1
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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 '15 at 18:03
1
$\begingroup$

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.)

$\endgroup$
  • 1
    $\begingroup$ 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 '15 at 18:03
  • 1
    $\begingroup$ @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 '15 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 '15 at 18:13
1
$\begingroup$

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.

$\endgroup$
-1
$\begingroup$

You want the corresponding matrix to have a nonzero determinant

$$\det\begin{bmatrix}2 & -5\\ 4 & a\end{bmatrix} \not = 0$$

$$\Rightarrow 2a+ 20 \not = 0$$ $$\Rightarrow 2a \not = -20$$ $$\Rightarrow a \not = -10$$

$\endgroup$
  • $\begingroup$ This is incorrect; the coefficients are wrong. I downvoted, not because of that error, but because this solution has little bearing on the OP's line of thinking. For all we know, he's never worked with matrices before, in which case this is arcane. $\endgroup$ – Théophile Jun 11 '15 at 18:18
  • $\begingroup$ The typo has been fixed. I think it's poor form to downvote alternative answers simply because they employs an alternative method. I find it unlikely that it is beyond him, as he is working at around a precalculus level - I recall matrices being employed to solve linear systems in Algebra 2. $\endgroup$ – TokenToucan Jun 11 '15 at 19:29
  • $\begingroup$ That's fair. I reconsidered my downvote and tried to remove it, but the site won't let me. In general, I'm all for alternative methods. In questions like this, however, where the OP's question shows that he was stuck halfway towards the answer, I find that it's potentially very confusing to give an alternative method with no explanation why, and without addressing the OP's attempt. $\endgroup$ – Théophile Jun 12 '15 at 3:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy