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# Three of them

$\begin{cases} x + 2y & = p + 6 \\ 2x - y & = 25 - 2p \end{cases}$

Let $$(x,y)$$ be positive real integers that satisfy the system of equations above. How many possible value of $$p$$? And for what value of $$p$$ ?

Note by Fidel Simanjuntak
8 months, 2 weeks ago

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Solving for $$x$$ and $$y$$ will give:

$x=\dfrac{56-3p}5 \text{ and } y= \dfrac{4p-13}5....(1)$ $\because x,y>0\implies \dfrac{13}{4}< p<\dfrac{56}3....(2)$

Looking at original first equation:$$x+2y-6=p$$ we can conclude $$p$$ is also an integer which along with $$(2)$$ gives $$p\in\{4,5,6,...,18\}$$.

From $$1$$, now since $$y$$ is an integer it's numerator must be a multiple of $$5$$ and for that $$4p$$ must end in $$8$$ or $$3$$ so when $$13$$ is subtracted, the numerator ends in $$5$$ or $$0$$ respectively and hence divisible by $$5$$. 4p cannot end in $$3$$ but ends in $$8$$ when $$p\equiv 5n+2$$ or $$p\in\{7,12,17\}$$. For these values of $$p$$, $$x$$ also assumes an integer.

$p=7,~~ (x,y)=(7,3)$ $p=12, ~~(x,y)=(4,7)$ $p=17,~~(x,y)=(1,11)$

PS: I might have missed some solutions bcoz I'm pretty good at it.

- 8 months, 2 weeks ago

Nice solution! It's important to notice that $$\dfrac{13}{4} < p < \dfrac{56}{3}$$. Thanks!

- 8 months, 2 weeks ago