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

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Answer 1 · 2015-06-02T17:44:54

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

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

$a=-8+(5b)/2$ ( we divide by $2$ on each side )

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

$5a+3b=-9$

$5(-8+(5b)/2)+3b=-9$

$-40+(25b)/2+3b=-9$

$(25b)/2+3b=-9+40$ ( we add 40 on each side )

$3b=(6b)/2$

Thus $(25b+6b)/2 = 31$

$(31b)/2=31$

$b/2=1$ ( we divide by 31 on each side )

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

We can thus now find $a$ :

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

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

How do you solve the system 5a + 3b=-95a + 3b=-9 and 2a - 5b= -16.2a - 5b= -16.?