# Solve the following systems of linear equations, showing all work: 2x+y-z = 14 x+2y-z = 12 2x-2y+z = 3

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*2x+y-z = 14*

*x+2y-z = 12*

*2x-2y+z = 3*

Both of the previous answers are helpful but they are also different answers. You have to decide which value of z is correct.

The best way to check your answers is to plug in the values and see which one works for ALL THREE EQUATIONS, not just one!

When I use 5, 3, and 1 I get...

2(5) + 3 - 1 = 14

10 + 3 - 1 = 14

12 does not equal 14

These values don't work for the equations.

On the other hand, when I try 5, 3, and -1, I get the following:

2(5) + 3 - (-1) = 14

10 + 3 + 1 = 14

14 = 14

5 + 2(3) -(-1) = 12

5 + 6 + 1 = 12

12 = 12

2(5) - 2(3) + (-1) = 3

10 - 6 - 1 = 3

3 = 3

Obviously, these are the correct answers.

To solve :

2x+y-z = 14...............(1)

x+2y-z = 12...............(2)

2x-2y+z = 3...............(3)

Solution:

We eliminate Z by the proces: eq(1)-eq(2) and eq(2)+eq(3):

eq(1)-eq(2) gives: 2x+y-z-(x+2y-z) = 14-12=2 Or x-y=2............(3)

eq(2)+eq(3) gives: x+2y-z+(2x-2y+z)=12+3. Or

3x=15. Or

2*5+3-z =14. So 13-12 =z Or z =1. So ,

x=5,

y=3 and

z=1 is the solution.

The given simultaneous equations are:

2x + y - z = 14 ... (1)

x + 2y - z = 12 ... (2)

2x - 2y + z = 3 ... (3)

Adding equation (2) and (3) we get:

x + 2x + 2y - 2y - z + z = 12 + 3

3x = 15

Therefore:

x = 15/3 = 5

Substituting this value of x in equations (1) we get:

10 + y - z = 14

y - z = 14 - 10 = 4 ... (4)

Substituting this value of x in equations (3) we get:

10 - 2y + z = 3

- 2y + z = 3 - 10 = - 7 ... (5)

Adding equations (4) and (5) we get:

y - 2y - z + z = 4 - 7

- y = - 3

y = 3

When we substitute value of x and y in equations (1), (2) and (3) we get different values of z.

Substituting values of x and y in equation (1) we get:

10 + 3 - z = 14

- z = 14 - 10 - 3 = 1

Substituting values of x and y in equation (2) we get:

5 + 6 - z = 12

-z = 1

z = -1

Substituting values of x and y in equation (3) we get:

10 - 6 + z = 3

z = 3 - 10 + 6 = -1

As we get different value of zwhen substituting in different e equation, thegiven equations are not consistent with each other. Thus there no unique values of x, y, and z can be determined using these three equations as simultaneous equations.

this is how I would do it

2x+y-z=14

2x+y-(lets say z=1)=14

2x+y-1=14

2x=15y

x=7.5y ok now we know x=7.5 so

2x+y-z=14

2(7.5y)+y-z=14

15y+y-z=14

16y=14z

y=0.875 or 7/8

2x+y-z=14

2(7.5Y)+(7/8)-z=14

2(7.5(7/8))+7/8-z=14

that is a far as i can go for now hope it helps