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.