# How do you solve 1/(t+1) = (2x-6)/(3x+8)?

Jun 24, 2015

$t = \frac{1}{2} + \frac{17}{2 x - 6}$

with exclusion $x \ne - \frac{8}{3}$

$x = \frac{17}{2 t - 1} + 3$

with exclusion $t \ne - 1$

#### Explanation:

First note that the values $x = - \frac{8}{3}$ and $t = - 1$ cause division by $0$, so are excluded values.

Multiply both sides of the equation by $\left(3 x + 8\right)$ to get:

$\frac{3 x + 8}{t + 1} = 2 x - 6$

Multiply both sides of this equation by $\left(t + 1\right)$ to get:

$3 x + 8 = \left(2 x - 6\right) \left(t + 1\right)$

$= \left(2 x - 6\right) t + \left(2 x - 6\right)$

Subtract $\left(2 x - 6\right)$ from both sides to get:

$\left(2 x - 6\right) t = \left(3 x + 8\right) - \left(2 x - 6\right)$

$= 3 x + 8 - 2 x + 6$

$= 3 x - 2 x + 8 + 6$

$= \left(3 - 2\right) x + 14$

$= x + 14$

Divide both ends by $\left(2 x - 6\right)$ to get:

$t = \frac{x + 14}{2 x - 6}$

$= \frac{x - 3 + 17}{2 \left(x - 3\right)}$

$= \frac{x - 3}{2 \left(x - 3\right)} + \frac{17}{2 \left(x - 3\right)}$

$= \frac{1}{2} + \frac{17}{2 \left(x - 3\right)}$

$= \frac{1}{2} + \frac{17}{2 x - 6}$

So:

$\textcolor{red}{t = \frac{1}{2} + \frac{17}{2 x - 6}}$

To find $x$ in terms of $t$, first subtract $\frac{1}{2}$ from both sides to get:

$t - \frac{1}{2} = \frac{17}{2 x - 6}$

Multiply both sides by $2$ to get:

$2 t - 1 = \frac{17}{x - 3}$

Multiply both sides by $\left(x - 3\right)$ and divide both sides by $\left(2 t - 1\right)$ to get:

$x - 3 = \frac{17}{2 t - 1}$

Add $3$ to both sides to get:

$\textcolor{red}{x = \frac{17}{2 t - 1} + 3}$