Question

Solve equations using substitution method: $x-y=3$ and $x+y=0$

• A. $\frac{3}{2}$ and $1$
• B. $\frac{-3}{2}$ and $-1$
• C. $\frac{3}{2}$ and $-\frac{3}{2}$
• D. $0$ and $\frac{3}{2}$

Answer 1

Answer: (C)

$\frac{3}{2}$ and $-\frac{3}{2}$

$x-y=3\dots \left(1\right)$
$x+y=0\dots \left(2\right)$
From equation $\left(1\right)$ :
$y=x-3$
Substitute the value of $y$ in equation $\left(2\right)$ :
$x+y=0$
$x+x-3=0$
$2x=3$
$x=\frac{3}{2}$

Now, Substitute $x=\frac{3}{2}$ in equation $\left(1\right)$:
$x-y=3$
$\frac{3}{2}-y=3$
$y=\frac{3}{2}-3$
$y=-\frac{3}{2}$
Therefore the solution is:

$x=\frac{3}{2}$ and $y=-\frac{3}{2}$