SOLUTION: Hello can you help me with this problem? It says you have to use the subsitution method to solve a system of equations that has three variables. 6x-4y+5z=31 5x+2y+2z=13 x+y+z=

Algebra ->  Coordinate Systems and Linear Equations  -> Lessons -> SOLUTION: Hello can you help me with this problem? It says you have to use the subsitution method to solve a system of equations that has three variables. 6x-4y+5z=31 5x+2y+2z=13 x+y+z=      Log On


   

Question 766025: Hello can you help me with this problem? It says you have to use the subsitution method to solve a system of equations that has three variables.
6x-4y+5z=31
5x+2y+2z=13
x+y+z=2
x=?
y=?
z=?
Found 2 solutions by rothauserc, Edwin McCravy:
Answer by rothauserc(4718) About Me  (Show Source):
You can put this solution on YOUR website!
6x-4y+5z=31
5x+2y+2z=13
x+y+z=2
========================================================================
x+y+z=2
solve for z
z=2-x-y
==================================================================
substitute for z in second equation
5x+2y+2*(2-x-y)=13
5x+2y+4-2x-2y=13
3x=9
x=3
====================================
substitute for z and x in equation 1
6*3 -4y +5*(2-3-y)=31
18 -4y +10 -15 -5y = 31
-9y +13 = 31
-9y = 18
y = -2
========================================
substitute x=3 and y=-2 into third equation
3-2+z=2
1+z=2
z=1
=============================================
check values in first, second and third equations
6*3 -4*(-2) +5*1 = 31
18 +8 +5 = 31
31 = 31 equation 1 checks
++++++++++++++++++++++++++++++++++++++++
5*3 +2*(-2) + 2*1 = 13
15 -4 +2 = 13
13 = 13 equation 2 checks
+++++++++++++++++++++++++++++++++++++++++
3 -2 + 1 = 2
2 = 2 equation 3 checks
===============================================
x=3, y=-2, z=1

Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
Use the substitution method to solve a
system of equations that has three
variables. 
[1]           6x - 4y + 5z = 31
[2]           5x + 2y + 2z = 13
[3]            x +  y +  z =  2

Solve [3] for one of the letters.
I'll pick z

[3]            x +  y +  z =  2
[4]                      z =  2 - x - y

Using [4], Substitute 2 - x - y 
for z in [1]

[1]           6x - 4y + 5z = 31
    6x - 4y + 5(2 - x - y) = 31
    6x - 4y + 10 - 5x - 5y = 31
               x - 9y + 10 = 31
[5]                 x - 9y = 21
 
Using [4], Substitute 2 - x - y 
for z in [2]

[2]           5x + 2y + 2z = 13   
    5x + 2y + 2(2 - x - y) = 13
     5x + 2y + 4 - 2x - 2y = 13
                    3x + 4 = 13
                        3x = 9
[6]                      x = 3
Using [6], Substitute 3 
for x in [5]

[5]                 x - 9y = 21
                    3 - 9y = 21
                       -9y = 18
[7]                      y = -2

Using [6] and [7], Substitute 3 
for x and -2 for y in [5] in [4]

[4]                      z = 2 - x - y
                         z = 2 - 3 - (-2)
                         z = 2 - 3 + 2
[8]                      z = 1

Answer: 

[6]                      x = 3
[7]                      y = -2
[8]                      z = 1

Edwin