-3a - 4b + 2c = 28
a + 3b - 4c = -31
2a + 3c = 11
I'm unsure of how to really work this because the very last equation doesn't have a b variable in it.
I do not know what methods your class is using. Do either of the names Substitution Method or Elimination Method sound familiar. You could use one or the other, or a combination of both.
We do not teach classroom material, so I'll give you some hints using the Elimination Method, and perhaps it will jog your memory. (Next time please say something about your situation. We like to know what you've tried or thought about thus far, or what your class examples look like.)
The Elimination Method involves adding equations in such a way as to eliminate one of the variables. We often need to first multiply one or both equations to be added by appropriate numbers to get coefficients to cancel when the resulting equations are added. Here's an example (using the last two equations) to obtain an equation in which variable a has been eliminated.
Code:
[FONT=courier new]
[COLOR=#3E3E3E] a + 3b - 4c = -31
[/COLOR]
2a + 3c = 11
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Now, if we were to add these two equations as is (left-hand side gets added to left-hand side, and right-hand side gets added to right-hand side), then we'd end up with 3a.
BUT, if we were to multiply the top equation by -2 first, then we'd have -2a + 2a when adding the equations, and so they'd cancel, and the resulting equation will have no symbol a (that's our goal -- to eliminate symbol a).
So, two steps.
(1) Multiply both sides of the top equation by -2
(2) Add the resulting equation to the bottom equation
[COLOR=#3E3E3E]-2a -6b + 8c = 62
[/COLOR]
2a + 3c = 11
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-6b + 11c = 73
[/FONT]
We got a new equation that contains only b and c.
Now, can you try similar steps to eliminate symbol a using the first and last equations?
You'll end up with another equation that contains only b and c.
Finally, solve your new system of two equations in b and c.
Once you know b and c, you can easily find a by substituting c into the last equation.
Please show your work, or ask a specific question, if you get stuck again.
Actually, I just realized that we'd save a step by eliminating b using the first two equation. The new system would be in a and c. Lot's of possible routes. See what you can do.
Cheers :cool:
