# Question #eda6e

Dec 27, 2017

$y = 26$

#### Explanation:

$\text{eliminate the fractions by multiplying ALL terms by}$
$\text{12 the lowest common multiple of 3 and 4}$

${\cancel{12}}^{4} \times \frac{1}{\cancel{4}} ^ 1 \left(y + 4\right) + 84 = {\cancel{12}}^{3} \left(3 y - 2\right) - 24$

$\Rightarrow 4 \left(y + 4\right) + 84 = 3 \left(2 y - 2\right) - 24$

$\text{distribute brackets on both sides and collect like terms}$

$\Rightarrow 4 y + 16 + 84 = 9 y - 6 - 24$

$\Rightarrow 4 y + 100 = 9 y - 30$

$\text{subtract 9y from both sides}$

$4 y - 9 y + 100 = \cancel{9 y} \cancel{- 6 y} - 30$

$\Rightarrow - 5 y + 100 = - 30$

$\text{subtract 100 from both sides}$

$- 5 y \cancel{+ 100} \cancel{- 100} = - 30 - 100$

$\Rightarrow - 5 y = - 130$

$\text{divide both sides by } - 5$

$\frac{\cancel{- 5} y}{\cancel{- 5}} = \frac{- 130}{- 5}$

$\Rightarrow y = 26$

$\textcolor{b l u e}{\text{As a check}}$

Substitute this value into the equation and if both sides are equal then it is the solution.

$\text{left } = \frac{1}{3} \left(30\right) + 7 = 10 + 7 = 17$

$\text{right } = \frac{1}{4} \left(76\right) - 2 = 19 - 2 = 17$

$\Rightarrow y = 26 \text{ is the solution}$