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There are infinitely many integer solutions for the equation $4x + 6y = 8 $

My work:

$2x+3y=4$ and $2x=4-3y$ so $x=2-(\frac 32)y$

Similarly $y=\frac 43 - \frac 23 x$ are integer solutions of equation.

But correct answer is $x = −4 + 3t, y = 4 − 2t$ for all integers $t$. Help me understand where I have gone wrong.

User8976
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Fawad
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  • You cannot just solve for $x$ and $y$ like $x$ and $y$ are real numbers. Thats the part which went wrong. – TheGeekGreek Mar 26 '17 at 17:07
  • Just try setting $y=1$ in your first equation and $x=1$ in your second - it should be apparent that something different is required. – Martin Rattigan Mar 26 '17 at 17:14
  • @MartinRattigan yes,they aren't integer solutions then. – Fawad Mar 26 '17 at 17:16
  • @MartinRattigan then what I need to do? – Fawad Mar 26 '17 at 17:22
  • I think you're being asked to find a description of all integer solutions and prove that it's valid. Since you already gave a description in your last sentence try showing the description always produces an integer solution and that any integer solution must be one of the solutions described. – Martin Rattigan Mar 26 '17 at 17:32
  • The correct answer can look different; many ways to describe the same pairs depending on your initial point you use. Find an integer pair on your line and use what you know about linear equations to find another integer pair or the rest of the integer pairs. (Hint use slope here; slope means rise over run.) – randomgirl Mar 26 '17 at 17:43
  • This should help. – projectilemotion Mar 26 '17 at 17:57
  • You haven't gone wrong. You just didn't finish. $4x+6y=8\iff x=2-3y/2$ . So for any even integer $y,$ if $x=2-3y/2$ then $x$ is an integer and $4x+6y=8.$ Therefore for every even integer $y$ there is an $x$ (namely, $x=2-3y/2$) such that $4x+6y=8.$ – DanielWainfleet Mar 26 '17 at 19:48

1 Answers1

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There are infinitely many solutions to the equation $4x+6y=8$ since $(4,6) =2$ and $2\mid8$. We use Euclidean Algorithm to determine $m,n$ such that $4m+6n=2$. Here $m=-1,n=1$ and $8=2.4$.

Thus $x_0 = 4(-1)$ and $y_0=4.1=4$ is a particular solution.

The solutions are given by $x=−4+3t,y=4−2t$ for all integers $t$.

User8976
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