Question

3y + 2z = 12 and y – z = 9.

Answer 1

If this is a system of equations, you can use substitution.
Solve the second equation for y:
y = z + 9
Now substitute y of the first equation with z + 9.

3y + 2z = 12
3(z + 9) + 2z = 12
3z + 27 + 2z = 12
5z + 27 = 12
5z = -15
z = -3

Now substitute -3 for z in the second equation:
y - (-3) = 9
y + 3 = 9
y = 6

Solution: y = 6 and z = -3

Answer 2

I'm guessing you're asking for how to solve for z and y using this equation.

y -z = 9 add z to each side
y = z + 9

We can substitute this (z + 9) into wherever y is in the first equation.

3y + 2z = 12
3( z + 9) + 2z = 12
3z + 27 + 2z = 12
5z + 27 = 12
5z = -15
z = -3
Now that we have z, we can plug this in to the smaller equation and solve for y.

y = z + 9
y = -3 + 9
y = 6

So z = -3 and y = 6

Hope this helps C: