SOLUTION: {𝑥−2𝑦+3𝑧=9−𝑥+3𝑦−𝑧=−62𝑥−5𝑦+5𝑧=1 elimination

Question 1191478: {𝑥−2𝑦+3𝑧=9−𝑥+3𝑦−𝑧=−62𝑥−5𝑦+5𝑧=1 elimination
Found 2 solutions by josgarithmetic, ankor@dixie-net.com:
Answer by josgarithmetic(39616) (Show Source): You can put this solution on YOUR website!
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𝑥−2𝑦+3𝑧=9−𝑥+3𝑦−𝑧=−62𝑥−5𝑦+5𝑧=1
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What you have is completely ambiguous.
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
Ambiguous is right, use commas, brackets, or at least separate the equations:
:
Assume you mean
𝑥 − 2𝑦 + 3𝑧 = 9
−𝑥 +3𝑦 − 𝑧 =−6
2𝑥 −5𝑦 + 5𝑧 = 1
;
Often with these problems their is one operation that greatly simplifies this
You can see this by multiplying the 1st equation by -1
-𝑥 + 2𝑦 - 3𝑧 =-9
−𝑥 +3𝑦 − 𝑧 =−6
2𝑥 −5𝑦 + 5𝑧 = 1
-----------------------adding eliminates x and y
0x + 0y + z = -14
:
Replace z with -14 in the first two equations and add eliminating x,
x - 2y + 3(-14) = 9
-x +3y - (-14) = -6
:
x - 2y = 9 + 42
-x +3y = -6 -14
:
x - 2y = 51
-x +3y = -20
------------------adding eliminates x, find y
0 + y = 31
:
find x using the first equation
x - 2(31) + 3(-14) = 9
x - 62 - 42 = 9
x = 104 + 9
x = 113
:
The solution: x=113, y=31, z=-14
:
You should confirm this in the 2nd and 3rd original equations

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