SOLVE THE EQUATION ELIMINATING ANY EXTRANEOUS SOLUTIONS

Example 1 :

Solve for x :

2x/(x - 1) + 1/(x - 3)  =   2/x² - 4x + 3

Solution :

2x/(x - 1) + 1/(x - 3)  =   2/x² - 4x + 3

Factoring the quadratic equation x² - 4x + 3, we get factors as (x - 1) (x - 3).

2x/(x - 1) + 1/(x - 3)  =   2/(x - 1) (x - 3)  --(1)

2x(x - 3) + 1(x - 1)  =  2

2x² - 6x + x - 1  =  2

2x² - 5x - 1 - 2  =  0

2x² - 5x - 3  =  0

(2x + 1) (x - 3)  =  0

To check whether above values are solution or not, we apply the values in (1).

If x  =  -1/2, we get the defined value.

If x  =  3, we get undefined value. So, only x  =  -1/2 is a solution.

Example 2 :

Solve for x :

(x - 3)/x  +  3/(x + 2)  +  6/(x² + 2x)  =  0

Solution :

(x - 3)/x  +  3/(x + 2)  +  6/(x² + 2x)  =  0

(x - 3)/x  +  3/(x + 2)  +  6/x(x + 2)  =  0

Least common multiple is x (x + 2).

(x - 3)(x + 2) + 3x  +  6  =  0

x² - x - 6 + 3x + 6  =  0

x² + 2x  =  0

x(x + 2)  =  0

x  =  0  and x + 2  =  0

x  =  0 and x  =  -2

Substituting x  =  0 or x  =  -2 into the original equation results in the division is zero. So both of these numbers are extraneous solution. So the original equation has no solution.

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