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First, find the least common denominator (LCD) of all fractions present.

Linear Equations With Fractions: Solving algebraically

Example: Solve .41

23

)1(4

xxx

Here the LCD is 4x(x – 1).

Then multiply each term on both sides by the LCD.

41

1)1(4

23

1)1(4

)1(41)1(4

xx

xxx

xxxx

x2 + 6(x – 1) = x (x – 1)

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x2 + 6(x – 1) = x (x – 1)

Linear Equations With Fractions: Solving algebraically

Slide 2

Next, solve the resulting equation. x2 + 6x – 6 = x2 – x

6x – 6 = - x

7x = 6,76x

Last, substitute the proposed solution(s) into the original equation to ensure it does not result in a denominator of zero in any of the fractions.

41

76

2

3

176

4

76

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Linear Equations With Fractions: Solving algebraically

Slide 3

Here no denominator of a fraction became zero, so the

solution set is (If a denominator had become zero,

6/7 would have been rejected as a solution.)

.76

.653

32

12

xx

xxx

Try to solve

The equation has no solution since when checking, x = 3 makes a denominator equal to zero.

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Linear Equations With Fractions: Solving algebraically