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Solving and Applications of Rational Equations

Key Concepts

  • Solve a rational equation
  • Solve a work-rate problem
  • Identify an extraneous solution
  • Solve problems with extraneous solutions
  • Solve a rate problem

Rational Equations 

Solve a Rational Equation 

What is the solution to each rational equation?  

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

1. 1/x+4 = 2  

(x + 4) (1/x+4) = 2(x+4)

1 = 2x + 8  

x = -7/2. 

The solution is x = -7/2.  

2. 1/x −3 = 5  

(x – 3) (1/x−3) = 5(x−3)

1 = 5x – 15  

x = 16/5  

The solution is x = 16/5.  

Solve a Work-Rate Problem 

Arthur and Cheyenne can paint a wall in 6 hours when working together. Cheyenne works twice as fast as Arthur. How long would it take Cheyenne to paint the wall if she were working alone? 

rational equations

Step 1:  

Determine the work rates of Arthur and Cheyenne.  

Let x represent the number of hours Arthur needs to paint the wall himself.  

Arthur can paint 1 wall in x hours, or 1/x of a wall in 1 hour.  

Cheyenne is twice as fast, so Cheyenne paints 2/x  of a wall in 1 hour.  

Together they paint 1 wall in 6 hours, or 1/6 of a wall in 1 hour.  

Step 2:  

Write the equation for their rates working together.  

1/x + 2/x = 1/6x

6x (1/x + 2/x)

= 6x (1/6)  

6 + 12 = x  

18 = x  

It took Arthur 18 hours to paint the wall alone. Since Cheyenne worked twice as fast as Arthur, it took her 9 hours to paint the wall alone.  

Extraneous Solutions of Rational Equations 

Identify an Extraneous Solution 

What is the solution of the equation 𝟏/𝒙 − 𝟓 + 𝒙/𝒙 − 𝟑 = 𝟐/𝒙𝟐 − 𝟖𝒙 + 𝟏𝟓 ? 

Step 1: Multiply each side of the equation by the common denominator, (x – 5) (x – 3).  

(x – 5) (x – 3) (1/x − 5 + x/x − 3) = 2(x − 5) (x − 5)/x2 − 8x 

Step 2: Continue to simplify.  

(x − 5)(x− 3)/x − 5 + x(x − 5)(x − 3)/x − 3 = 2(x − 5) (x − 3)/x2− 8x + 15 

Step 3: Divide out common factors in the numerator and the denominator. 

Step 3:

You can divide out common factors under the assumption that (x – 5)/(x – 5) = 1 and  

(x – 3)/ (x – 3) = 1. This is only true if x ≠ 5 or 3.  

Step 4: Solve the equation.  

(x – 3) + x (x – 5) = 2  

x – 3 + x2 −5x = 2 

x2 – 4x – 3 = 2  

x2 – 4x – 5 = 0 

(x – 5) (x + 1) = 0  

The solution x = 5 is an extraneous solution because it makes the value of a denominator in the original equation equal to 0.  

The solution of the equation

1/x − 5 + x/x −3 = 2/x2 −8x +15   is -1.  

Confirm with a graph.  

Consider the graphs of 1/x − 5 + x/x −3 and  2x2 −8x +15

graph

Note that each graph has a vertical asymptote at x = 3 and x = 5.  

Therefore, neither graph has a value at x = 5. The graphs only intersect at one point at x = -1.  

Solve Problems with Extraneous Solutions 

What are the solutions to the following equations?  

1. 5x/x−2 = 7 + 10/x−2

5x/x−2 = 7 + 10/x−2 ——————–Write the original equation 

 (x – 2) (5x/x−2) = (7+ 10/x −2x−2)=(7+ 10 (x – 2) ——–Multiply by the LCD.  

5x = 7(x – 2) + 10 —————————–Distributive Property  

5x = 7x – 14 + 10 —————————–Distributive Property  

-2x = -4 —————————————–Collect terms and simplify.  

x = 2 ———————————————Solve for x.  

Check the solution in the original equation. The value 2 is an extraneous solution because it would cause the denominator in the original equation to be equal to 0. This equation has no solution.  

2.  3/x−3 =  x/x−3 – x/4  

3/x−3 =  x/x−3 – x/4  ——————–Write the original equation 

 (4)(x – 3)(3/x-3) = (x/x-3 – x/4) (4)(x – 3) —————Multiply by LCD.  

4(3) = 4(x) – x(x – 3) ——————Distributive Property  

12 = 4x – x2 + 3x ———————–Simplify.  

x2 – 7x + 12 = 0 ————————Write in standard form.  

(x – 3)(x – 4) = 0 ———————–Factor.  

x – 3 = 0 or x – 4 = 0 ——————Solve using the zero product property.  

x = 3 or x = 4 —————————Solve for x.  

Check the solutions in the original equation. The value 3 is an extraneous solution because it would cause the denominator of the original equation to be equal to zero. The only solution to the equation is x = 4.  

Application  

Solve a Rate Problem  

Paddling with the current in a river, Jake traveled 16 miles. Even though he paddled upstream for an hour longer than the amount of time he paddled downstream, Jake could only travel 6 miles against the current. In still water, Jake paddles at a rate of 5 mph. What is the speed of the current in the river?  

Solve a Rate Problem  

Solution:  

Step 1 – Formulate:  

Let be the rate of the river’s current.  

Recall that distance = rate × time so time = distance/rate. 

Formulate

Step 2 – Compute:  

Solve the equation for c.   

Compute

Step 3 – Interpret: 

The solution c = -25 is extraneous because the speed of the current cannot be negative. 

The speed of the current is 3 mph.   

Questions  

Question 1 

It takes 12 hours to fill a pool with two pipes, where the water in one pipe flows three times as fast as the other pipe. How long will it take the slower pipe to fill the pool by itself?  

Solution: 

Let’s say pipe 1 fills the pool in x hours. So, it can fill 1/x of a pool in 1 hour.  

Since water in pipe 2 flows three times as fast as pipe 1, pipe 2 can fill 3/x of a pool in 1 hour.  

Together they fill 1 pool in 12 hours. So together they fill 1/12 of the pool in 1 hour.  

1/x + 3/x = 1/12  

12x(1/x + 3/x) = 12x(1/12)  

48 = x  

So, the slower pipe, i.e., pipe 1 fills the pool in 48 hours.  

Question 2 

What is the solution to the equation?  

1/x + 2 + 1/x −2 = 4(x+ 2) (x −2)  

Solution:  

Solution:  

Question 3 

Three people are planting tomatoes in a community garden. Marta takes 50 minutes to plant the garden alone, Benita takes x minutes and Tyler takes x + 15 minutes. If the three of them takes 20 minutes to finish the garden, how long would it have taken Tyler alone?  

Solution:  

Marta takes 50 minutes to plant the garden alone. So, she plants 1/50 of the garden in 1 minute. 

Benita takes x minutes to plant the garden alone. So, she plants 1/x of the garden in 1 minute.  

Tyler takes x + 15 minutes to plant the garden alone. So, she plants 1/(x + 15) of the garden in 1 minute. 

Together they take 20 minutes to plant the garden. So, they plant 1/20 of the garden together in 1 minute.  

1/50 + 1/x + 1/(x + 15) = 1/20  

1/x + 1/(x + 15) = 1/20 – 1/50 = 3/100  

[x(x + 15)] ( 1/x + 1/(x + 15) ) = 3x(x + 15)/100  

x + 15 + x = (3x2 + 45x)/100  

100(2x + 15) = 3x2 + 45x 

 3x2 – 155x – 1500 = 0  

3x2 – 180x + 25x – 1500 = 0  

(3x + 25)(x – 60) = 0

So, x = -25/3 is an extraneous solution because time can’t be negative.  

The solution is x = 60.  

Tyler will take 75 minutes to plant the garden alone.  

Key Concepts Covered  

Key Concepts Covered

Exercise:

Solve the following:

  1. 5/y— 5/6 = 5/3
  2. 7/y+2/9 = -(5/3y)
  3. (6/(2-3)) +9 =-3/(z-3)
  4. 2/(y+2)—1/2 = —-3/4
  5. (22+ 6)/3z+ 1/6 = 1/2
  6. (y-3)/7 = (4y + 12)/7
  7. 3/(yt+2) – 1/y = 1/Sy
  8. 10/(z+4) = 15/(4(2+1))

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