SOLUTION: PLEASE HELP!!!!!!!!!!!!!!!!!!! SOLVE FOR X 2x + y + z = 0 3x - y + z = 3 7x - 5y - 3z = 15 thanks.

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Question 709251: PLEASE HELP!!!!!!!!!!!!!!!!!!!
SOLVE FOR X
2x + y + z = 0
3x - y + z = 3
7x - 5y - 3z = 15
thanks.
Found 2 solutions by KMST, hheuler:
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
system%282x+%2B+y+%2B+z+=+0%2C3x+-+y+%2B+z+=+3+%2C7x+-+5y+-+3z+=+15%29
I do not like the problem either, but no one answered yesterday, so I'll try today. (Warning: I make stupid mistakes in easy calculations, so watch for my mistakes)
Probably you want to use the elimination method, and not matrices.
You want to combine equations to get 2 combination equations with 2 variables.
(We still have a 3-variable-3-equation system, but we can set aside one of the original equations, with all 3 variables, as our third equation, and not worry about it for now).
Then, we would be able to solve the set of combination equations as a two-variable-two-equation system.

It looks like the easiest way would be to use the simpler first equation,
combining it with each of the other equations, to eliminate z.
Subtracting the first equation from the second one you get
x-2y=3
Adding 3 times the first equation to the third one you get
13x-2y=15
Now we solve
system%28x-2y=3%2C13x-2y=15%29
I would subtract the new system's first equation from the other equation to get
12x=12 --> highlight%28x=1%29
Substituting into x-2y=3 (rather than into 13x-2y=15 ) looks easy enough.
1-2y=3 --> -2y=3-1 --> -2y=2 --> -2y%2F%28-2%29=2%2F%28-2%29 --> highlight%28y=-1%29

Then, I would go back to one of the original equations and substitute x=1 and y=-1 to find z.
I chose 2x+%2B+y+%2B+z+=+0 as the equation that looks easiest.
2%281%29+%2B+%28-1%29+%2B+z+=+0 --> 2-1%2Bz=0 --> 1%2Bz=0 --> highlight%28z=-1%29
I verified on my own on scrap paper and it seems to work,
but remember, I warned you that I'm error prone.

Answer by hheuler(6) About Me  (Show Source):
You can put this solution on YOUR website!
eq 1 2x + y + z = 0
eq 2 3x - y + z = 3
Eq 3 7x - 5y - 3z = 15
Add eq 1 and eq 2: 5x + 2z = 3 and z = 1/2[3 - 5x] eq 4
Subtract eq 1 from eq 2; x - 2y = 3 and y = 1/2[x - 3] eq 5
Substitute eq 4 and eq 5 into eq 3 giving:
7x - 5[1/2(x - 3)] - 3[1/2(3 - 5x)] = 15
Multiply both sides of this equation by 2 to eliminate the fractions and simplify the resulting equation by adding like terms. Isolate the variable x to the left side of the equation.
14x - 5x + 15 - 9 + 15x = 30
24x = 24
x = 1
From eq 4 we have z = 1/2(3 - 5(1)] = -1
From eq 5, we have y = 1/2[1 - 3] = -1
The solution is (x, y, z) = (1, -1, -1).