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

May 28, 2016

$x \in \left\{1 , 9\right\}$

#### Explanation:

First, note that $2 x - 3$ and $x + 1$ are in the denominators of rational terms, meaning that to avoid dividing by $0$ we must have $x \ne \frac{3}{2}$ and $x \ne - 1$.

With that in mind, first we eliminate the denominators by multiplying by $\left(2 x - 3\right) \left(x + 1\right)$

$\frac{x}{2 x - 3} + \frac{4}{x + 1} = 1$

$\implies \left(2 x - 3\right) \left(x + 1\right) \left(\frac{x}{2 x - 3} + \frac{4}{x + 1}\right) = \left(2 x - 3\right) \left(x + 1\right) \cdot 1$

$\frac{x \cancel{\left(2 x - 3\right)} \left(x + 1\right)}{\cancel{2 x - 3}} + \frac{4 \left(2 x - 3\right) \cancel{\left(x + 1\right)}}{\cancel{x + 1}} = \left(2 x - 3\right) \left(x + 1\right)$

$\implies x \left(x + 1\right) + 4 \left(2 x - 3\right) = \left(2 x - 3\right) \left(x + 1\right)$

$\implies {x}^{2} + x + 8 x - 12 = 2 {x}^{2} - 3 x + 2 x - 3$

$\implies {x}^{2} + 9 x - 12 = 2 {x}^{2} - x - 3$

$\implies {x}^{2} - 10 x + 9 = 0$

$\implies \left(x - 1\right) \left(x - 9\right) = 0$

$\implies x = 1$ or $x = 9$

As neither of these conflict with our initial conditions of $x \ne \frac{3}{2}$ and $x \ne - 1$, we have the final solution set $x \in \left\{1 , 9\right\}$