Question

Solve the system of equation:

$\frac{1}{5}(x-2)=\frac{1}{4}(1-y)$
$26x+3y+4=0$

Answer 1

Consider the given equation,

$\frac{1}{5}(x-2)=\frac{1}{4}(1-y)$

$4x-8=5-5y$

$4x+5y-13=0$ ...... $\left(1\right)$

$26x+3y+4=0$ ...... $\left(2\right)$

In equation $1$ multiply by $3$, we get

$12x+15y-39=0$ ...... $\left(3\right)$

In equation 2 multiply by $5$, we get

$130x+15y+20=0$ ...... $\left(4\right)$

On subtarcting equations $3$ from equation $4$ we get,

$x=-2$

put the value of $x$ in equation $1$ we get,

$4(-2)+5y-13=0$

$5y=21$

$y=\frac{21}{5}$

Hence,the value of $x=-2$ and $y=\frac{21}{5}$