Solve the following system of equations:

$7(y\ +\ 3)\ –\ 2(x\ +\ 2)\ =\ 14$
$4(y\ –\ 2)\ +\ 3(x\ –\ 3)\ =\ 2$

Given:

The given system of equations is:

$7(y\ +\ 3)\ –\ 2(x\ +\ 2)\ =\ 14$

$4(y\ –\ 2)\ +\ 3(x\ –\ 3)\ =\ 2$

To do:

We have to solve the given system of equations.

Solution:

The given system of equations can be written as,

$7(y+3)-2(x+2)=14$

$\Rightarrow 7y+21-2x-4=14$

$\Rightarrow 7y-2x=14-17$

$\Rightarrow 7y-2x=-3$

$\Rightarrow 2x-7y=3$---(i)

$4(y-2)+3(x-3)=2$

$\Rightarrow 4y-8+3x-9=2$

$\Rightarrow 3x+4y=2+17$

$\Rightarrow 3x=19-4y$

$\Rightarrow x=\frac{19-4y}{3}$----(ii)

Substitute $x=\frac{19-4y}{3}$ in equation (i), we get,

$2(\frac{19-4y}{3})-7y=3$

$\frac{2(19-4y)}{3}-7y=3$ 

Multiplying by $3$ on both sides, we get,

$3(\frac{38-8y}{3})-3(7y)=3(3)$

$38-8y-21y=9$

$-29y=9-38$

$-29y=-29$

$y=\frac{-29}{-29}$

$y=1$

Substituting the value of $y=1$ in equation (ii), we get,

$x=\frac{19-4(1)}{3}$

$x=\frac{19-4}{3}$

$x=\frac{15}{3}$

$x=5$

Therefore, the solution of the given system of equations is $x=5$ and $y=1$.

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Updated on: 10-Oct-2022

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