Lesson 4-6 Rational
Equations and Partial
Fractions
Objective: To solve rational equations and inequalities
To decompose a fraction into partial fractions
Rational Equations
 Rational Equation – has 1 or more rational
expressions.
 Solve by multiplying each side by the LCD
Rational Equations
To solve a rational equation:
1. Find the LCM of the denominators.
2. Clear denominators by multiplying both sides of the
equation by the LCM.
3. Solve the resulting polynomial equation.
4. Check the solutions.
Examples: 1. Solve:
1
x 1 .

x 3 x 3
LCM = x – 3.
Find the LCM.
1=x+1
Multiply by LCM = (x – 3).
x=0
(0)
Solve for x.
1
1
 (0)
(0)
3
3
1
1

3 3
Check.Substitute 0.
Simplify. True.
1
2 . LCM = x(x – 1).

Find the LCM.
x x 1
1
 2 
  x( x  1)  
 x( x  1) Multiply by LCM.
 x
 x 1 
2. Solve:
x – 1 = 2x
x = –1
Simplify.
Solve.
Example: Solve:
x
6 .
 2
x  3 x  8 x  15
x2 – 8x + 15 = (x – 3)(x – 5) Factor.The LCM is (x – 3)(x – 5).
x
6
 2
x  3 x  8 x  15
Original Equation.
6
 x 


(
x

3
)(
x

5
)

 x  3
 x 2  8 x  15  ( x  3)( x  5)




x(x – 5) = – 6
x2 – 5x + 6 = 0
(x – 2)(x – 3) = 0
x = 2 or x = 3
Polynomial Equation.
Simplify.
Factor.
Check. x = 2 is a
solution.
Check. x = 3 is not a solution
since both sides would be
undefined.
Decomposing a fraction into
Partial Fractions.
Sometimes we need more tools to
help with rational expressions…
We will learn to perform a process
known as partial fraction
decomposition…
To find partial fractions for an
expression, we need to reverse the
process of adding fractions.
To find the partial fractions, we start with
5x 1

( x  2)( x  1)
A
B

x 2 x 1
The expressions are equal for all values of x so
we have an identity.
The identity will be important for finding the
values of A and B.
To find the partial fractions, we start with
5x 1

( x  2)( x  1)
A
B

x 2 x 1
Multiply by the LCD
A
5x 1

 ( x  2)( x  1) 
( x  2)( x  1) 
( x  2)( x  1)
x2
B
( x  2)( x  1) 
x 1
So,
5x 1  A( x  1)  B( x  2)
If we understand the cancelling, we can in future go
straight to this line from the 1st line.
5 x  1  A( x  1)  B( x  2)
This is where the identity is important.
The expressions are equal for all values of x, so I can
choose to let x = 2.
Why should I choose x = 2 ?
ANS: x = 2 means the coefficient of B is zero, so
B disappears and we can solve for A.
5 x  1  A( x  1)  B( x  2)
This is where the identity is important.
The expressions are equal for all values of x, so I can
choose to let x = 2.
x2 
5( 2)  1  A( 2  1)  B( 2  2)
9  3A  A  3
What value would you substitute next ?
ANS: x =  1 so that the first term becomes 0.
5 x  1  A( x  1)  B( x  2)
This is where the identity is important.
The expressions are equal for all values of x, so I can
choose to let x = 2.
x2 
5( 2)  1  A( 2  1)  B( 2  2)
9  3A  A  3
x  1  5( 1)  1  A( 1  1)  B( 1  2)
 6  3B  B  2
So,
5x 1

( x  2)( x  1)
A  B
x 2 x 1
Example 2 Express the following as 2 partial
fractions.
Solution: Let
Multiply by
1
( x  3)( x  1)
1
A
B


( x  3)( x  1)
x  3 x 1
( x  3)( x  1) :
1  A( x  1)  B( x  3)
A
x  3  1  A( 2) 
x 1 
1  B ( 2 ) 
1
2
B   12
Decomposing Fractions
6x  2
 Decompose 2
into partial fractions
x  3 x  10
Rational Inequalities
 To solve rational inequalities:
 Find the zeros and mark on a number line.
 Find any exclusions (restrictions) and mark on
a number line.
 Test a value on each interval.
Rational Inequalities
 Solve 1  2  1
b0
3b 5b
1
2
 Set to 0
LCD=15b
 1  0
3b 5b
 Find the zeros
5  6  15b  0
1
b
15
Rational Inequalities
-1/15 0
Test b<-1/15
try b=-1
Test -1/15<b<0
try b=-1/30
Test b>0
try b=1
1 2
  1
3 5
1
2


1
3
5
30 30
1 2
True
 1
3 5
True
False
1
b , b0
15
Rational Inequalities
 So
1
b , b0
15
Rational Inequalities
 Solve
( x  2)( x  3)
0
2
( x  1)( x  1)
Sources
 www.collinsville.k12.ok.us/webpages/documents/HSSparksR/PC4.6.ppt, Jan 11, 2014.