To solve a using the , one of the terms needs to be eliminated when one equation is added to or subtracted from the other equation. In this exercise, this means that either the
a-terms or the
b-terms must cancel each other out.
{9a−2b=-8-7a+3b=12(I)(II)
Currently, none of the terms in this system will cancel out. Therefore, we need to find a common multiple between two variable in the system. If we multiply Equation (I) by
3 and multiply Equation (II) by
2, the
b-terms will have opposite .
{3(9a−2b)=3(-8)2(-7a+3b)=2(12)⇒{27a−6b=-24-14a+6b=24
We can see that the
b-terms will eliminate each other if we add Equation (I) to Equation (II).
{27a−6b=-24-14a+6b=24
{27a−6b=-24-14a+6b+(27a−6b)=24+(-24)
{27a−6b=-24-14a+6b+27a−6b=24+(-24)
{27a−6b=-2413a=0
{27a−6b=-24a=0
Now we can now solve for
b by substituting the value of
a into either equation and simplifying.
{27a−6b=-24a=0
{27(0)−6b=-24a=0
{0−6b=-24a=0
{-6b=-24a=0
{b=4a=0
The solution, or , of the system of equations is
(0,4).