# Stuck on this Algebra 2 question!

what is the solution of the system of equations?
2x+2y+3z=-6
3x+5y+4z=3
2x+3y+4z=-10

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2. 👎 0
3. 👁 63
1. 2x+2y+3z=-6 times 3
3x+5y+4z=3 times 2

6 x + 6 y + 9 z = -18
6 x + 10y + 8 z = 6
-----------------------subtract
0 x - 4 y + 1 z = -24

3x+5y+4z=3 times 2
2x+3y+4z=-10 times 3

6 x + 10y + 8 z = 6
6 x + 9 y + 12z = -30
------------------------subtract
0 x + 1 y - 4 z = 36
so

- 4 y + 1 z = -24 times 4
+ 1 y - 4 z = 36

-16y + 4 z = - 96
+ 1y - 4 z = 36
-15 y = -60
y = 4
now go back and get z and x

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2. 👎 0
posted by Damon
2. solve it using matrix inversion method.
AX=B, where A is the coefficient matrix
2 2 3
A= 3 5 4
2 3 4
X=[x y z]' and B =[-6 3 -10]'
then X=(A^-1)B,that is A inverse*B

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2. 👎 0
posted by gg

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