How do you solve #3/(x-2) = 1/(x-1) + 7/(x^2-3x+2)#?

1 Answer
Jun 7, 2017

Answer:

#x=4#

Explanation:

Factorise the denominator of the last term by inspection:

#3/(x-2)=1/(x-1)+7/((x-2)(x-1))#

We need to get the #x#'s off the denominator to solve. I see lots of common terms, so to get rid of them, multiply both sides of the equation by #(x-2)(x-1)#:

#(3(x-2)(x-1))/(x-2)=(1(x-2)(x-1))/(x-1)+(7(x-2)(x-1))/((x-2)(x-1))#

Now we can cancel out a whole bunch of terms:

#(3cancel(x-2)(x-1))/cancel(x-2)=(1(x-2)cancel(x-1))/cancel(x-1)+(7cancel(x-2)cancel(x-1))/(cancel(x-2)cancel(x-1))#

And we are left with this:

#3(x-1)=(x-2)+7#

Expand the brackets, and put all the #x# terms on the left, and the constants on the right:

#3x-3=x+5#

#2x=8#