SOLVING RATIONAL EQUATIONS
 A rational equation is solved by rearranging to make the
equation equal to zero.
 We can then find the common denominator and simplify
the numerator.
 Set the numerator equal to zero to solve. A rational
equation is zero when the numerator is zero.

1.Solve the following equation:
This equation is so simple that I can solve it just by looking at it: since I have two-thirds equal to
x-thirds, clearly x = 2. The reason this was so easy to solve is that the denominators were the
same, so all I had to do was solve the numerators.
x=2

2.Solve the following equation:
To solve this, I can convert to a
common denominator of 15:
Now I can compare the
numerators:
x–1=6
x=7
Note, however, that I could also
have solved this by multiplying
through on both sides by the common denominator:
x – 1 = 2(3)
x–1=6
x=7

3.Solve the following equation:
First, check for the restrictions: they tell me that x cannot equal zero or –2 (since these values
would cause division by zero). I'll re-check at the end, to make sure any solutions I find are
"valid".
There are two ways to proceed with solving this equation. I could convert everything to the
common denominator of 5x(x + 2) and then compare the numerators:
At this point, the denominators are the same. All I really need to do now is solve the numerators:
15x – (5x + 10) = x + 2
10x – 10 = x + 2
9x = 12
x = 12/9 = 4/3
Since x = 4/3 won't cause any division-by-zero problems in the fractions in the original equation,
then this solution is valid.
Copyright © Elizabeth Stapel 2003-2011 All Rights Reserved
x = 4/3
Another method is to find the common denominator but, rather than converting everything to that
denominator, I'll take advantage of the fact that I have an equation here, and multiply through on both
sides by that common denominator. This will get rid of the denominators:
3(5x) – 1(5(x + 2)) = 1(x + 2)
15x – 5x – 10 = x + 2
10x – 10 = x + 2
9x = 12
x = 12/9 = 4/3

4.Solve the following equation:
The common denominator here will be x(x – 2), and x cannot be zero or 2. I can solve this
equation by multiplying through on both sides of the equation by this denominator:
10 + 4(x – 2) = 5(x)
10 + 4x – 8 = 5x
4x + 2 = 5x
2=x
Or I can solve by converting to the common denominator and then solving the numerators:
10 + (4x – 8) = 5x
10 + 4x – 8 = 5x
4x + 2 = 5x
2=x
Either way, the answer I get is:
x = 2 Copyright © Elizabeth Stapel 2003-2011 All Rights Reserved
However, I need to check this solution with the original equation. This value would cause division
by zero in the original equation! Since the only possible solution causes division by zero, then this
equation really has no solution.
no solution

5.Solve the following equation:
First I'll need to factor that quadratic, so I can tell what factors I'll have in my common
denominator.
x2 – 6x + 8 = (x – 4)(x – 2)
That worked out nicely: the factors of the quadratic are duplicates of the other denominators.
(This often happens for these problems.) So the common denominator will be (x – 4)(x – 2).,
and I'll need to remember (at the end) that x cannot be 2 or 4.
I can convert everything to the common denominator and then solve the numerators:
(x2 – 4x) + (x – 2) = 2
x2 – 4x + x – 2 = 2
x2 – 3x – 4 = 0
(x – 4)(x + 1) = 0
x = 4 or x = –1
Checking these solutions against the denominators of the original equation, I see that " x
would cause division by zero, so I throw that solution out. Then the answer is:
= 4"
x = –1

6.Solve the following equation:
There is only one fraction, so the common denominator is just x, and the solution cannot be
0. To solve, I can start by multiplying through on both sides by x:
x=
x2 + x = 72
x2 + x – 72 = 0
(x + 9)(x – 8) = 0
x = –9 or x = 8
...or I can convert to the common
denominator and solve the
numerators:
ADVERTISEMENT
x2 + x = 72
x2 + x – 72 = 0
(x + 9)(x – 8) = 0
x = –9 or x = 8
Either way, the solution is x = –9 or x= 8. Since neither solution causes a division-by-zero
problem in the original equation, both solutions are valid.
x = –9 or x = 8

7.Solve the following equation:
First, I note that I cannot have x = –1 or x = –4. Then I notice that this equation is a proportion:
the equation is of the form "one fraction equals another fraction". So all I need to do here is crossmultiply.
10(4(x + 1)) - 15(x + 4)
40x + 40 - 15x + 60
25x + 40 = 60
25x = 20
x = 20/25 = 4/5
Since this solution won't cause any division-by-zero problems, it is valid:
x = 4/5
WHAT DID I LEARN TODAY?
EXIT CARD.
ASSESSMENT FOR YOUR LEARNING
21. It takes Frank 2 hours longer than Jane to carpet a certain type of room. Together they can carpet that
type of room in 1
a.
b.
c. 3 hours
1 hours
d. 5 hours
3 hours
____ 22. Solve
a.
hours. How long would it take for Frank to do the job alone?
for x.
c. x = 2
x=
b. x = 2, 2
d. x = 2
____ 23. How many solutions does the equation
have?
a. 0
b. 1
c. 2
d. 3
____ 24. What value(s) of x makes
a.
b.
true?
c.
0,
d.
0,
____ 25. If x
3 which of the following is equivalent to
a.
8+
c.
b. 5x
8 + 12
d. 5x
8(x 3) + 12
____ 26. Two pipes are used to fill the water for the swimming pool in the park. Pipe A can fill the pool alone in 5
hours, Pipe B can fill the pool alone in 4 hours. If Pipe A is turned on and 30 minutes later Pipe B is also
turned on, how long will it take to fill the pool?
a.
c.
2 hours
hour
b.
d.
2 hours
____ 27. Solve
a. x = 2, 1
b. x = 1
hour
for x.
c. x = 1
d. x = 0
____ 28. An airplane has a speed of 400 km/h with no wind. The airplane flies 2140 km with the wind. The
airplane can only fly 1860 km against the wind in the same time. If w equals the speed of the wind, which
equation would be used to find w?
a.
c.
b.
d.
____ 29. The harmonic mean of two numbers, a and b, is a number m such that the reciprocal of m is the average of
the reciprocals of a and b. Which of the following is a formula for the harmonic mean of a and b?
a.
c.
m
m
b.
m
d.
m
____ 30. Brad has just enough money to rent a paddleboat for 2
hours. How far out into the lake can he paddle
and return on time if he paddles out at 3km/h and back at 2 km/h?
a. 6 km
c. 6.25 km
b. 3 km
d. 1.5 km
21.
22.
23.
24.
25.
26.
27.
28.
29.
30.
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
D
C
C
D
D
B
D
C
A
B