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Question 977830: Using elimination, what are the steps to figure out the unknowns in this linear equation:
3a+5b-c=47
2a-b+3c=-2
4a+b-2c=30
Answer by Cromlix(4381)   (Show Source):
You can put this solution on YOUR website! Hi there,
3a + 5b - c = 47 ....(1)
2a - b + 3c = -2 ....(2)
4a + b - 2c = 30 ....(3)
.............
Add (2) + (3)
2a - b + 3c = -2 ....(2)
4a + b - 2c = 30 ....(3)
6a + c = 28 ......(4)
.................
Multiply (2) x 5 and add to (1)
3a + 5b - c = 47 .....(1)
10a - 5b + 15c = -10...(2)
Add (1) + (2)
13a +14c = 37.....(5)
...............
Now consider Equations (4) and (5)
6a + c = 28 .....(4)
13a + 14c = 37 ...(5)
Multiply (4) by 14 and subtract (5)
84a + 14c = 392 ...(4)
13a + 14c = 37 ....(5)
..
71a = 355
a = 5 (Solution)
...
Substitute a = 5 into Equation(4)
6a + c = 28 ....(4)
6(5) + c = 28 ...(4)
30 + c = 28 .....(4)
c = 28 - 30
c = -2 (Solution)
..
Substitute a = 5 , c = -2 into
Equation (1)
3a + 5b - c = 47 ...(1)
3(5) + 5b - (-2) = 47....(1)
15 + 5b + 2 = 47 ....(1)
Collect like terms
5b = 47 - 17 .....(1)
5b = 30 .....(1)
b = 6 (Solution)
a = 5 : b = 6 : c = -2
Hope this helps:-)
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