How do you solve #x/(2x-6)=2/(x-4)#?

1 Answer
Apr 14, 2017

Answer:

Restrict the domain so that the solutions do not cause division by zero in the original equation.
Multiply both sides by both denominators.
Solve the resulting quadratic.
Check.

Explanation:

Given: #x/(2x-6)=2/(x-4)#

Restrict the values of x so that any solutions that would cause division by zero are discarded:

#x/(2x-6)=2/(x-4);x!=3,x!=4#

Multiply both sides of the equation by #(2x-6)(x-4)#

#(2x-6)(x-4)x/(2x-6)=(2x-6)(x-4)2/(x-4);x!=3,x!=4#

Please observe how the factors cancel:

#cancel(2x-6)(x-4)x/cancel(2x-6)=(2x-6)cancel(x-4)2/cancel(x-4);x!=3,x!=4#

Here is the equation with the cancelled factor removed:

#(x-4)x=(2x-6)2;x!=3,x!=4#

Use the distributive property on both sides:

#x^2-4x=4x-12;x!=3,x!=4#

We can put the quadratic into standard form by adding #12-4x# to both sides:

#x^2-8x+12=0;x!=3,x!=4#

This factors and the restrictions can be dropped:

#(x - 2)(x-6)=0#

#x = 2 and x = 6#

Check:

#2/(2(2)-6)=2/(2-4)#
#6/(2(6)-6)=2/(6-4)#

#2/(-2)=2/(-2)#
#6/6=2/2#

#-1=-1#
#1=1#

This checks.