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

1 Answer
Oct 13, 2015

Answer:

#x = -1" "# or #" "x = 4#

Explanation:

Your equation looks like this

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

The first thing to do here is find the common denominator of the three fractions. This will allow you to get rid of the denominators altogether,

The least common multiple for the three expressions that serve as denominators will be #3 * (x-1)#.

This means that the first fraction must be multiplied by #1 = 3/3#, the second # 1 = (x-1)/(x-1)#, and the third by #1 = 3/3#.

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

#6/(3(x-1)) + ((x-2)(x-1))/(3(x-1)) = 12/(3(x-1))#

This is equivalent to

#6 + (x-2)(x-1) = 12#

Expand the parantheses, group like terms, and move all the terms on one side of the equation to get

#6 + x^2 - 2x - x + 2 - 12 = 0#

#x^2 - 3x - 4 = 0#

You can factor this quadratic by writing

#x^2 + x - 4x - 4 = 0#

#x * (x+1) - 4(x+1) = 0#

#(x+1)(x-4) = 0#

This means that you must have

#x + 1 = 0 implies x = color(green)(-1)#

or

#x - 4 = 0 implies x = color(green)(4)#

Do a quick check to make sure that the calculations are correct

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

#-1 -1 = -2color(white)(x)color(green)(sqrt())#

and

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

#2/3 + 2/3 = 4/3 color(white)(x)color(green)(sqrt())#