$0 = 9a + 3b + c$
$2 = 25a + 5b + c$
$6 = 49a + 7b + c$
These are my 3 equations that have 3 unknowns. How do I solve for these unknowns?
algebra-precalculus
$0 = 9a + 3b + c$
$2 = 25a + 5b + c$
$6 = 49a + 7b + c$
These are my 3 equations that have 3 unknowns. How do I solve for these unknowns?
Best Answer
(1) $9a+3b+c=0$
(2) $25a+5b+c=2$
(3) $49a+7b+c=6$
Using (2)-(1) we have:
(4) $16a+2b=2$
Using (3)-(2) we have:
(5) $24a+2b=4$
Using (5)-(4) we have:
(6) $8a=2$
It follows that $a=\frac{1}{4}$, $b=-1$ and $c=\frac{3}{4}$.