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 = [tex]\frac{4}{3}[/tex] , so the value of y = [tex]\frac{4}{3}[/tex]
Now you can substitute this value into one of the equations:
2x + 2y = 8
2x + 2([tex]\frac{4}{3}[/tex]) = 8
2x + [tex]\frac{8}{3}[/tex] = 8
2x = 8 - [tex]\frac{8}{3}[/tex]
2x = [tex]\frac{16}{3}[/tex]
x = [tex]\frac{16}{3}[/tex] ÷ 2
x = [tex]\frac{8}{3}[/tex]
So that means x = [tex]\frac{8}{3}[/tex] and y = [tex]\frac{4}{3}[/tex]
Hope this helps!
