Solution

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:

Example Solve: 1

.5 2 4

x x

20 201

5 2 4

x x

20 205 4

201

2

x x

4 10 5x x

6 5x

5

6x

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.

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

Checking Answers

Since a variable expression could represent 0, multiplying both sides of an equation by a variable expression does not always produce an equivalent equation. Thus checking each solution in the original equation is essential.

Solution – Algebraic Note that x cannot equal 0. The LCM is 15x.

Example Solve: 1 1 4

.3 15x x

1 1 4

3 1515 15

x xx x

515x1

3x 15 x

1

x 15

4

15x

5 15 4x

20 4x5x

Solution – Graphically

The solution is x = 5.

SolutionNote that x cannot equal 1 or 1. Multiply both sides of the equation by the LCD = (x - 1)(x + 1) .

Example Solve:

2

2 5 4.

1 1 1x x x

( 1)( 1) ( 1)( 12 5 4

1 1 ( 1)( 1)

)x x x x

x 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.

SolutionNote that x cannot equal 3 or 3. We multiply both sides of the equation by the LCM = (x - 3)(x + 3) .

Example Solve:

2

5 3 2.

9 3 3y y y

5 3 2

( 3)( 3( 3)( 3) ( 3 ( 3

) 3 3) )y

y y y yy y y

( 3y )( 3y )5

( 3y )( 3y )

( 3y

)( 3)3

3

y

y

( 3)( 3y y

)2

3y

5 3( 3) 2( 3)y y

3 9 25 6y y

5 15y

20y

Solution

Example Find all values of a for which f(a) = 7 12

Let ( ) .f x xx

12(7)a

aa a

a aa 12

a 7a

2 12 7a a

2 7 12 0a a

( 3)( 4) 0a a

12( ) thusf a a

a 12

7. Note: 0a aa

3 or 4a a