Solve a system in three variables

Question

2x-4y-2z=-6

3x+2y-z=-17

4x-3y-2z=-14

Answer 1

If you haven't learned to use matrices for this type of problem, here's another way to solve it.

 

2x - 4y - 2z = -6
3x + 2y - z = -17
4x - 3y - 2z = -14

 

 

First, solve that first equation for x:

 

2x = -6 + 4y + 2z

x = -3 + 2y + z

 

 

Plug that expression for x into the second and third equations:

 

3(-3 + 2y + z) + 2y - z = -17

-9 + 6y + 3z + 2y - z = -17

8y + 2z = -8

4y + z = -4

z = -4y - 4

 

4(-3 + 2y + z) - 3y - 2z = -14

-12 + 8y + 4z - 3y - 2z = -14

5y + 2z = -2

 

 

Now you've essentially got a new system of two equations:

 

z = -4y - 4

5y + 2z = -2

 

 

Plug that expression for z into the new second equation to solve for y:

 

5y + 2(-4y - 4) = -2

5y - 8y - 8 = -2

-3y - 8 = -2

-3y = 6

y = -2

 

 

Plug that value for y back into the new first equation to solve for z:

 

z = -4y - 4

z = -4(-2) - 4

z = 8 - 4

z = 4

 

 

Finally, plug those values for y and z back into an original equation to solve for x:

 

x = -3 + 2y + z

x = -3 + 2(-2) + (4)

x = -3 - 4 + 4

x = -3

Answer 2

2X - 4Y -2Z = -6

3X + 2Y -Z = -17

4X - 3Y -2Z = -14

 

Let's set up a matrix using the coefficients and constants (Gaussian elimination) remembering that the last column is the far side of the equal sign and the other columns are the coefficients of the variables X, Y, Z

 

2  -4   -2    -6

3   2   -1    -17

4  -3   -2    -14

 

We want to get a matrix that looks like

1  0   0   C₁

0  1   0   C₂

0  0   1   C₃

where the C_n are just constants (regular numbers)

We can do this by multiplying or dividing rows by numbers and adding and subtracting rows;

 

I can see that I can eliminate the Z variable from the 1st row if I multiply row 1 by -1 and add it to row 3

 

-2    4   2     6       [row 1 times -1]

 3    2  -1   -17

 4   -3  -2   -14

 

Now add row 1 and row 3 and replace the 1st row with that

 

2    1    0   -8     [row 1 + row 3]

3    2   -1  -17

4   -3  -2  -14

 

Let's multiply row 2 by negative 2 so that we can add it to row 3 and again eliminate the Z variable

 

  2    1    0    -8

-6   -4    2     34     [row 2 times 2]

 4   -3   -2    -14

 

Now replace row 2 with row 2 plus row 3

 

 2    1    0    -8

-2   -7   0    20  [row 2 + row 3]

 4   -3  -2   -14

 

Now add row 1 and row 2 to eliminate the x variable

 

 2    1    0   -8

 0   -6    0   12   [row 1 + row 2]

 4   -3   -2  -14

 

Divide row 2 by -6 to solve for Y

 

2    1    0    -8

0    1    0    -2    [row 2 divided by -6 showing that 1Y = -2 or Y = -2]

4   -3   -2   -14

 

We can use the value of Y from row 2 to solve for X in row 1

 

2X + 1Y + 0 = -8

2X + Y = -8

2x -2 = -8

2X = -6

X = -3

 

We now have the value of X and Y both giving us

 

1   0    0    -3

0   1    0    -2

4  -3   -2   -14

 

The only equation left is 4X - 3Y = 14 so substitute the values X = -3 and Y = -2 in giving

 

4(-3) -3(-2) -2Z = -14

-12 + 6 -2Z = -14

-6 -2Z = -14

-2Z = -8

Z = 4

 

So we have X = -3, Y = -2, Z = -4