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
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
2 answers
I did make a typo lol
And thanks a ton, very nicely structured too! Thank you for the help.
And thanks a ton, very nicely structured too! Thank you for the help.