Question

Explain how to solve the following system of equations. What is the solution to the new system?

2x + 2y + z = -5

3x + 4y + 2z = 0

x + 3y + 2c = 1

Answer 1

I will assume you have a typo in the last equation, and that 2c is 2z

double the first:
4x + 4y + 2z = -10 ----> *
subtract the end:

x = -10 , well , that was lucky

4x + 4y + 2z = -10
x + 3y + 2c = 1
subtract them:
3x + y = -11
sub in x = -10
-30 + y = -11
y = 19

back in the original first:
-20 + 38 + z = -5
z = -23

I tested the values, they work in all 3 equations

Answer 2

I did make a typo lol

And thanks a ton, very nicely structured too! Thank you for the help.