# How do you solve for u in y= (u+1)/(u+2)?

Mar 18, 2016

color(blue)(u = (1 -2y)/(y-1)

#### Explanation:

y= (u+1) / color(green)((u+2)

$y \cdot \textcolor{g r e e n}{\left(u + 2\right)} = \left(u + 1\right)$

$y \cdot \textcolor{g r e e n}{\left(u\right)} + y \cdot \textcolor{g r e e n}{\left(2\right)} = \left(u + 1\right)$

$y u + 2 y = u + 1$

Isolating $u$ on the L.H.S

$y \textcolor{b l u e}{u} - \textcolor{b l u e}{u} = 1 - 2 y$

$u$ is common to both terms of the L.H.S:

$\textcolor{b l u e}{u} \left(y - 1\right) = 1 - 2 y$

color(blue)(u = (1 -2y)/(y-1)

Mar 18, 2016

$u = \frac{- 2 y + 1}{y - 1}$

#### Explanation:

Another (more difficult) method:

Rewrite $u + 1$ as $u + 2 - 1$.

$y = \frac{u + 2 - 1}{u + 2}$

Split up the numerator.

$y = \frac{u + 2}{u + 2} - \frac{1}{u + 2}$

$y = 1 - \frac{1}{u + 2}$

Rearrange the terms.

$y - 1 = - \frac{1}{u + 2}$

Multiply both sides by $\left(u + 2\right)$.

$\left(y - 1\right) \left(u + 2\right) = - 1$

Distribute the $\left(y - 1\right)$ into $u$ and $2$:

$u \left(y - 1\right) + 2 \left(y - 1\right) = - 1$

$u \left(y - 1\right) + 2 y - 2 = - 1$

$u \left(y - 1\right) = - 2 y + 1$

After this algebraic rearrangement, divide both sides by $\left(y - 1\right)$ to isolate $u$.

$u = \frac{- 2 y + 1}{y - 1}$