Solve the following system of equations:
$3x\ –\ 7y\ +\ 10\ =\ 0$
$y\ –\ 2x\ –\ 3\ =\ 0$

Given:

The given system of equations is:

$3x\ –\ 7y\ +\ 10\ =\ 0$

$y\ –\ 2x\ –\ 3\ =\ 0$

To do:

We have to solve the given system of equations.

Solution:

The given system of equations can be written as,

$3x-7y+10=0$---(i)

$y-2x-3=0$

$\Rightarrow 2x=y-3$

$\Rightarrow x=\frac{y-3}{2}$----(ii)

Substitute $x=\frac{y-3}{2}$ in equation (i), we get,

$3(\frac{y-3}{2})-7y+10=0$

$\frac{3(y-3)}{2}-7y+10=0$ 

Multiplying by $2$ on both sides, we get,

$2(\frac{3y-9}{2})-2(7y)+2(10)=2(0)$

$3y-9-14y+20=0$

$-11y+11=0$

$11y=11$

$y=\frac{11}{11}$

$y=1$

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

$x=\frac{1-3}{2}$

$x=\frac{-2}{2}$

$x=-1$

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

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Updated on: 10-Oct-2022

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