# How do you solve the system s+r=0 and r-s=6?

$r = 3$, $s = - 3$

#### Explanation:

Given equations:

$s + r = 0 \setminus \ldots \ldots \ldots . . \left(1\right)$

$r - s = 6 \setminus \ldots \ldots \ldots . . \left(2\right)$

adding (1) & (2) as follows

$s + r + r - s = 0 + 6$

$2 r = 6$

$r = \frac{6}{2} = 3$

setting $r = 3$ in (1), we get

$s + 3 = 0$

$s = - 3$