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

Question

Eva Abramson · Accepted 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.

Given: #x/(2x-6)=2/(x-4)# Limit the values of x such that no solutions are found that would result in division by zero: #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 note how the variables cancel out: #cancel(2x-6)(x-4)x/cancel(2x-6)=(2x-6)cancel(x-4)2/cancel(x-4);x!=3,x!=4# The equation with the cancelled factor subtracted is as follows: #(x-4)x=(2x-6)2;x!=3,x!=4# On both sides, apply the distributive property: #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# These elements and the limitations are reversible: #(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 verifies.

Jace Anderson · Answer

undefined To solve the equation x/(2x-6) = 2/(x-4), we can start by cross-multiplying. This means multiplying the numerator of the first fraction by the denominator of the second fraction, and vice versa.

After cross-multiplying, we get x(x-4) = 2(2x-6).

Expanding both sides of the equation, we have x^2 - 4x = 4x - 12.

Combining like terms, we get x^2 - 8x + 12 = 0.

This quadratic equation can be factored as (x-6)(x-2) = 0.

Setting each factor equal to zero, we have x-6 = 0 or x-2 = 0.

Solving for x, we find x = 6 or x = 2.

Therefore, the solutions to the equation x/(2x-6) = 2/(x-4) are x = 6 and x = 2.