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

Jun 7, 2017

$x = 4$

#### Explanation:

Factorise the denominator of the last term by inspection:

$\frac{3}{x - 2} = \frac{1}{x - 1} + \frac{7}{\left(x - 2\right) \left(x - 1\right)}$

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 $\left(x - 2\right) \left(x - 1\right)$:

$\frac{3 \left(x - 2\right) \left(x - 1\right)}{x - 2} = \frac{1 \left(x - 2\right) \left(x - 1\right)}{x - 1} + \frac{7 \left(x - 2\right) \left(x - 1\right)}{\left(x - 2\right) \left(x - 1\right)}$

Now we can cancel out a whole bunch of terms:

$\frac{3 \cancel{x - 2} \left(x - 1\right)}{\cancel{x - 2}} = \frac{1 \left(x - 2\right) \cancel{x - 1}}{\cancel{x - 1}} + \frac{7 \cancel{x - 2} \cancel{x - 1}}{\cancel{x - 2} \cancel{x - 1}}$

And we are left with this:

$3 \left(x - 1\right) = \left(x - 2\right) + 7$

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

$3 x - 3 = x + 5$

$2 x = 8$