How do you find the 3 variables?

Question

-3x - 2y + 3z = 36
-3x + 3y + 2z = 31
1x - 1y + 2z = 3

Answer 1

Add first two equations to elimitate x

-(-3x-2y+3z=36)+
-3x+3y+2z=31

to get

y - z = -5

add eq 2 + 3* eq 3 to get
-3x+3y+2z=31
3x-3y+6z=9

to get

8z = 40 ==> z = 40/8 = 5

then from y - z = -5 ==> y - 5 = -5 ==> y = 0

then choose any equation to get x

1x -1y + 2z = 3

x - 0 + 10 = 3 ==> x = - 7

x= -7, y = 0, z = 5

Answer 2

From the third equation you get:
x = 3 + y - 2z and you put it into second one:

-3 (3 + y - 2z) + 3y + 2z = 31
- 9 - 3y + 6z + 3y + 2z = 31
8z - 9 = 31
8z = 31+9
8z = 40
z = 40/8
z = 5

Than, you put everything in the first equation:

- 3 (3 + y - 2*5) - 2 y + 3*5 = 36
- 3 ( y -7) - 2y + 15 = 36
- 3y + 21 - 2y + 15 = 36
- 5y = 36 - 21 - 15
- 5y = 0
y = 0

And since x = 3 + y - 2z, than x = -7

Answer 3

Let's start with the first two equations:
-3x-2y+3z=36
-3x+3y+2z=31

Multiply the second equation by -1:
-3x-2y+3z=36
3x-3y-2z=-31
===========
-5y+z=5

Let's do the next two equations:
-3x+3y+2z=31
x-y+2z=3

Multiply the second equation by 3:
-3x+3y+2z=31
3x-3y+6z=9
===========
8z=40
z=5

-5y+5=5
-5y=0
y=0

x-0+2(5)=3
x+10=3
x=-7

The solution set is (-7,0,5)

Check:
-3(-7)-2(0)+3(5)=36
21+15=36
36=36

-3(-7)+3(0)+2(5)=31
21+10=31
31=31

-7-0+2(5)=3
-7+10=3
3=3

I hope this helps!

Answer 4

The same way(s) as you would find two if you had a system with two unknowns and two equations.

In this case, reduce from 3 to 2:

equation 1 minus equation 2:

-3x - 2y + 3z = 36
+3x - 3y - 2z = -31
_____________
0 x - 5y + z = 5

equation 1 plus 3 times equation 3:

-3x - 2y + 3z = 36
3x - 3y + 6z = 9
___________
0x - 5y +9z = 45

We are left with 2 unknowns (y and z) and two equations (which I have rewritten):

-5y = 5 - z
-5y = 45 - 9z

-5y = -5y, therefore

5 - z = 45 - 9 z
8z = 40
z = 5

From there, work backwards to find the other variables.

Answer 5

-3x-2y+3z+36.....(1)
-3x+3y+2z=31.....(2)
1x-1y+2z=3..........(3)
subtracting (2) from (1),we get
-5y+z=5.........(4)
Multiplying equation 3 by 3 and adding it toeqn 2
8z=40
z=5
Putting the value of z in eqn 4
-5y+5=5
=> y=0
Putting the values of y and z in eqn 3
x-0+10=3
=>x= -7