Question

Solve the following pair of equations:

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

• A. $x=\frac{16}{7};y=-\frac{4}{5}$
• B. $x=-\frac{4}{5};y=-\frac{17}{8}$
• C. $x=\frac{18}{5};y=-\frac{4}{5}$
• D. $x=\frac{26}{3};y=-\frac{8}{3}$

Answer 1

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

We pick either of the equations and write one variable in terms of the other.

Let us consider the equation $3-(x-5)=y+2$ and write it as

$3-x+5=y+2$

or $-x-y=-6$

or $x+y=6$

or $y=6-x...........\left(1\right)$

Now the equation $2(x+y)=4-3y$ can be rewritten as:

$2x+2y=4-3y$

or $2x+5y=4$

Substitute the value of $y$ in the equation $2x+5y=4$. We get

$2x+5(6-x)=4$

i.e. $2x+30-5x=4$

i.e. $-3x=-26$

i.e. $x=\frac{26}{3}$

Therefore, $x=\frac{26}{3}$

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

$y=6-\frac{26}{3}=\frac{18-26}{3}=-\frac{8}{3}$

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