2. A rational equation is an equation that contains one or more rational expressions. The time t in hours that it takes to travel d miles can be determined by using the equation t = , where r is the average rate of speed. This equation is a rational equation. d r

3. To solve a rational equation strategy:
Find the least common denominator (LCD)
Multiply each term of the equation by the LCD for all of the expressions in the equation.
This step eliminates the denominators (reduces) of the rational expression and results in an equation you can solve by using algebra.
Factor and solve.
Check for extraneous solutions.

4. Example 1: Solving Rational Equations18 x
Solve the equation x – = 3.
Find the LCD.
x(x) – (x) = 3(x) 18 x
Multiply each term by the LCD, x.
x2 – 18 = 3x
Simplify. Note that x ≠ 0.
x2 – 3x – 18 = 0
Write in standard form.
(x – 6)(x + 3) = 0
Factor.
x – 6 = 0 or x + 3 = 0
Apply the Zero Product Property.
x = 6 or x = –3
Solve for x.

5. An extraneous solution is a solution of an equation derived from an original equation that is not a solution of the original equation. When you solve a rational equation, it is possible to get extraneous solutions. These values should be eliminated from the solution set. Always check your solutions by substituting them into the original equation.

6. Example 2B: Extraneous Solutions
Solve each equation.
2x – 5
x – 8 x 2 11
x – 8 =
Find the LCD.
Multiply each term by the LCD, 2(x – 8).
2x – 5
x – 8
2(x – 8) (x – 8) = (x – 8) 11 x 2
Divide out common factors.
2x – 5
x – 8
2(x – 8) (x – 8) = (x – 8) 11 x 2
2(2x – 5) + x(x – 8) = 11(2)
Simplify. Note that x ≠ 8.
Use the Distributive Property.
4x – 10 + x2 – 8x = 22

7. Example 2B Continued
x2 – 4x – 32 = 0
Write in standard form.
(x – 8)(x + 4) = 0
Factor.
x – 8 = 0 or x + 4 = 0
Apply the Zero Product Property.
x = 8 or x = –4
Solve for x.
The solution x = 8 is extraneous because it makes the denominator of the original equation equal to 0. The only solution is x = –4.

8. Check It Out! Example 3
On a river, a kayaker travels 2 mi upstream and 2 mi downstream in a total of 5 h. In still water, the kayaker can travel at an average speed of 2 mi/h. Based on this information, what is the average speed of the current of this river? Round to the nearest tenth.

9. Understand the Problem
Check It Out! Example 3 Continued 1
Understand the Problem
The answer will be the average speed of the current.
List the important information:
The kayaker spent 5 hours kayaking.
She went 2 mi upstream and 2 mi downstream.
Her average speed in still water is 2 mi/h.
Remember t=d/r

10. Check It Out! Example 3 Continued2
Make a Plan
Let c represent the speed of the current. When the kayaker is going upstream, her speed is equal to her speed in still water minus c. When the kayaker is going downstream, her speed is equal to her speed in still water plus c.
Distance (mi)
Average Speed (mi/h)
Time (h) Up 2
2 – c
Down
2 + c
2 – c 2
2 + c 2
total time =
time up- stream +
time down- stream 5
2 – c 2
2 + c = +

11. Use the Distributive Property.
Solve 3
The LCD is (2 – c)(2 + c).
(2 + c)(2 – c)
= (2 + c)(2 – c)
5(2 + c)(2 – c)
2 – c 2
2 + c
Simplify. Note that x ≠ ±2.
5(2 + c)(2 – c) = 2(2 + c) + 2(2 – c)
Use the Distributive Property.
20 – 5c2 = 4 + 2c + 4 – 2c
20 – 5c2 = 8
Combine like terms.
–5c2 = –12
Solve for c.
c ≈ ± 1.5
The speed of the current cannot be negative. Therefore, the average speed of the current is about 1.5 mi/h.