2x+y=9 x-2z=-3 2y+3z=15
Forgot how to do This. Problems
Solving systems of this type requires eliminating
all but one unknown so you can solve for that unknown.
Then, plug that value into one of the equations to solve
for another of the unknowns. Finally, plug both of those
values into an equation to solve for the last unknown.
1) 2x + y = 9
2) x - 2z = -3
3) 2y + 3z = 15
Multiply equation 2 by 2 so we can subtract it from equation 1.
2 * (x - 2z) = -3 * 2
4) 2x - 4z = -6
2x + y = 9
-(2x - 4z = -6)
---------------------
y + 4z = 15
5) y + 4z = 15
We now have two equations with y and z. Multiply
equation 5 by 2...
2 * (y + 4z) = 15 * 2
6) 2y + 8z = 30
...and subtract equation 6 from equation 3. Then, solve for z.
2y + 3z = 15
-(2y + 8z = 30)
-------------------
- 5z = -15
-5z = -15
z = 3 <<<<<<<<<<<<<<<<<<<<
Plug that into equation 3 to solve for y.
2y + 3z = 15
2y + 3(3) = 15
2y + 9 = 15
2y = 6
y = 3 <<<<<<<<<<<<<<<<<<<<
Also, plug the z value into equation 2 to solve for x.
x - 2z = -3
x - 2(3) = -3
x - 6 = -3
x = 3 <<<<<<<<<<<<<<<<<<<<
Plug the values for x and y into equation 1 to verify that they are correct.
2x + y = 9
2(3) + 3 = 9
6 + 3 = 9
9 = 9
Usually, all three equations have all three unknowns, requiring more
manipulation and more elimination, but the process is the same.
For this problem, x = 3, y = 3 and z = 3.