How do you solve #(2(x-2))/(x^2-10x+16)=2/(x+2)#?

2 Answers
Jun 1, 2017

Answer:

#x in {cancel"O"}#

(no solutions)

Explanation:

First, let's factor #x^2-10x+16#.

We need two factors of 16 which add up to -10.

#-2 and -8# work, so our factored polynomial is:

#(x-2)(x-8)#

Now we can solve the equation:

#(2(x-2))/(x^2-10x+16)=2/(x+2)#

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

#(color(blue)cancel(color(black)2)color(red)cancel(color(black)((x-2))))/(color(red)cancel(color(black)((x-2)))(x-8)) = color(blue)cancel(color(black)2)/(x+2)#

#1/(x-8) = 1/(x+2)#

Now take the reciprocal of both sides:

#x-8 = x+2#

#-8 = 2#

This is impossible, so there are no solutions.

Jun 1, 2017

Answer:

No solution

Explanation:

#(2(x-2))/(x^2-10x+16)=2/(x+2)#

#:.(2(cancel(x-2)^1))/((cancel(x-2)^1)(x-8))=2/(x+2)#

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

multiply both sides by (x-8)(x+2)

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

#:.2x+4=2x-16#

#:.2x-2x=-16-4#

#:.0=-20#

No solution