# 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

 2x + 1  =  0x  =  -1/2 x - 3  =  0 x  =  3

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. Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here.

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