# Equations With Two Linear Congruences

Is there a known way to isolate a variable when the equation has two linear congruences each containing the variable?

$$200=\left(\frac{x}{4}-(x \mod 17)+\frac{3}{4} \right)\cdot 3 \cdot \left(\frac{x}{3}-(x \mod 17)+\frac{4}{3} \right) \cdot 7$$

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It seems to me that if $x$ is an integer then the right side's a multiple of $7$ but the left side isn't, so there can't be a solution. Are you prepared to accept fractional values of $x$? – Gerry Myerson Feb 12 '13 at 11:47
Yes, any value is fine. But I am wholly uninterested in the 'answer'... I am really interested in the method, or steps one takes to isolate the unknown. – Perry Feb 12 '13 at 12:01
Well, if there isn't an answer, then there isn't a method. – Gerry Myerson Feb 12 '13 at 12:06
Are you saying there isn't a way to isolate the $x$ from that equation? – Perry Feb 12 '13 at 12:16
If there were a way to isolate $x$, you'd get $x=\rm something$, and that would be a solution. Try to isolate $x$ from $x=x+1$, or from $2x\equiv1\pmod4$. – Gerry Myerson Feb 12 '13 at 23:26

If you replace $\rm\ x\ mod\ 17\$ by $\rm\ x-17\,n\$ then you obtain the quadratic equation

$$\rm 42\, x^2 - (2023\, n + 126)\, x + 24276 n^2\! + 2975\, n - 716\, =\, 0$$

This has no integer roots, else reducing it $\rm\, mod\ 7\,$ yields $\rm\, 5\equiv 0,\,$ a contradiction.

The roots of the quadratic are

$$\rm x = \frac{2023\,n+126 \pm \sqrt{7}\sqrt{2023\, n^2 + 1428\, n + 19452}}{84}$$

e.g. one of the simplest is when $\rm\: n = 2:$

$$\rm x = \frac{1043\pm 10 \sqrt{133}}{21}$$

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This seems to be a restatement of my comment that the right side's a multiple of $7$ and the left side isn't. But OP is happy to accept fractional values of $x$. If, say, $x=1/7$, we can interpret $x\mod{17}=5$, and now $7$ doesn't divide the right side any more, so the arguments presented so far don't work. But now there's too much $7$ in the denominators on the right side, so the no-solution conclusion still holds. – Gerry Myerson Feb 12 '13 at 23:24
@Gerry The above works for any value of $\rm\,x,\,$ assuming only that $\rm\, x\ mod\ 17\:$ denotes some "residue" $\rm\:x - 17\,n,\,$ for some $\rm\,n\in\Bbb Z.\,$ It parametrizes all the possible solutions by $\rm\,n.\ \$ – Math Gems Feb 13 '13 at 0:04
You have proved it has no integer roots, and you have done that by reduction modulo $7$. You haven't proved it has no rational roots, in particular, roots with denominator divisible by $7$. When $x=1/7$, I don't see $5\equiv0$. – Gerry Myerson Feb 13 '13 at 0:30
@Gerry It answer's the OP's question on how to isolate $\rm\,x,\,$ using a very liberal interpretation of $\rm\,x\ mod\ 7,\,$ for non-integral $\rm\,x.$ – Math Gems Feb 13 '13 at 0:45
@Math Gems ... I appreciate the work you've done here, and that's why I feel bad asking for more, but I am having trouble following some things I would ask you to elaborate on... Can you explain this in more detail 'This has no integer roots, else reducing it mod 7 yields 5≡0, a contradiction.'? Also, how do you move from the first $x=$ eq with the n variable to the simplest root answer without it? Thanks. – Perry Feb 13 '13 at 0:53