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

Oct 7, 2017

$x = 4$

#### Explanation:

Multiplying the numerators of the fractions by
the common denominator, $\left(x - 2\right) \left(x - 3\right)$,

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

Multiplying both sides by $\left(x - 2\right) \left(x - 3\right)$,

$\implies \left({x}^{2} - 3 x - x + 2\right) = {x}^{2} - 5 x + 6$

$\implies - 4 x + 5 x = 6 - 2$

$\implies x = 4$

$x = 4$

#### Explanation:

The first thing I'd do is combine the two fractions on the left:

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

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

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

$\frac{{x}^{2} - 4 x + 2}{{x}^{2} - 5 x + 6} = 1$

And now multiply through by the denominator of the fraction:

${x}^{2} - 4 x + 2 = {x}^{2} - 5 x + 6$

We can now simplify:

${x}^{2} \textcolor{red}{- {x}^{2}} - 4 x \textcolor{red}{+ 5 x} + 2 \textcolor{red}{- 2} = {x}^{2} \textcolor{red}{- {x}^{2}} - 5 x \textcolor{red}{+ 5 x} + 6 \textcolor{red}{- 2}$

$x = 4$