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$$3xy + 14x + 17y + 71 = 0$$

Need to find both $x$ and $y$. If there was only one variable then this is easy problem.
Have tried:

$$\begin{align}3xy &= -14x - 17y - 71 \\ x &= \frac{-14x - 17y - 71}{3y}\end{align}$$

Then tried to put this expression everywhere instead of $x$ but it tooks forever to find both $x$ and $y$.
I don't even know how to get on right track. Please give any advice. Thanks.

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    $\begingroup$ The solution is not unique. $\endgroup$
    – azarel
    Jul 26, 2012 at 16:10
  • $\begingroup$ Solve how? One solution in $\mathbb{Z}$ is $x = -4, y = -3$. $\endgroup$
    – Cocopuffs
    Jul 26, 2012 at 16:10
  • $\begingroup$ Where is the linear algebra here? $\endgroup$ Jul 26, 2012 at 16:14

4 Answers 4

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There will be infinitely many pairs $(x,y)$ of real numbers that satisfy the equation. If you are looking for integer solutions, that is another matter.

I would rewrite the equation as $9xy+42x+51y+213=0$, and then as $$(3x+17)(3y+14)-(17)(14)+213=0,$$ which turns into the attractive $$(3x+17)(3y+14)=25.$$ Note that we did an analogue of completing the square. We get a hyperbola.

Now for integer solutions the analysis becomes quite simple. We take all ordered pairs $(s,t)$ of integers (both positive or both negative) such that $st=25$. There are only $6$ such pairs.

For some but not all of these pairs, it turns out that $x$ and $y$ are integers. Let's start. Look at $s=1, t=25$. No good, there is no integer $x$ such that $3x+17=1$. Look at $s=-1,t=-25$. That gives $3x+17=-1$, $3y+14=-25$, which has the integer solution $x=-6, y=-13$. Continue. There is not far to go!

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If $x$ and $y$ can be real numbers, with one equation in two unknowns you will have one dimension of freedom. Solving for $x$, for example, $x=- \frac {17y+71}{3y+14}$. You can substitute in any value for $y$ you want except $\frac {-14}3$ and find $x$. If $x$ and $y$ are integers you can use divisibility testing to restrict the options.

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    $\begingroup$ I think you've made an algebra error (or typo), and ended up solving for -x instead of x . . . $\endgroup$
    – ernie
    Jul 26, 2012 at 19:49
  • $\begingroup$ @ernie: Thanks. pickhu has fixed. $\endgroup$ Jul 26, 2012 at 21:00
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First, I will assume that you want to solve for $x$ in terms of $y$ and vice-versa. To do this, you first need to keep all the terms with $x$ on one side of the equation and move all the others to the other side:

$$3xy + 14x = -17y - 71$$

Next, factor out the $x$:

$$x(3y + 14) = -17y - 71$$

Now divide by $3y + 14$:

$$x = \frac{-17y - 71}{3y + 14}$$

The steps are similar to solve for $y$ in terms of $x$. I will leave that as an exercise for the reader.

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This is another way to find the integer solutions.
Define $d=y-x$ and the equation becomes $3x(x+d)+14x+17(x+d)+71=0$ which boils down to $$3x^2+(3d+31)x+17d+71=0.$$ The quadratic formula gives the roots:$$6x=-(3d+31)\pm\sqrt{9d^2-18d+109},$$ and we want the discriminant, $9(d-1)^2+100 = (3(d-1))^2+10^2$ to be a square.
$d=1$ is obviously one candidate, leading to final values $x=-4, y=-3$.

Otherwise we are just looking for Pythagorean triples. A quick look at a list of Pythagorean triples shows that there is just one triple with $10$ as one of the lower values: (10, 24, 26).
Thus $9(d-1)^2=24^2$, giving $d=9$ or $d=-7$, from which the appropriate values of $x$ and $y$ can be found.

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