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

##### 1 Answer

#### 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

This means that the first fraction must be multiplied by

#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())#