How can we solve an equation with two variables, when only given one equation?

Question

I have been looking at the following problem:

I initially tried using simultaneous equation, but with no real progress. I decide to give in and check the answer:

Now, I understand this solution, except for the line where it says:

’The only positive-integer solutions of this equation are $w=2, d=10$ and $w=5, d=3$’

This is surely where most of the challenge is with this problem, and yet they gloss over it in the answer. So my question is this: how are you meant to figure out these values without the use of trial and error, and if you do have to use trial and error, how can we be sure these are the only solutions?

Answer 1

$7w+3d=44 \implies d = \dfrac{44-7w}{3}=\dfrac{42-6w}{3}+\dfrac{2-w}{3} =14-2w+\dfrac{2-w}{3}$

The numbers that give integer solutions are of the form $w=3x+2$.

Letting $w=3x+2$, we get $d = 10-7x$

$x < 0$ makes $w$ negative. $x>1$ makes $d$ negative.

$x=0$ gives $(w,d)=(2,10)$ and $x=1$ gives $(w,d)=(5, 3)$.

Response to comment.

I chose $d = \dfrac{44-7w}{3}$ because you only have to try $w=0,1,2$ to find a value of $d$ that makes $w$ an integer. If you felt like writing

$$w=\dfrac{44-3d}{7}= 6 - \dfrac{3d-2}{7}$$ then $d=3$ gives us $w=5$.

For small numbers, this works just fine. If you have to, the best way is to use modular arithmetic.

For example, to search for an initial value of $d$,

\begin{align} 7w+3d=44 &\implies 3d \equiv 44 \pmod 7 \\ &\implies 3d \equiv 2 \pmod 7 \\ &\implies 5 \cdot 3d \equiv 5 \cdot 2 \pmod 7 \\ & \implies d \equiv 3 \pmod 7 \end{align}

You can then use $d=3+7t$ and you will get all integer values for $w$ and $d$.

Answer 2

If $7w+3d=44$, then $7w+3d-44$ is divisible by $3$,

so $7w+3d-44-3(2w+d-14)=w-2 $ is,

so $w\in\{2,5,8,11,14,17,...\}$,

but in that set only $2$ and $5$ can be multiplied by $7$ to get a result not more than $44$.

Answer 3

If $w$ and $d$ are positive integers, you need $7 w < 44$ so $w < 44/7 < 7$. So you only need to try $w = 1, 2, 3, 4, 5, 6$.

Answer 4

Equations that have only integer solutions (or, in this case, positive integer solutions) are called Diophantine equations. A linear Diophantine equation like $7w+3d=44$ is relatively easy to solve about a quarter of the way through an introductory number theory course. So I'll do some handwaving here at the basic concepts and you know where you can study more deeply if you like.

So first we're going to find all of the solutions in integers, even though that makes no sense in the context of the word problem. We'll start by noting that $7(1)+3(-2)=1$. So, we can multiply both sides by 44 to get $7(44)+3(-88)=44$. So $w=44, d=-88$ is an integer solution.

Our next step is to find all integer solutions. Obviously, $7(-3)+3(7)=0$, so if we have a specific solution for $w$ and $d$, then subtracting $3$ from $w$ and adding $7$ to $w$ will result in another solutions to the general equation. It turns out that this is basically the only way to generate new solutions, so all of the solutions are of the form $w=44-3t$ and $d=7t-88$ where $t$ ranges over the set of integers.

Finally, we need to tie this back to the context of the problem, where $w$ and $d$ are positive integers and $w>d$. We can do that by solving the simultaneous inequalities $44-3t>0$ and $7t-88>0$, which has the solution set $12<t<15$. $t=13$ leads to $w=5, d=3$, and $t=14$ leads to $w=2, d=10$, and only the first solution satisfies the second condition. Therefore, a win is worth $5$ points, a draw is worth $3$ points, and therefore your sisters team must have scored $31$ points.

Answer 5

To solve for integer solutions a linear equation, you could use plotting and inspection among other methods.

An easy way for this case would be to draw the line (very easy to draw even by hand):

$$7w+3d=44 $$

Using (x,y) instead of (w,d) because the plotting program I use only uses (x,y)! We get:

$$z=7x+3y=44$$

The plot for the this relationship is shown below. You can inspect the line to verify that the only 2 points where $x$ is an integer and $y$ is an integer are: $(2,10), (5,3)$

EDIT: In the case at hand we are looking for positive values for $x,y and z$. The relation $z=7x+3y$ shows to be positive before the intersection with the $x-axis$ at approximately $x=6.286$. Even if other integer solution would exist, the other solution would involve a negative values for at least on of: $z, x$ or $y$.