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

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
2 months, 1 week 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. Rishabh Cool · 2 months, 1 week ago

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@Rishabh Cool Nice solution! It's important to notice that \( \dfrac{13}{4} < p < \dfrac{56}{3} \). Thanks! Fidel Simanjuntak · 2 months, 1 week ago

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