Question 174522: Solve for X, Y, and Z in the following equations:
c. 22X + 5Y + 7Z = 12
10X + 3Y + 2Z = 5
9X + 2Y + 12Z = 14
I added all three to get 41x + 10y + 21z=31..then subtracted equation 2 from 1 to get 12x + 2y + 5z=7...then multiplied by 7 getting 84x + 14y + 35z=49...and now I'm stuck...any help would be appreciated:)
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! Solve for X, Y, and Z in the following equations:
22X + 5Y + 7Z = 12
10X + 3Y + 2Z = 5
9X + 2Y + 12Z = 14
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To use elimination methods, you have to pick one of the 3 variables to be eliminated that will give you 2 eqns in 2 unknowns. In these 3, y has the smallest coefficients, so I'd pick it and use the 3rd eqn to subtract from the other two, since its coeff is 2.
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Multiply eqn 3 by 5 and eqn 1 by 2
44x + 10y + 14z = 24 from eqn 1
45x + 10y + 60z = 70 from eqn 3. Subtract 1 from 3
x + 46z = 46 That's one.
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Now multiply eqn 3 x3 and eqn 2 x2
20x + 6y + 4z = 10 from eqn 2
27x + 6y + 36z = 42 from eqn 3 Subtract 2 from 3 (to give pos coeffs)
7x + 32z = 32 That's the 2nd eqn in x and z only.
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7x + 32z = 32 Multiply the other one by 7
7x + 322z = 322
Subtract
-290z = -290
z = 1
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Sub into either of the 2 variable eqns
7x + 32*1 = 32
7x = 0
x = 0
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Now sub into the original eqn 3
9X + 2Y + 12Z = 14
9*0 + 2y + 12*1 = 14
2y = 2
y = 1
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So it's 0, 1, 1 for x, y, z
The important thing to remember is to eliminate one variable first, then another, etc.
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