Martin-Gay, Beginning Algebra, 5ed 22

Martin-Gay, Beginning Algebra, 5ed 33

Solve the following rational equation.EXAMPLE1

5 2 4x x

Because no variable appears in the denominator, no restrictions exist. The LCM of 5, 2, and 4 is 20, so we multiply both sides by 20:

20 20 15 2 4x x

20 205 4

20 12

x x

Using the multiplication principle to multiply both sides by the LCM. Parentheses are important!

Using the distributive law. Be sure to multiply EACH term by the LCM.

Martin-Gay, Beginning Algebra, 5ed 44

4 10 5x x

6 5x

56

x

Simplifying and solving for x. If fractions remain, we have either made a mistake or have not used the LCM of the denominators.

We should check our solution, but no need to since we never make a mistake

Martin-Gay, Beginning Algebra, 5ed 55

SolveEXAMPLE

Restrictions: x cannot equal 0 LCD = 6x

671

35

x

671

35

x

xx

x 6671

356

xx 7610

x10

Martin-Gay, Beginning Algebra, 5ed 66

Restrictions: x cannot equal 3 LCD = 5(p + 3)

5( 3)p 5( 3)p

Martin-Gay, Beginning Algebra, 5ed 77

Restrictions: x cannot equal 0 LCD = x

Martin-Gay, Beginning Algebra, 5ed 88

Restrictions: x cannot equal 6 LCD = x 6

Martin-Gay, Beginning Algebra, 5ed 99

EXAMPLE Solve

2

2 5 41 1 1x x x

Restrictions: x cannot equal 1 or 1 LCD = (x 1)(x + 1)

( 1)( 1) ( 1)( 12 5 41 1 ( 1)( 1

))

x x x xx x x x

2( 1) 5( 1) 4x x

2 2 5 5 4x x

3 7 4x

3 3x

1x Because of the restriction above, 1 must be rejected as a solution. This equation has no solution.

Martin-Gay, Beginning Algebra, 5ed 1010

= (x 2)(x + 2) Restrictions: x cannot equal 2 or 2 LCD = (x 2)(x + 2)

Martin-Gay, Beginning Algebra, 5ed 1111

Restrictions: x cannot equal 1 or 3

Martin-Gay, Beginning Algebra, 5ed 1212