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

2 Answers
Sep 7, 2016

Answer:

#=> x=0" and "x=1#

Explanation:

Configuring the equation so that both sides end up with the same denominator.

Write #1" as " 1=x/x# giving

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

But #x^2-x = x(x-1)#

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

Consider just the left hand side

#2/(x-1)+x/x" " ->" "(2+x(x-1))/(x(x-1)) #

Putting it all together

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

As the denominators are the same on both sides the equation is equally true if you just equate the numerators to each other and totally forget about the denominators.

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

#x(x-1)=0#

#=> x=0" and "x=1#

Sep 7, 2016

Answer:

#x = -1, 2#

Explanation:

We have: #(2) / (x - 1) + 1 = (2) / (x^(2) - x)#

Let's simplify the left-hand side of the equation:

#=> (2 + (x - 1)) / (x - 1) = (2) / (x^(2) - x)#

#=> (x + 1) / (x - 1) = (2) / (x^(2) - x)#

Then, let's cross-multiply:

#=> (x - 1) (x^(2) - x) = 2 (x - 1)#

We can expand the parentheses to get:

#=> x^(3) - 2 x^(2) + x = 2 x - 2#

Let's move all terms to the left-hand side to form a cubic equation:

#=> x^(3) - 2x^(2) - x + 2 = 0#

#=> (x^(3) - 2 x^(2)) + (- x + 2) = 0#

Now, let's factorise:

#=> x^(2) (x - 2) - (x - 2) = 0#

#=> (x - 2) (x^(2) - 1) = 0#

The difference of squares can be simplified to:

#=> (x - 2) (x + 1) (x - 1) = 0#

#=> x = -1, 1, 2#

However, using #x = 1# yields an undefined result.

Therefore, the solutions to the equation are #x = -1# and #x = 2#.