# How do you solve 6y - 8 = 12 + 10y?

Nov 1, 2015

Expanded explanation

#### Explanation:

$\textcolor{red}{\text{Objective is to have y, on its own, on one side of the = and everything else on the other side.}}$

Brackets are used to show previous part to help make things clearer about what is happening.

.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$\textcolor{b l u e}{\text{Add 8 to both sides to remove it from the left.}}$

$\left(6 y - 8\right) + \textcolor{red}{8} = \left(12 + 10 y\right) + \textcolor{red}{8}$

$6 y - 8 + 8 = 12 + 8 + 10 y$

$6 y + 0 = 20 + 10 y$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Subtract 10y from both sides to remove it from the right.}}$

$\left(6 y\right) - \textcolor{red}{10 y} = \left(20 + 10 y\right) - \textcolor{red}{10 y}$

$6 y - \textcolor{red}{10 y} = 20 + 10 y - \textcolor{red}{10 y}$

$- 4 y = 20$

We need the y part to be positive so multiply both sides by negative 1

$\textcolor{red}{- 1} \times \left(- 4 y\right) = \textcolor{red}{- 1} \times \left(20\right)$

$4 y = - 20$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Divide both sides by 4 so that the 4 in front of 4y becomes 1. As 1 times anything is itself.}}$

$\textcolor{red}{\frac{1}{4}} \times \left(4 y\right) = \textcolor{red}{\frac{1}{4}} \times \left(- 20\right)$

$1 \times y = \frac{- 20}{4}$

$y = - 5$