3x + 2y = 9 (1)
2x + 6y = 6 (2)
We need to eliminate one of them
So let us at first divide equation (2) by 2 to simplify it
[tex]\begin{gathered} \frac{2x}{2}+\frac{6y}{2}=\frac{6}{2} \\ x+3y=3 \end{gathered}[/tex]x + 3y = 3 (3)
Now we can multiply equation (3) by -3 to eliminate x
-3(x) + -3(3y)=-3(3)
-3x - 9y = -9 (4)
Now add equations (1) and (4)
(3x - 3x) + (2y - 9y) = (9 -9)
0 - 7y = 0
-7y = 0
Divide both sides by -7
[tex]\frac{-7y}{-7}=\frac{0}{-7}[/tex]y = 0
Now substitute the value of y in equation (1) or 2 to find x
I will take (1)
3x + 2(0) = 9
3x + 0 = 9
3x = 9
Divide both sides by 3 to find x
[tex]\frac{3x}{3}=\frac{9}{3}[/tex]x = 3
The solution of this system of equations is (3, 0)