solve system of 3 equations with elimination

Question

-4x-5y-z=18 -2x-5y-2z=12 -2x+5y+2z=4

Answer 1

Let -4x-5y-z=18 ··· (1), -2x-5y-2z=12 ··· (2) and -2x+5y+2z=4 ··· (3)

Each of 3 equations has its own 5y-term, so we eliminate y-terms first.

1. Subtract (2) from (1): -4x-(-2x)-5y-(-5y)-z-(-2z)=18-12.  We have: -2x+z=6 ··· (4)

2. Add (3) to (1): -4x+(-2x)-5y+(5y)-z+(2z)=18+4.  We have: -6x+z=22 ··· (5)

3. Subtract (5) from (4) to eliminate z: -2x-(-6x)+z-(z)=6-22, that is: 4x=-16.  We have: x=-4

4.Substitute x=-4 for x in (4): -2(-4)+z=6, that is: 8+z=6.  We have: z=-2

5. Plug x=-4 and z=-2 into (1): -4(-4)-5y-(-2)=18, that is: 16-5y+2=18, so -5y=0.  We have: y=0

Here, we have: x=-4, y=0 and z=-2

6. CK: Plug these values into the left side of (2): -2(-4)-5(0)-2(-2)=12, so LHS=RHS  CKD.

The answer is: (x,y,z)=(-4,0,-2)

 

 

Answer 2

Let -4x-5y-z=18···(1), -2x-5y-2z=12···(2), and -2x+5y+2z=4···(3)

Each of 3 equations has its own 5y-term, so we eliminate y-terms first.

1. Subtract (2) from(1): -4x-(-2x)-5y-(-5y)-z-(-2z)=18-12.  We have -2x+z=6···(4)

2. Add (3) to (1): -4x+(-2x)-5y+(5y)-z+(2z)=18+4,  We have: -6x+z=22···(5)

3. Subtract (5) from (4) to eliminate z: -2x-(-6x)+z-(z)=6-22, that is: 4x=-16.  We have: x=-4

4. substitute x=-4 for x in (4): -2(-4)+z=6, that is 8+z=6.  We have: z=-2

5. Plug x=-4 and z=-2 into (1): -4(-4)-5y-(-2)=18, that is: -5y=0.  We have: y=0

Here, we have: x=-4, y=0 and z=-2

CK: Plug these values into the left side of (2): -2(-4)-5(0)-2(-2)=12, so LHS=RHS  CKD.

The answer is: (x,y,z)=(-4,0,-2)

Answer 3

Given -4x-5y-z=18.....(1)
-2x-5y-2z=12......(2)
-2x+5y+2z=4......(3)

Subtract equation(2) from equation(1)
-2x+z=6···(4)

Add equation(3) to equation(1)
-6x+z=22···(5)

Subtract equation(5) from equation(4) to eliminate z
4x=-16
x=-4

Substitute x=-4 for x in equation(4)
8+z=6
z=-2
and y=0

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