# How do you solve the system 5a + 3b=-9 and 2a - 5b= -16.?

Jun 2, 2015

We are first going to find $a$ in function of $b$ :

$2 a - 5 b = - 16$
$2 a = - 16 + 5 b$ ( we add $5 b$ on each side )

$a = - 8 + \frac{5 b}{2}$ ( we divide by $2$ on each side )

Now that we have $a$, we can find $b$ with the first equation :

$5 a + 3 b = - 9$

$5 \left(- 8 + \frac{5 b}{2}\right) + 3 b = - 9$

$- 40 + \frac{25 b}{2} + 3 b = - 9$

$\frac{25 b}{2} + 3 b = - 9 + 40$ ( we add 40 on each side )

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

Thus $\frac{25 b + 6 b}{2} = 31$

$\frac{31 b}{2} = 31$

$\frac{b}{2} = 1$ ( we divide by 31 on each side )

$b = 2$ ( we multiply by 2 on each side )

We can thus now find $a$ :

$2 a - 5 b = - 16$
$2 a - 5 \cdot 2 = - 16$
$2 a - 10 = - 16$
$2 a = - 16 + 10$ ( we add 10 on each side )
$2 a = - 6$
$a = - 3$ ( we divide by 2 on each side )

And thus :
$a = - 3$
and $b = 2$