x=0 and y=4 in the second problem

X+y=13
22x + 36y=356

-3x + 2y =8
3x + 2y=8

y=4

hold, on i'm trying to find x

x=2 & y=1 on the last one :)

To find the value of x in the last equation, let's solve the system of equations:

-3x + 2y = 8
3x + 2y = 8

To eliminate the y terms, you can add the two equations together:

(-3x + 2y) + (3x + 2y) = 8 + 8

This simplifies to:

-3x + 3x + 2y + 2y = 16

Combine like terms:

4y = 16

Now, we can solve for y by dividing both sides of the equation by 4:

4y/4 = 16/4

y = 4

So, y is indeed 4. Now, substitute this value back into one of the original equations:

3x + 2y = 8

3x + 2(4) = 8

Simplify:

3x + 8 = 8

Subtract 8 from both sides:

3x = 0

Divide both sides by 3:

x = 0

Therefore, the value of x in the last equation is 0 and y is 4.