How do you solve the rational equation $(2x) / (x+2) - 2 = (x-8) / (x-2)$ ?

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Answer 1 · 2016-12-24T04:52:06

The solution for the above rational equation is $x=6,-4$

$(2x)/(x+2)-2=(x-8)/(x-2)$

or, $(2x-2(x+2))/(x+2)=(x-8)/(x-2)rarr$ Take LCM on LHS.

or, $(2x-2x-4)/(x+2)=(x-8)/(x-2)rarr$ Simplify LHS.

or, $(-4)/(x+2)=(x-8)/(x-2)$

or, $(-4)*(x-2)=(x+2)(x-8)rarr$ Cross Multiply.

or, $-4x+8=x(x-8)+2(x-8)rarr$ Simplify.

or, $-4x+8=x^2-8x+2x-16rarr$ Simplify.

or, $-4x+8=x^2-6x-16rarr$ Simplify

or, $-4x+4x+8-8=x^2-6x+4x-16-8rarr$ Adding $4x$

and subtracting $8$ from both sides.

or, $x^2-2x+24=0rarr$ By simplifying RHS.

or, $x^2+4x-6x-24=0rarr$ Factoring new LHS.

or, $x(x+4)-6(x+4)=0$

or, $(x-6)(x+4)=0rarr$ Factored new LHS.

Either

$x-6=0$

or, $x=6$

Or,

$x+4=0$

or, $x=-4$

Hence, the solution for this equation is $x=6,-4$

How do you solve the rational equation (2x) / (x+2) - 2 = (x-8) / (x-2)(2x) / (x+2) - 2 = (x-8) / (x-2)?