# How do you solve for y in k(y+3z)=4(y-5)?

May 23, 2018

See a solution process below:

#### Explanation:

First, expand the terms in parenthesis on each side of the equation by multiplying each term within the parenthesis by the term outside the parenthesis:

$\textcolor{red}{k} \left(y + 3 z\right) = \textcolor{red}{4} \left(y - 5\right)$

(color(red)(k) xx y) + (color(red)(k) xx 3z) = (color(red)(4) xx y) - ((color(red)(4) xx 5)

$k y + 3 k z = 4 y - 20$

Next, subtract $\textcolor{red}{3 k z}$ and $\textcolor{b l u e}{4 y}$ from each side of the equation to isolate the $y$ terms while keeping the equation balanced:

$k y - \textcolor{b l u e}{4 y} + 3 k z - \textcolor{red}{3 k z} = 4 y - \textcolor{b l u e}{4 y} - 20 - \textcolor{red}{3 k z}$

$k y - 4 y + 0 = 0 - 20 - 3 k z$

$k y - 4 y = 20 - 3 k z$

The factor out the common term on the left side of the equation:

$k \textcolor{red}{y} - 4 \textcolor{red}{y} = 20 - 3 k z$

$\left(k - 4\right) y = 20 - 3 k z$

Now, divide both sides of the equation by $\textcolor{red}{\left(k - 4\right)}$ to solve for $y$ while keeping the equation balanced:

$\frac{\left(k - 4\right) y}{\textcolor{red}{\left(k - 4\right)}} = \frac{20 - 3 k z}{\textcolor{red}{\left(k - 4\right)}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{\left(k - 4\right)}}} y}{\cancel{\textcolor{red}{\left(k - 4\right)}}} = \frac{20 - 3 k z}{k - 4}$

$y = \frac{20 - 3 k z}{k - 4}$