How do you solve #2/(x-1) - 2/3 =4/(x+1)#?

2 Answers
Apr 6, 2018

Answer:

#x=-5# and #x=2# are both solutions to this equation.

Explanation:

#2/(x-1)-2/3=4/(x+1)#

Lots of ways to solve this. All of them work! Here's one way.

Divide both sides of the equation by 2.

#1/(x-1)-1/3=2/(x+1)#

Put the two terms on the left-hand side of the equation under a common denominator.

#(3-(x-1))/(3(x-1))=2/(x+1)#

Simplify the numerator of the left-hand side.

#(4-x)/(3(x-1))=2/(x+1)#

Invert both sides of the equation.

#(3(x-1))/(4-x)=(x+1)/2#

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

#6(x-1)=(x+1)(4-x)#

Expand both sides of the equation using the distributive property.

#6x-6=-x^2+3x+4#

Put this quadratic equation in standard form.

#x^2+3x-10=0#

The left-hand side of this equation factors nicely.

#(x+5)(x-2)=0#

#x=-5# and #x=2# are both solutions to this equation.

I bet you can come up with another way to solve this. What I like about THIS way is that you don't have to deal with quadratics AND quotients in the same equation.

Apr 6, 2018

Answer:

#x_1=-5# and #x_2=2#

Explanation:

#2/(x-1)-2/3=4/(x+1)#

#2/(x-1)-4/(x+1)=2/3#

#(2*(x+1)-4*(x-1))/((x-1)*(x+1))=2/3#

#(6-2x)/(x^2-1)=2/3#

#3*(6-2x)=2*(x^2-1)#

#18-6x=2x^2-2#

#2x^2+6x-20=0#

#x^2+3x-10=0#

#(x+5)*(x-2)=0#

So #x_1=-5# and #x_2=2#