Solving Rational Equations and Applications
Learning Objective(s)
· Solve rational equations.
· Check for extraneous solutions.
· Solve application problems involving rational equations.
Introduction
Equations that contain rational expressions are called rational equations. 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. In particular, they are quite good for describing distance-speed-time relationships and for modeling work problems that involve more than one person.
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.
Example | |||
Problem | Solve the equation | ||
4 = 2 • 2 8 = 2 • 2 • 2 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 x. | ||
| Check the solution by substituting 9 for x in the original equation. | ||
Answer | | ||
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:
Example | |||
Problem | Solve the equation | ||
| Multiply the right side of the equation by | ||
| Since the denominators are the same, the numerators must be equal for the equation to be true. Solve for x. | ||
Answer | | ||
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.
Example | |||
Problem | Solve the equation | ||
| 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. | ||
Answer | | ||
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.
Example | |||
Problem | Solve the equation | ||
| Both fractions in the equation have a denominator of 3. Multiply both sides 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 x. | ||
Answer | | ||
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.
Example | |||
Problem | Solve the equation | ||
5 is an excluded value because it makes the denominator x - 5 equal to 0. | Determine any values for x 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 x. | ||
| Check the solution in the original equation. | ||
Answer | | ||
Give the excluded values for . Do not solve.
A)
B) 2
C) −2, 2
D) −2, 2, 4
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Let’s look at an example with a more complicated denominator.
Example | ||
Problem | Solve the equation | |
3 is an excluded value because it maxes x – 3 and −3 is an excluded value because it makes x + 3 and | Determine any values for x that would make the denominator 0. | |
| Since x2 – 9 or (x ‒ 3)(x + 3) is a common multiple of x ‒ 3 and x + 3, you can multiply both sides of the equation by (x ‒ 3)(x + 3) to clear the denominator from the equation. Solve for x. | |
| Check the solution in the original equation. | |
Answer | |
Solve the equation , m
0 or 2
A) m = 2
B) no solution
C) m = 8
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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.
Example | |||||||||
Problem | Solve the equation | ||||||||
−4 is an excluded value because it makes m + 4 equal to 0. | Determine any values for m that would make the denominator 0. | ||||||||
16 = m2
m = 4, −4 | 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. | ||||||||
Answer | m = 4 | ||||||||
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.
W = rt
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.
Example | ||
Problem | 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: Francis: | Think about how many bulbs each person can plant in one hour. This is their planting rate. | |
Myra and Francis together: | Combine their hourly rates to determine the rate they work together. | |
| Use one of the work formulas to write a rational equation, for example | |
t = | Solve the equation by multiplying both sides by the common denominator, then isolating t. | |
Answer | 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.
Example | ||
Problem | 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 3x = time it takes John to complete the job | Choose variables to represent the unknowns. Since it takes John 3 times as long as Joe to paint the house, his time is represented as 3x. | |
Joe’s rate: John’s rate: | The work is painting 1 house or 1. Write an expression to represent each person’s rate using the formula
| |
combined rate: | Their combined rate is the sum of their individual rates. Use this rate to write a new equation using the formula W = rt. | |
| The problem states that it takes them 24 hours together to paint a house, so if you multiply their combined hourly rate | |
| Now solve the equation for x. (Remember that x represents the number of hours it will take Joe to finish the job.) | |
| Check the solutions in the original equation. | |
The solution checks. Since x = 32, it takes Joe 32 hours to paint the house by himself. John’s time is 3x, so it would take him 96 hours to do the same amount of work. | ||
Answer | It takes 32 hours for Joe to paint the house by himself and 96 hours for John the paint the house himself. | |
As shown above, many work problems can be represented by the equation , 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.
Mari and Liam can each wash a car and vacuum its interior in 2 hours. Zach needs 3 hours to do this same job alone. If Zach, Liam, and Mari work together, how long will it take them to clean a car?
A) 20 minutes
B) 45 minutes
C) 1.2 hours
D) 1 hour
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Summary
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.