[tex]3x-2y=14\\5x+y=32[/tex]
Multiply 2 in the 5x+y=32 so that we can eliminate the "y"
[tex]3x-2y=14\\5x(2)+(2)y=32(2)[/tex]
From 5x(2)+(2)y=32(2), [tex]10x+2y=64[/tex]
[tex]3x-2y=14\\10x+2y=64[/tex]
Then start eliminating the y,
[tex]13x=78[/tex] [3x+10x = 13x] [-2y+2y = 0] [14+64 = 78]
[tex]x=\frac{78}{13}[/tex]
Therefore, [tex]x=6[/tex] but we are not done yet, we need to find the value of y.
Substitute x = 6 in any equations but only one equation so I'll substitute in 5x+y=32 instead.
[tex]5(6)+y=32\\30+y=32\\y=32-30\\y=2[/tex]
Therefore, y = 2.
So the answer is x = 6, y = 2 or (6,2)
