How do you solve #(8(x-1))/(x^2-4)=4/(x-2)#?

Question

2129 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-06T02:48:23

#x=4#

Rewrite this as follows

#(8(x-1))/(x^2-4)=4/(x-2)#

#[8(x-1)]/[(x-2)(x+2)]-4/(x-2)=0#

#1/(x-2)*[(8(x-1))/(x+2)-4]=0#

#1/[(x-2)*(x+2)][8(x-1)-4(x+2)]=0#

#[8x-8-4x-8]/[(x-2)(x+2)]=0#

#(4(x-4))/[(x-2)(x+2)]=0#

From the last equation we get that #x=4#

Footnote

For the (initial) equation to hold must be #x!=2# and #x!=-2#