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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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Looks pretty good, except you can also have $y=18$. –  Gerry Myerson May 22 '12 at 7:30
    
Thanks Gerry. Yeah, I didn't think that one. –  alik May 22 '12 at 7:33
    
Do you also allow negative solutions? Then you will get more. I am not sure about the word "whole". I've never seen it in this context when talking english. Translating it back into German yields the word for integer. –  Simon Markett May 22 '12 at 7:37
    
@Simon: In English whole number is equivalent either to positive integer or to non-negative integer, depending on the person using it; alik is evidently using it in the latter sense. It’s not equivalent to ganze Zahl. –  Brian M. Scott May 22 '12 at 7:39
    
@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 ;) –  Simon Markett May 22 '12 at 8:05

1 Answer 1

up vote 1 down vote accepted

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. –  alik May 22 '12 at 7:47

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