Algebraic Fractions and Algebraic Fractions and Rational EquationsRational Equations

In this discussion, we will look at examples of simplifying Algebraic Fractions using the 4 rules of fractions.

AdditionAddition

bd

bcad

d

c

b

a

xx

2

31Simplify

)2(

)(3)2(1

xx

xx

)2(

32

xx

xx

)2(

22

xx

x

AdditionAddition

bd

bcad

d

c

b

a

y

x

x 2

3Simplify

)2(

))(()2(3

yx

xxy

xy

xy

2

6 2

SubtractionSubtraction

bd

bcad

d

c

b

a

p

t

p 2

2Simplify

€

(2)(2p) − (p)(t)

(p)(2p)

22

4

p

ptp

22

)4(

p

tp

p

t

2

)4(

SubtractionSubtraction

bd

bcad

d

c

b

a

4

33

2

12

xxSimplify

)4)(2(

)33(2)12(4

xx

8

6648

xx

8

22 x

8

)1(2

x4

4

1x

MultiplicationMultiplication

bd

ac

d

c

b

a

5

5

3

x

xxSimplify

)5(3

)5)((

x

xx

)5(3

52

x

xx

MultiplicationMultiplication

bd

ac

d

c

b

a

62

5

15

3

x

xSimplify

)62)(15(

)5)(3(

x

x

)3)(2)(15(

)5)(3(

x

xFactoring (2x-6)Factoring (2x-6)

3

)2)(3(

1

6

1

DivisionDivision

bc

ad

d

c

b

a

Simplify

14

4

7

2 yx

y

x

4

14

7

2

)4)(7(

)14)(2(

y

x

2

2

y

x

EquationsEquationsOnce you know how to simplify algebraic fractions, youCan solve equations containing them

For example

743

2xx

35

2

2

1

xx

Click an equation to see it solved

743

2xx

Solve

7)4)(3(

))(3()2)(4(

xx

712

38

xx

712

5x

845 x5

84 x

Solve 35

2

2

1

xx

3)5)(2(

)2)(2()1)(5(

xx

310

4255

xx 310

97

x

3097 x

217 x

3x

The equation in the previous example is called a rational equation.

and are rational equations.

3

21

xx xx

x

x

x

2

1

1

33

2

4. Check the solutions.

3. Solve the resulting polynomial equation.

2. Clear denominators by multiplying both sides of the equation by the LCM.

1. Find the LCM of the denominators.

To solve a rational equation:

Examples: 1. Solve: .3

1

3

1

x

x

xMultiply by LCM = (x – 3).

Solve for x.

1 = x + 1

x = 0

Check.

Simplify.

3

1

3

1

(0)

(0)

(0)

3

1

3

1

True.

2. Solve: .1

21

xx

x – 1 = 2x

Find the LCM.LCM = x(x – 1).

Multiply by LCM.

Simplify.

x = –1 Solve.

1

2

1

xx)1( )1( xxxx

After clearing denominators, a solution of the polynomial equation may make a denominator of the rational equation zero.

Since x2 – 1 = (x – 1)(x + 1),

Since – 1 makes both denominators zero, the rational equation has no solutions.

Example: Solve: .1

1

1

132

xx

x

2x = – 2 x = – 1

3x + 1 = x – 1

Check.

It is critical to check all solutions.

In this case, the value is not a solution of the rational equation.

LCM = (x – 1)(x + 1).

1

1

1

132 xx

x)1)(1( )1)(1( xxxx

16

Example: Solve: .158

6

3 2

xxx

x

Factor.

Polynomial Equation.

Simplify.

Factor.

The LCM is (x – 3)(x – 5).x2 – 8x + 15 = (x – 3)(x – 5)

x(x – 5) = – 6

x2 – 5x + 6 = 0

(x – 2)(x – 3) = 0

x = 2 or x = 3Check. x = 2 is a solution.

Check. x = 3 is not a solution since both sides would be undefined.

158

6

3 2

xxx

xOriginal Equation.

158

6

3 2

xxx

x)5)(3( )5)(3( xxxx