SOLUTION: I need help working this problem out. I have no idea what I am doing. 4a + 2b - 6c = 2 6a + 3b - 9c = 3 8a + 4b - 12c = 6

Algebra ->  Coordinate Systems and Linear Equations  -> Lessons -> SOLUTION: I need help working this problem out. I have no idea what I am doing. 4a + 2b - 6c = 2 6a + 3b - 9c = 3 8a + 4b - 12c = 6      Log On


   



Question 538167: I need help working this problem out. I have no idea what I am doing.
4a + 2b - 6c = 2
6a + 3b - 9c = 3
8a + 4b - 12c = 6

Answer by jpg7n16(66) About Me  (Show Source):
You can put this solution on YOUR website!
The idea behind a more complicated problem like this is to make things as easy as possible. For instance, start reducing everything. Equation 1 is all multiples of 2. #2 is multiples of 3. And #3 is multiples of 4 (except the 6) - but it's worth reducing anyways.
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The ultimate goal is to eliminate one/two of the variables to solve for the rest.
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Step 1: reduce all the coefficients down by dividing by 2, 3 and 4 respectively
1) 2a%2Bb-3c=1
2) 2a%2Bb-3c=1
3) 2a%2Bb-3c=3%2F2
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And something's wrong with this problem. 2 equations are NEVER supposed to be identical after reducing. And you'll notice #1 and #2 are equal. (So one of them is irrelevant)
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But worse than that, we know that if a=b and a=c then b=c, right? Well according to your problem:
Since 2a%2Bb-3c=1 and 2a%2Bb-3c=3%2F2, then 1=3%2F2 - which is def not true.
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So there you have it. There's something wrong with the problem! It's not your fault :-)