For this case we have a system of two equations with two unknowns:
[tex]3x-2y = 14\\5x + y = 32[/tex]
We multiply the second equation by 2:
[tex]3x-2y = 14\\10x + 2y = 64[/tex]
We add the equations:
[tex]13x = 78\\x = \frac {78} {13}\\x = 6[/tex]
We substitute the value of "x" in the following equation, and we clear the value of "y":
[tex]5 (6) + y = 32\\30 + y = 32\\y = 32-30\\y = 2[/tex]
So, the solution is: (6,2)
ANswer:
(6,2)
