Three variable equations

Question

5x+3y+z=0

 

x-3y+2z=9

 

14x-2y+3z=1

Answer 1

We will use the same elimination and substitution technique when dealing with two variables.

 

 

  5x + 3y + z = 0              eq1

    x - 3y + 2z = 9             eq2

14x - 2y + 3z = 1             eq3

 

We will solve for z from eq1.

 

z = -5x - 3y

 

Substitute this z value into eq2 and eq3.

 

x - 3y + 2z = 9                            eq2

x - 3y + 2(-5x - 3y) = 9

x - 3y -10x - 6y = 9

-9x - 9y = 9

 

14x - 2y + 3z = 1                         eq3

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

14x - 2y -15x - 9y = 1

-x - 11y = 1

 

Our new eq2 and eq3 is:

 

-9x - 9y = 9            eq2

  -x - 11y = 1          eq3

 

Next, we multiply eq3 by 9.

 

-9x - 9y = 9         

-9x - 99y = 9

 

Now we can subtract these new equations to eliminate the x terms.

 

90y = 0

   y = 0

 

Lets substitute this y value into eq2.

 

-9x - 9y = 9

-9x - 9(0) = 9

-9x = 9

   x = -1

 

Now substitute these x and y values into eq1.

 

z = -5x - 3y

z = -5(-1) - 3(0)

z = 5

 

Your solutions are

 

x = -1

y = 0

z = 5