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Solve the following system of equations:
$\frac{x}{3}\ +\ \frac{y}{4}\ =\ 11$
$\frac{5x}{6}\ −\ \frac{y}{3}\ =\ −7$
Given:
The given system of equations is:
$\frac{x}{3}\ +\ \frac{y}{4}\ =\ 11$
$\frac{5x}{6}\ −\ \frac{y}{3}\ =\ −7$
To do:
We have to solve the given system of equations.
Solution:
The given system of equations can be written as,
$\frac{x}{3}+\frac{y}{4}=11$
$\Rightarrow \frac{4(x)+3(y)}{12}=11$
$\Rightarrow 4x+3y=11(12)$ (On cross multiplication)
$\Rightarrow 4x+3y=132$---(i)
$\frac{5x}{6}-\frac{y}{3}=-7$
$\Rightarrow \frac{5x-2(y)}{6}=-7$
$\Rightarrow 5x-2y=-7(6)$ (On cross multiplication)
$\Rightarrow 5x=2y-42$
$\Rightarrow x=\frac{2y-42}{5}$----(ii)
Substitute $x=\frac{2y-42}{5}$ in equation (i), we get,
$4(\frac{2y-42}{5})+3y=132$
$\frac{4(2y-42)}{5}+3y=132$ 
Multiplying by $5$ on both sides, we get,
$5(\frac{8y-168}{5})+5(3y)=5(132)$
$8y-168+15y=660$
$23y=660+168$
$23y=828$
$y=\frac{828}{23}$
$y=36$
Substituting the value of $y=36$ in equation (ii), we get,
$x=\frac{2(36)-42}{5}$
$x=\frac{72-42}{5}$
$x=\frac{30}{5}$
$x=6$
Therefore, the solution of the given system of equations is $x=6$ and $y=36$.