For this case we have the following system of equations:[tex]-2x + y = -3\\-x + 2y = 3[/tex]
We multiply the first equation by -2:
[tex]4x-2y = 6[/tex]
We add the equations:
[tex]-x + 2y = 3\\4x-2y = 6[/tex]
Adding we have:
[tex]4x-x-2y + 2y = 3 + 6\\3x = 9\\x = \frac {9} {3}\\x = 3[/tex]
We look for the value of "y":
[tex]-x + 2y = 3\\-3 + 2y = 3\\2y = 3 + 3\\2y = 6\\y = \frac {6} {2}\\y = 3[/tex]
So, the solution is [tex](x, y) :( 3,3)[/tex]
Answer:
[tex](x, y) :( 3,3)[/tex]
