1

I understand how to solve linear congruence if the equation is in the form of, example: $4x \equiv 5 \pmod 9$ or $x+5\equiv 2 \pmod {11}.$

My problem is, I don't know exactly what to do with the $-7$ part of the $(5x - 7)$ equation.

Michael Hardy
  • 1
Johnny James
  • 39
  • 1
    Take it to the other side of equivalence. That is, $5x \equiv 2+7=9 \mod{17}$. – Math Lover Oct 11 '17 at 17:06
  • You can add or subtract the same number of both sides of the equivalence "$\equiv$ " sign, just like elementary algebra. – Doug M Oct 11 '17 at 17:07
  • Hi! For $a,b,c\in Z, n\in Z^+$, we have $a\equiv b\pmod n\Rightarrow a+c\equiv b+c\pmod n$ –  Oct 11 '17 at 17:08
  • How did you get $12x$? –  Oct 11 '17 at 17:12
  • Check out https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference to properly format your posts –  Oct 11 '17 at 17:13
  • @JeremyJackson It can become $-12x\equiv 9\pmod{17}$ but not $12x\equiv 9\pmod{17}$ – kingW3 Oct 11 '17 at 17:14

1 Answers1

0

write $$x\equiv \frac{9}{5}\mod 17$$ we will adding $17$ to the numerator untill it is divisible by $5$ $$x\equiv \frac{9}{5}\equiv \frac{26}{5}\equiv \frac{43}{5}\equiv \frac{60}{5}\equiv 12 \mod 17$$

Dr. Sonnhard Graubner
  • 95,283