SOLVE THE EQUATION ELIMINATING ANY EXTRANEOUS SOLUTIONS

Extraneous Solutions :

When we multiply or divide an equation by an expression containing variables, the resulting equation may have solutions that are not solutions of the original equation. These are extraneous solutions.

For this reason we must check each solution of the resulting equation in the original equation.

Example 1 :

Solve the equation

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 

(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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