Question

Solve the following pair of equations :

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

• A. $x=-\frac{1}{2};y=3$
• B. $x=-\frac{1}{6};y=1$
• C. $x=-\frac{7}{2};y=4$
• D. $x=-\frac{7}{5};y=7$

Answer 1

The correct option is A $x=-\frac{1}{2};y=3$

On simplifying $\frac{(x-2)}{5}=\frac{(1-y)}{4}$, we get $4x+5y=13$---- (1)
And equation $26x+3y=-4$ ---- (2)
Multiplying equation $\left(1\right)$ with $3$ we get, $12x+15y=39$ ----- equation $\left(3\right)$
Multiplying equation $\left(2\right)$ with $5$ we get, $130x+15y=-20$ ----- equation $\left(4\right)$

subtracting equation $\left(3\right)$

from $\left(4\right)$, we get $118x=-59=>x=-\frac{1}{2}$

Substituting $x=-\frac{1}{2}$ in the equation $\left(1\right)$, we get $4(-\frac{1}{2})+5y=13$
$⟹y=3$