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| SOLVING
A SYSTEM OF EQUATIONS WITH THREE UNKNOWNS |
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Step 5: Choose two equations
to work with and substitute back for z, and solve for y to get:
| (E1) |
x + 4y – 2z = 3 → |
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x + 4y – 2(0) = 3 → |
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x + 4y = 3 |
| (E2) |
x + 3y + 7z = 1 → |
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-x + 3y – 7(0) = -1 → |
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x + 3y = 1 → |
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-x – 3y = -1 |
x + 4y = 3
-x – 3y = -1
y = 2
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