SOLUTION: x + 2y − z = −5 y − 3z = −13 x − 3y + 2z = 12

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Question 1197422: x + 2y − z = −5
y − 3z = −13
x − 3y + 2z = 12
Found 5 solutions by josgarithmetic, MathLover1, greenestamps, Alan3354, MathTherapy:
Answer by josgarithmetic(39618) About Me  (Show Source):
You can put this solution on YOUR website!
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x + 2y − z = −5
y − 3z = −13
x − 3y + 2z = 12
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What skill or method are you required to use?
You could use E1 and E3 to find an equation in variables y and z, and then you would have a system in just two variables, y and z.

--

system%28x=-2y%2Bz-5%2Cx=3y-2z%2B12%29

-2y%2Bz-5=3y-2z%2B12
-5y%2B3z=17
5y-3z=-17

System in two variables is
system%28y-3z=-13%2C5y-3z=-17%29

system%28y-3z=-13%2C-5y%2B3z=17%29

Add corresponding members, eliminating z.
-4y=4
highlight%28y=-1%29

Using from the simpler system OR second equation of original system, y-3z=-13
-1-3z=-13
-3z=-12
highlight%28z=4%29

and you can find x.

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

x+%2B+2y+-+z+=+-5 .........eq.1
y+-+3z+=-13..............eq.2
x+-+3y+%2B+2z+=+12............eq.3
________________________________
start with
y+-+3z+=-13..............eq.2, solve for y

y++=3z-13..............eq.2a
go to
x+%2B+2y+-+z+=+-5 .........eq.1, substitute y
x+%2B+2%283z-13%29-+z+=+-5
x+%2B+6z-26-+z+=+-5
x+%2B+5z+=+-5%2B26 ........., solve for x
x+=+21-5z...........eq.1a
go to
x+-+3y+%2B+2z+=+12............eq.3, substitute x and y
21-5z+-+3%283z-13%29+%2B+2z+=+12........., solve for z
21-5z+-+9z%2B39%2B+2z+=+12
-12z%2B60+=+12
-12%2B60+=+12z
48+=+12z
z=48%2F12
z=4
go to
y++=3z-13..............eq.2a, substitute z
y++=3%2A4-13
y++=12-13
y++=-1
go to
x+=+21-5z...........eq.1a, substitute z
x+=+21-5%2A4
x+=+21-20
x+=+1
solution:
x+=+1
y++=-1
z=4


Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


There are many different ways to solve a system of three equations in three variables....

Given this particular system, I would solve the second equation for y: y-3z=-13 --> y=3z-13

Then substitute "3z-13" for "y" in the other two equations to get a system of two equations in two variables.

Then solve that pair of equations by your favorite method.

Note that in your post you didn't actually ask a question -- although it is obvious what was to be done.

But, because there are many different methods for solving systems of equations, we don't know if any response we give helps you. Your post should either specify a method for solving the system, or else specifically say that any method can be used.

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
You didn't ask a question or make a request.

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!
x + 2y − z = −5
y − 3z = −13
x − 3y + 2z = 12
     x + 2y −  z = −  5 ----- eq (i)
          y − 3z = − 13 ----- eq (ii)  
     x − 3y + 2z =   12 ----- eq (iii)

Looking at the equations it's clear that there're only 2 variables (y & z) in eq (ii),
It's also clear that x in eqs (i) & (iii) can be 
eliminated by subtracting 1 equation from the other.
x + 2y −  z = −  5 --- eq (i)
x − 3y + 2z =   12 --- eq (iii)
    5y - 3z = - 17 --- Subtracting eq (iii) from eq (i) ----- eq (iv)
     y − 3z = − 13 --- eq (ii)  
   5y - 15z = - 65 --- Multiplying eq (ii) by 5 ----- eq (v)
        12z = 48 ----- Subtracting eq (v) from eq (iv)
         highlight_green%28matrix%281%2C5%2C+z%2C+%22=%22%2C+48%2F12%2C+%22=%22%2C+4%29%29
     
   y - 3(4) = - 13 ----- Substituting 4 for z in eq (ii)
     y - 12 = - 13
         y = - 13 + 12 = - 1

x + 2(- 1) - 4 = - 5 --- Substituting 4 for z and - 1 for y in eq (i)
     x - 2 - 4 = - 5
            x = - 5 + 6 = 1