Solve the following system of equations:

$3x\ –\ \frac{(y\ +\ 7)}{11}\ +\ 2\ =\ 10$
$2y\ +\ \frac{(x\ +\ 11)}{7}\ =\ 10$

Given: The system of equations are:

$3x\ –\ \frac{(y\ +\ 7)}{11}\ +\ 2\ =\ 10$ ; $2y\ +\ \frac{(x\ +\ 11)}{7}\ =\ 10$

To do: Solve the system of equation.

Solution:

Given system of equation is:

$3x\ –\ \frac{(y\ +\ 7)}{11}\ +\ 2\ =\ 10$..............i)

$2y\ +\ \frac{(x\ +\ 11)}{7}\ =\ 10$...........ii)

From i) we get

$3x\ –\ \frac{(y\ +\ 7)}{11}\ +\ 2\ =\ 10$

$\Rightarrow \frac{33x-y-7+22}{11}=10$

$\Rightarrow 33x-y+15=10\times11$

$\Rightarrow 33x-y=110-15$

$\Rightarrow 33x-y=95$

$\Rightarrow 33x-95=y$

From ii) we get

$2y\ +\ \frac{(x\ +\ 11)}{7}\ =\ 10$

$\Rightarrow \frac{14y+x+11}{7}=10 $

$\Rightarrow 14y+x+11=10\times7$ 

$\Rightarrow 14y+x=70-11$

$\Rightarrow 14y+x=59$................iii)

Substituting $33x-95=y$ in iii) we get

$14(33x-95)+x=59$

$462x-1330+x=59$

$461x=1330+59$

$x\frac{1389}{461}$

$x=3$

Now, put $x= 3$ in $33x-95=y$ we get,

$33(3)-95=y$

$y=99-95$

$y=4$

Hence, solution of the given system of equation is $x=3,y=4$