics, chemistry, biology, medicine, finance, history, and more. Khan Academy is a nonprofit with the mission of providing a free, worldclass education for anyone, anywhere.
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
Study with Quizlet and memorize flashcards containing terms like Solve. Remember to check for extraneous solutions. 1/x = 6/(5x) + 1, Solve. Remember to check for extraneous solutions. 1/(m²m) + 1/m = 5/(m²m), Solve. Remember to check for extraneous solutions. 1/(x2) + 1/(x²7x+10) = 6/(x2) and more.
7.7 Practice  Solving Rational Equations Solve the following equations for the given variable: 1) 3x − 1 2 − 1 x =0 3) x + 20 x − 4 = 5x x−4 − 2 5) x + 6 x − 3
Practice Rational Equations, receive helpful hints, take a quiz, improve your math skills. Solutions; Graphing; Calculators; Geometry; Practice; Notebook; Groups; Cheat Sheets ... Submit Assignment Start Over Back. Summary. Start Again Back. Prev Student Next Student. in 01:12:13 hours
Students solve rational equations, monitoring for the creation of extraneous solutions. Lesson Notes In the preceding lessons, students learned to add, subtract, multiply, and divide rational expressions so that in this lesson we can solve equations involving rational expressions (AREI.A.2). The skills developed in this lesson will be required to
Practice Solutions. pc_6.1_practice_solutions.pdf. File Size: 664 kb. File Type: pdf. Download File. Application solutions are available for purchase! click here.
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. Solve the equation .
A student solves the following equation for all possible values of x:His solution is as follows:Step 1: 8 (x  4) = 2 (x + 2)Step 2: 4 (x  4) = (x + 2)Step 3: 4x  16 = x + 2Step 4: 3x = 18Step 5: x = 6He determines that 6 is an extraneous solution because the difference of the numerators is 6, so the 6s cancel to 0.Which best describes the ...
Solving Rational Equations Assignment Solve each equation and check your work. 4 points each 1. ? 5 5 9 L 6 9 2. 5 6 T E 7 5 4 L 5 9 3. 9 à . L : à 1 4. 5 6 ì L 5 ; ì 5. < ë > 7 L 5 ë 6. 8 7 : ê > 8 ; 1 L 6 ê ê > 8. Student Name: Solving Rational Equations Assignment 7. 7 á ? 5
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No, the student did not check the solution to the derived equation in the original equation. No, a possible solution is a = 2, but it does not check in the original equation because it makes two of the denominators equal to zero. No, a = 2 is an extraneous solution, and there is no solution to the original rational equation.
Rational equations (advanced) Solve. The equation has no solutions. The equation has no solutions. Learn for free about math, art, computer programming, economics, physics, chemistry, biology, medicine, finance, history, and more. Khan Academy is a nonprofit with the mission of providing a free, worldclass education for anyone, anywhere.