Greetings!
To solve this, you can multiply one of the equations by a number so that one of the unknown values are the same as the other equation. Or simply in these two equations you can multiply the second equation by a minus (-1):
(2x + 2y = 8 ) x -1
= -2x - 2y = -8
Now, you can add or minus the two equations, but seeing as both values before the x are negative we can add the two equations to cancel out the x:
-2x + 4y = 16 +
-2x = 2y = -8
= 0x + 6y = 8
6y = 8
Simply divide 8 by 6 to find y:
8 ÷ 6 =
, so the value of y =
![(4)/(3)](https://img.qammunity.org/2019/formulas/mathematics/high-school/2ef7qw0k27nks1lxds61rbq54v3mvv6vjw.png)
Now you can substitute this value into one of the equations:
2x + 2y = 8
2x + 2(
) = 8
2x +
= 8
2x = 8 -
![(8)/(3)](https://img.qammunity.org/2019/formulas/mathematics/high-school/rabhxjqykgve8iayuwi9hull5mmg7v2v4w.png)
2x =
![(16)/(3)](https://img.qammunity.org/2019/formulas/mathematics/middle-school/d85lb0g70393nzjk204yilffpx2jjbtrq1.png)
x =
÷ 2
x =
![(8)/(3)](https://img.qammunity.org/2019/formulas/mathematics/high-school/rabhxjqykgve8iayuwi9hull5mmg7v2v4w.png)
So that means x =
and y =
![(4)/(3)](https://img.qammunity.org/2019/formulas/mathematics/high-school/2ef7qw0k27nks1lxds61rbq54v3mvv6vjw.png)
Hope this helps!