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. $x=-\frac{6}{5};y=9$
• C. $x=-\frac{8}{5};y=8$
• D. $x=-\frac{2}{3};y=7$

Answer 1

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

The equation $\frac{1}{5}(x-2)=\frac{1}{4}(1-y)$ can be rewritten as:

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

$\Rightarrow 4(x-2)=5(1-y)\Rightarrow 4x-8=5-5y\Rightarrow 4x+5y=5+8\Rightarrow 4x+5y=13.........\left(1\right)$

The equation $26x+3y+4=0$ can be rewritten as:

$26x+3y=-\mathrm{4..........}\left(2\right)$

Multiply the equation 1 by $3$ and equation 2 by $5$ to make the coefficients of $y$ equal. Then we get the equations:

$12x+15y=\mathrm{39.........}\left(3\right)$

$130x+15y=-\mathrm{20.........}\left(4\right)$

Subtract Equation 4 from equation 3 to eliminate $y$, because the coefficients of $y$ are the same. So, we get

$(12x-130x)+(15y-15y)=39+20$

i.e. $-118x=59$

i.e. $x=-\frac{1}{2}$

Substituting this value of $x$ in the equation 1, we get

$-2+5y=13$

i.e. $5y=13+2=15$

i.e. $y=3$

Hence, the solution of the equations is $x=-\frac{1}{2},y=3$.