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I am solving a problem and I reached the following system: $$\begin{cases}2y=x+z\\z^2=yt\\x+t=37\\y+z=36\end{cases}.$$ I have to find only the integer solutions, if that helps. I would be very grateful if you could give me some tips on solving it.

Math Student
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  • Begin by solving the linear system formed by equations $1$, $3$, and $4$, find the integer solutions, and see which one satisfy equation $2$. – TheSilverDoe Nov 19 '20 at 17:07
  • If we take the three equations $\begin{cases}2y=x+z\x+t=37\y+z=36\end{cases},$ we would have $3$ equations for $4$ variables, so I don't think we will be able to find the integer solutions of the system. Do I get it wrong? Thank you in advance! – Math Student Nov 19 '20 at 17:11
  • Of course you will be able to. Every linear system can be solved explicitely, even if it has an infinity number of solutions. – TheSilverDoe Nov 19 '20 at 17:12
  • I really don't see what I am supposed to do to solve the system. I am sorry! – Math Student Nov 19 '20 at 17:15
  • There are lots and lots of methods to solve linear systems : have a look at https://en.wikipedia.org/wiki/System_of_linear_equations#Solving_a_linear_system (but if I may give you an friendly advice : begin by learning how to solve linear systems, before searching about non-linear ones !) – TheSilverDoe Nov 19 '20 at 17:16
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    (eq. 1+3+4) $\implies t = 73-3y$, (eq. 4) $\implies z = 36 - y$. Substitute that into (eq. 2), you get a quadratic equation in $y$.... – achille hui Nov 19 '20 at 17:19
  • @achillehui, nice! What was the idea behind this? A very clever approach, but how am I supposed to find it? :)) – Math Student Nov 19 '20 at 17:24
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    (eq. 1,3,4) are 3 linear equations in 4 unknowns. if you treat any one of the unknowns as parameter (in this case $y$), you get 3 equations in the $x,z,t$. Solving those equations allow you to express $x,z,t$ in terms of $y$. It really didn't matter which unknown you use as parameter, I choose $y$ because corresponding coefficients will be integers instead of fractions. – achille hui Nov 19 '20 at 17:44

1 Answers1

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I started as so:

We have $$\begin{cases}2y=x+z\\z^2=yt\\x+t=37\\y+z=36\end{cases}.$$ and I labeled equation the equations from 1-4. By equation 1, we have that $z=2y-x$ and by equation 4 we have that $z=36-y$. We can now equate the both z equations and we have that $36-y=2y-x$. Solving this for $y$ we have that $y=12+\frac{x}{3}$. By equation 3 we have that $t=37-x$. As stated earlier we know what Fromm equation 1 that $z=2y-x$, so by equation 2 we have that $z^2 = (2y-x)^2 = (4y^2-4yx+x^2)$. Where $z^2=yt$, where $t=37-x$. Thus we have that $(4y^2-4yx+x^2)=(37-x)(y)$ where $y=12 +\frac{x}{3}$. By plugging this into the equation we have now, we have $\frac{x^2-144x}{9} +576 = \frac{-x^2}{3} + \frac{x}{3} +444$. To get rid of all of the fractions we can multiply both sides by 9 and get $x^2-144x +5184 = -3x^2+3x+3996$, and by simplifying this we have $4x^2-147x+1188=0$ which we can solve with the quadratic formula. We then get that $x=99/4, x=12$ since we only want integer solutions we take the $x=12$ solution and plug it into our $y=12+\frac{x}{3}$ and get that $y=16$. From earlier we had that $t=37-x$ so $t=25$. Lastly since $z=2y-x$ we have that $z=20$. So your integer solutions are $x=12,y=16,t=25,z=20$. I hope this process makes sense!

Joey
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