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I need to solve for $x$ in this congruence:

$$17x+7 \equiv 3 \pmod 6$$

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    $\begingroup$ Welcome! What have you tried, and where are you stuck? :) $\endgroup$ – Siddharth Bhat Mar 28 '19 at 18:11
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    $\begingroup$ I do not know where to start $\endgroup$ – Samantha Barr Mar 28 '19 at 18:18
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    $\begingroup$ How would you solve $17x+7=3$? $\endgroup$ – J. W. Tanner Mar 28 '19 at 18:20
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$$17x+7 \equiv 3 \pmod 6$$

$$17x \equiv 3-7 \pmod 6$$

$$-1x \equiv -4 \pmod 6$$

$$x \equiv 4\pmod 6$$

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$$17x\equiv -4\equiv 2 \mod{6}$$ But $$17x\equiv5x\mod{6}$$ $$\therefore 5x\equiv 2\mod{6}$$ $$x\equiv 4 \mod{6}$$

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As $17$ and $6$ are coprime, $17$ is a unit mod. $6$, and $17\equiv 5 \mod 6$, so $$ 17^{-1}\equiv 5^{-1}\equiv 5\mod 6\quad\text{since }\; 5^2\equiv 1\mod 6.$$Now the given congruence is equivalent to $$17x\equiv 3-7=-4\equiv 2\iff x\equiv 5\cdot 2\equiv 4\mod 6.$$

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You can avoid the congurences. We have $$6\mid (17x+7)-3 = 17x+4$$

Since $6\mid 18x+6$ we have $$6\mid (18x+6)-(17x+4) = x+2$$ so $x+2 = 6k$ or $$x\equiv-2 \pmod 6$$

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