SOLUTION: 1.Solve the system of equations. x-2y+3z=7 2x+y+z=4 -3x+2y-2z=-10

Algebra ->  Coordinate Systems and Linear Equations -> SOLUTION: 1.Solve the system of equations. x-2y+3z=7 2x+y+z=4 -3x+2y-2z=-10      Log On


   

Question 1193249: 1.Solve the system of equations.
x-2y+3z=7
2x+y+z=4
-3x+2y-2z=-10
Found 2 solutions by MathLover1, MathTherapy:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

x-2y%2B3z=7........eq.1
2x%2By%2Bz=4........eq.2
-3x%2B2y-2z=-10........eq.3
_____________________

x-2y%2B3z=7........eq.1.......multiply by 2
2x%2By%2Bz=4........eq.2
_________________then subtract eq.2 from eq.1

2x-4y%2B6z-%282x%2By%2Bz%29=14-4
2x-4y%2B6z-2x-y-z=10
-5y%2B5z=10.........divide by 5
-y%2Bz=2...........solve for z
z=y%2B2..........eq.1a


2x%2By%2Bz=4........eq.2.......multiply by 2
-3x%2B2y-2z=-10........eq.3
_________________then subtract eq.3 from eq.2
4x%2B2y%2B2z-%28-3x%2B2y-2z%29=8-%28-10%29
4x%2B2y%2B2z%2B3x-2y%2B2z=8%2B10
7x%2B4z=18........solve for z
4z=-7x%2B18
z=-7x%2F4%2B18%2F4
z=-7x%2F4%2B9%2F2..........eq.2a

from eq.1a and eq.2a we have

y%2B2=-7x%2F4%2B9%2F2.......solve for y
y=-7x%2F4%2B9%2F2-2
y=-7x%2F4%2B5%2F2........eq.3a


go to

x-2y%2B3z=7........eq.1, substitute y and z from eq.3a and eq.2a
x-2%28-7x%2F4%2B5%2F2%29%2B3%28-7x%2F4%2B9%2F2%29=7.........solve for x
x%2B7x%2F2-10%2F2-21x%2F4%2B27%2F2=7
x%2B7x%2F2-10%2F2-21x%2F4%2B27%2F2=7..........multiply by 4 to get a rid of fractions
4x%2B14x-20-21x%2B54=28
-3x=28%2B20-54
-3x=-6
x=2

go to
y=-7x%2F4%2B5%2F2........eq.3a, substitute x
y=-%287%2A2%29%2F4%2B5%2F2

y=-7%2F2%2B5%2F2
y=-2%2F2
y=-1

go to

z=y%2B2..........eq.1a, substitute y
z=-1%2B2
z=1
solutions:
x=2
y=-1
z=1

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!

1.Solve the system of equations.
x-2y+3z=7
2x+y+z=4
-3x+2y-2z=-10
There she goes again!!
For your sanity you need to IGNORE that woman's attempt to solve this system.


   4x + 2y + 2z = 8 ------ Multiplying eq (ii) by 2 ------ eq (iv)
        5x + 5z = 15 ----- Adding eqs (iv) & (i) 
       5(x + z) = 5(3)_____x + z = 3 ------ eq (v)
       - 2x + z = - 3 -----Adding eqs (i) & (iii) ------ eq (vi)
             3x = 6 ------ Subtracting eq (vi) from eq (v)
             highlight_green%28matrix%281%2C5%2C+x%2C+%22=%22%2C+6%2F3%2C+%22=%22%2C+2%29%29

          2 + z = 3 ------- Substituting 2 for x in eq (v)
             z = 3 - 2 = 1
   
   2(2) + y + 1 = 4 ------- Substituting 2 for x and 1 for z in eq (ii)
      4 + y + 1 = 4
          y + 5 = 4
             y = 4 - 5 = - 1
 
   You can do the CHECKS!!