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2x + 5y = 90

x and y are whole numbers. how many (x,y) can this equation get?

I thought it like this;

2x = 90-5y

x = (90-5y)/2

x is a whole number so must be divisible by 2. So I thought--- by giving y the values {0, 2, 4, 6, 8, 10, 12, 14, 16} it can be divided by 2.

There can be 9 (x,y)s

If it is true, is there a better way to do it? And If not where did I make the mistake?

Thanks!

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    @BrianM.Scott Thanks for the clarification! I guess I never came across it since people tend to use the less ambiguous terms...nothing better than learning new vocabulary at 9 in the morning ;)2012-05-22

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What you did is correct, except that you stopped short: $y=18$ is also possible. You could also have rewritten the equation as $5y=90-2x$, so $y=\frac{90-2x}5=18-\frac{2x}5\;.$ This implies that $x$ must be a multiple of $5$, so you get $x=0,5,10,15,20,25,30,35,40$, and $45$.

You can reduce the work by noticing that in the equation $x=\frac{90-5y}2=45-\frac{5y}2\;,$ $x$ decreases when $y$ increases. The smallest permissible value of $y$ is of course $0$, and as you saw, $y$ must be even. With just a little more work we determine directly what the largest possible value of $y$ is, and then we don’t have to try values one by one. The smallest permissible value for $x$ is $0$, so we must have $\begin{align*}&45-\frac{5y}2\ge 0\;,\\\\ &45\ge\frac{5y}2\;,\\\\ &5y\le90\;,\text{ and finally}\\\\ &y\le18\;. \end{align*}$

Thus, we know that $y$ can only be one of the numbers $0,2,4,\dots,16,18$.

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    Many thanks Brian.2012-05-22