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This is most likely a pretty simple problem although my textbook doesn't quite explain how to solve it. I have a linear system with two equations and two variables (x and y) below:

2x - 5y = 9

4x + ay = 5 ('a' being the unknown coefficient of y)

The question asks to find the value of a that will make this system have a single solution. At first this seemed trivial, although I tried 1) multiplying the top equation by -2 and then adding the two equations to eliminate x. I then got stuck at 10y + ay = -4. And 2) I tried only multiplying the top by 2 to get the 4x terms in both equations, in which case I thought that I could somehow remove the x terms out of both equations entirely but I don't think that's correct (unless I do it to both sides which doesn't help me much?). Thanks in advance for any help!

I also wasn't able to think of any suitable tags for this question except "systems of equations" (which is supposed to be used on conjunction with a more specific tag). Couldn't find "linear equations" or other more suited tags.

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    $\begingroup$ Both your methods work, by the way. In your second attempt, after multiplying by $2$, you then have to subtract one equation from the other to get rid of the $x$, rather than adding. This effectively amounts to the same thing as multiplying by $-2$ and adding. $\endgroup$
    – Théophile
    Jun 11, 2015 at 18:03

2 Answers 2

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You're on the right track. One small mistake so far: you forgot to multiply the $9$ in the first equation by $-2$, so you should end up with $10y+ay=-13$. Now, if there were a unique solution, then we would be able to solve for $y$, thus: $$y(10+a) = -13\\ y = \frac{-13}{10+a}$$

What does that say about restrictions on $a$? (You should be able to see that there is one forbidden value. As an experiment, try letting $a$ equal that value, then graph both equations.)

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    $\begingroup$ Ah, thanks! So the coefficient a can be anything but -10? So there is no way to find a unique solution for this system? Ah cool, just graphed them, a value of -10 for will make the slopes of the lines equal so there is no solution, anything else results in a unique solution. $\endgroup$
    – user90572
    Jun 11, 2015 at 18:03
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    $\begingroup$ @MagnusQ.: Good work! The restriction is that $a \neq -10$. But for any other value of $a$, you can find a unique solution, because you have $y=\frac{-13}{10+a}$, and from there you can find $x$ using either of the initial equations. $\endgroup$
    – Théophile
    Jun 11, 2015 at 18:10
  • $\begingroup$ @MagnusQ.: Try, for example, setting $a=3$ (this will make the fraction cancel out nicely) and solving algebraically for $x$ and $y$. $\endgroup$
    – Théophile
    Jun 11, 2015 at 18:13
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The only thing that would keep the system from yielding a single value is if $2a+20=0$ or $a=-10$. In that case the loci of the two equations would be parallel. No intersections means no solutions.

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