Question

Quick help: 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

Answer 1

-3x - 4y - 2z = 0
x + 3y + 2z = 1

-2x - y = 1

-4x - 4y - 2z = 10
3x + 4y +2z= 0

-x = 10. x = -10

-2(-10) - y = 1
20 - y = 1
-y = -19
y = 19

-10 + 3(19) + 2z = 1
-10 + 57 + 2z = 1
47 + 2z = 1
2z = -46
z= -23

check: 2(-10)+2(19)-23=-5
-20+38-23=-5
-43+38=-5
-5=-5

Answer 2

Given equations:
2x+2y+z = -5 -----------(1)
3x+4y+2z = 0 ----------(2)
x+3y+2z = 1 ---------(3)
Subtract equation (3) from equation (2) to eliminate z
3x+4y+2z =0
-x-3y-2z=-1
__________
2x+y=-1 -------------------(4)
Multiply equation (1) by 2 and subtract it from equation (3) to eliminate z
x+3y+2z=1
-4x-4y-2z=10 ____________
-3x-y=11 --------(5)
Add equation (4) and (5) to eliminate y
2x+y=-1
-3x-y=11 _______
-x=10
X=-10
Substitute the value of x in equation (4) to find y
2(-10)+y=-1
-20+y=-1
y=-1+20
y=19
Substitute the values of x and y in any one of the 3 equations, to find z.
Let’s substitute in equation (1)
2(-10)+2(19)+z=-5
-20+38+z=-5
18+z=-5
z=-5-18
z=-23
Therefore the solution is: x=-10, y=19, z=-23