Question

Solve the following pair of equations:

$3-(x-5)=y+2$, $2(x+y)=4-3y$

• A. $x=\frac{7}{2},y=-\frac{9}{5}$
• B. $x=\frac{1}{6},y=-\frac{4}{3}$
• C. $x=\frac{26}{3},y=-\frac{8}{3}$
• D. $x=\frac{6}{5},y=-\frac{7}{3}$

Answer 1

The correct option is C $x=\frac{26}{3},y=-\frac{8}{3}$

The equation $3-(x-5)=y+2$ can be solved as:

$3-x+5=y+2\Rightarrow 8-x=y+2\Rightarrow x+y=6.........\left(1\right)$

The equation $2(x+y)=4-3y$ can be solved as:

$2x+2y=4-3y\Rightarrow 2x+2y+3y=4\Rightarrow 2x+5y=4.........\left(2\right)$

Multiply the equation 1 by $2$. Then we get:

$2x+2y=\mathrm{12.........}\left(3\right)$

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

$(2x-2x)+(5y-2y)=4-12$

i.e. $3y=-8$

i.e. $y=-\frac{8}{3}$

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

$x+(-\frac{8}{3})=6\Rightarrow 3x-8=18\Rightarrow 3x=18+8=26\Rightarrow x=\frac{26}{3}$

Hence, the solution of the equations is $x=\frac{26}{3},y=-\frac{8}{3}$.