How do you solve #(2x)/ (x+2) - 2 = (x-8) / (x-2)#?

2 Answers
Jun 10, 2015

Answer:

I would find a common denominator and then solve for #x#.

Explanation:

Have a look:
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Jun 10, 2015

Answer:

Multiply through by #(x+2)# and #(x-2)#, expand and simplify to get:
#0 = x^2-2x-24 = (x-6)(x+4)#, hence #x = 6# or #x=-4#

Explanation:

Given:

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

Multiply both sides by #(x+2)(x-2)# to get:

#2x(x-2)-2(x+2)(x-2) = (x-8)(x+2)#

Note that when we multiply by #(x+2)(x-2)# we could in theory introduce spurious extra solutions #x=+-2#, but these are excluded values of the original equation anyway.

Multiplying out:

#cancel(2x^2)-4x-cancel(2x^2)+8 = x^2-6x-16#

Add #4x-8# to both sides to get:

#0 = x^2-2x-24 = (x-6)(x+4)#

To get this factorization, I looked for a pair of factors of #24# whose difference is #-2#, since:

#(x - a)(x + b) = x^2-(a-b)x-(axxb)#

Since #(x-6)(x+4) = 0#, the solutions are #x=-4# and #x=6#.