Question

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

Systems of equations
x-y+2x=3
2x+z=1
3x+2y+z=4

2020-11-10
We are given the system:
x—y+2z=3
2x+z=1
3x+2y+z=4
First, we eliminate y. Multiply (1) by 2 to obtain (4):
2x-2y+4z=6
Add (3) and (4) to obtain (5): 5x+5z= 10
or by dividing by 5, x+z=2
Next, we eliminate 2. Subtract (2) and (5) then solve for x:
x=-1
Solve for 2 using (5):
-l+z=2
z=3
Solve for y using (1):
—1-y+2(3)=3
-y+5=3
-y=-2
y=2
So, the solution of the system is:
(-1,2,3)