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Let $a=\pmatrix{1\\1}, b=\pmatrix{1\\2}, c=\pmatrix{2\\1}$. Solve $xa + yb = c$ for $x, y \in \mathbb{C}$.

Does this mean I do $$1x+1y=2$$ $$1x+2y=1$$ and would be the final answer?

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    For some reason you have switched from $1x+1y=2$ to $1x+2y=1$. Why?2012-01-30
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    It's just basic algebra. If $xa+yb=c$, then $yb=c-xa$, then $y=\dfrac{c-xa}{b}$. So you have a straight line assuming b isn't zero.2012-01-30
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    I have changed the tag of Linear Algebra to Algebra-Precalculus. I don't see how this applies to Linear Algebra given the little information.2012-01-30
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    Its a matrix. That is why.2012-01-30
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    It is linear algebra but I dont know how to put it in matrix form form in the post I put2012-01-30
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    @KjX: I assume that the way I've rewritten your question is what you intended?2012-01-30
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    *and would be the final answer?* No. The answer would be the set of solutions. Here, there is exactly one solution (x,y) and your task is to compute it.2012-01-30

2 Answers 2

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Assuming Zev's edit is correct, then your equations $$1x+1y=2$$ $$1x+2y=1$$ are correct. To complete the problem, you need to find the values of $x$ and $y$ that make both of those equations true at the same time.

Hint: Subtract the first equation from the second equation.

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    Someone is reading meta these days :)2012-12-28
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$$1x+1y=2\cdots\cdots(1)$$ $$1x+2y=1\cdots\cdots(2)$$

Multiply (1) by -1 $$-1x-1y=-2 $$ Adding above and (2),

$$1x+2y -1x-1y=1-2$$ $$y=-1$$ Put it in (1) $$1x+1y=2$$ $$1x-1=2$$ $$x=3$$ Solution Set = (3,-1)

Verify this by putting in the 2 equations.