Solve the following system of equations:

$11x\ +\ 15y\ +\ 23\ =\ 0$
$7x\ –\ 2y\ –\ 20\ =\ 0$

Given:

The given system of equations is:

$11x\ +\ 15y\ +\ 23\ =\ 0$  

$7x\ –\ 2y\ –\ 20\ =\ 0$

To do:

We have to solve the given system of equations.

Solution:

The given system of equations can be written as,

$11x+15y+23=0$---(i)

$7x-2y-20=0$

$\Rightarrow 7x=2y+20$

$\Rightarrow x=\frac{2y+20}{7}$----(ii)

Substitute $x=\frac{2y+20}{7}$ in equation (i), we get,

$11(\frac{2y+20}{7})+15y+23=0$

$\frac{11(2y+20)}{7}+15y+23=0$ 

Multiplying by $7$ on both sides, we get,

$7(\frac{22y+220}{7})+7(15y)+7(23)=7(0)$

$22y+220+105y+161=0$

$127y+381=0$

$127y=-381$

$y=\frac{-381}{127}$

$y=-3$

Substituting the value of $y=-3$ in equation (ii), we get,

$x=\frac{2(-3)+20}{7}$

$x=\frac{-6+20}{7}$

$x=\frac{14}{7}$

$x=2$

Therefore, the solution of the given system of equations is $x=2$ and $y=-3$.

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Updated on: 10-Oct-2022

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