I would use elimination.
4x + 4y + 2z = -10 ----- 1st times 2
3x + 4y + 2z = 0 ----- #2 as is
subtract them
x = -10 ------ that was lucky
3x + 4y + 2z = 0 --- #2
x + 3y + 2z = 1 ----#3
subtract them
2x + y = -1
sub x = -10 into that
-20 + y = -1
y = 19
back into #1
2x+2y+z = -5
-20 + 38 + z = -5
z = -23
Explain how to solve the following system of equations. What is the solution to the system?
2x+2y+z=-5
3x+4y+2z=0
x+3y+2z=1
1 answer
1 answer