# Solving $17x+7 \equiv 3 \pmod 6$ [duplicate]

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

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

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

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

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

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

$$x \equiv 4\pmod 6$$

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

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 $$17x\equiv 3-7=-4\equiv 2\iff x\equiv 5\cdot 2\equiv 4\mod 6.$$

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 $$x\equiv-2 \pmod 6$$