Linear systems with 3 variables

Question

3x + y - 4z = -22

2x + 3y + 4z = 32

x - y + 2z = 16

Answer 1

Hi Alyssa

Linear systems with 3 variables

3x + y - 4z = -222x + 3y + 4z = 32x - y + 2z = 16

 Write and number each equation.

Eq 1        3x+y-4z = 16

Eq 2   -222x+3y+4z = 16

Eq 3      32x-y+2z = 16

    Eliminate one variable by adding  equations 1, 3.  Number the result Eq 4

 Eq 1      3x+y-4z=16

Eq 2   + 32x-y+2z=16

          ---------------------        

Eq 4      35x-2z= 32

     Eliminate the same variable using equations 1, 2. You shall have to multiply Eq 1

by -3 to eliminate the same variable y; number this result equation 5. Now add Eq 2, Eq5, number the result Eq 6.

 Eq 1      3x+y-4z=16         (-3)    Eq 5   -9x-3y+12z=-48

Eq 2   -222x+3y+4z = 16             Eq 2   -222x+3y+4z = 16

                                                     --------------------------------

                                                     Eq 6 -231x+16z=-32

 We now have two equations with 2 variables; Eq 4, 6, we shall now eliminate one variable by multiplying Eq 4 by 8, then adding Eq 4, 6. 

Eq 4      35x-2z= 32   (8)   280x-16z = 256        x = 224/49 

Eq 6     -231x+16z=-32      -231x+16z= -32             

                                             --------------------

                                             49x = 224

  We use the value of x in equation 4 to calculate the value of z

Eq 4      35x-2z= 32   35(224/49) – 2 z = 32    7840/49- 2 z = 32    -2z = -7840/49 + 32

-2z = -160 + 32     z = (-128/-2)   z = 64

We now use Eq 1 to calculate the value of y

Eq 1        3x+y-4z = 16   3(224/49)+y-4(64) = 16    672/49+y-256 = 16    y = 256 + 16 -672/49

y = 272 – 672/49     y = (13328 -672)/49     y= 12656/49

Both x, y are improper fractions that cannot be simplified, but verification shows they are indeed correct (you should always verify your answers)

Verification

3x + y - 4z = -222x + 3y + 4z = 32x - y + 2z = 16

3(224/49)+12656/49–4(64) = -222(224/49) + 3(12656/49)+4(64)=32(224/49)-12656+2(64)=16

(672+12656)/49-256 = (-49728+37968)/49+256 = 7168/49-12656/49+128 =16

13328/49-256 = -11760/49+256 =-5488/49+128 = 16

272-256 = -240+256 = -112+128=16         

Sincerely Pete

Answer 2

Alyssa,

When you look at problems with more than one variable, your first instinct should be to put one of the variables in terms of the other(s).

My first reaction was to notice how easy it would be to do this with the third equation.

x - y + 2z = 16

x = y - 2z + 16

Now that I have x in terms of y and z, I can rewrite the first equation:

3(y - 2z + 16) + y - 4z = -22

3y - 6z + 48 + y - 4z = -22

4y - 10z = 26

And I will rewrite my second equation:

2(y - 2z + 16) + 3y + 4z = 32

2y - 4z + 32 + 3y + 4z = 32

5y = 0

y = 0

Now, I can go back to my first equation, knowing that y is 0.

4(0) - 10z = 26

z = -2.6

Last thing, solve for x:

x = y - 2z + 16 = 0 - 2(-2.6) + 16 = 21.2

Again, the first thing you do should always be to get one of the variables in terms of the others. Then rewrite the equations plugging in your new way of writing that variable. This will bring you to two equations with 2 variables, which is easier to solve. In this case, y was 0, which made things easier than usual.

Hope this helps! :)