number theory - Solving Linear Congruences.

Question is -:

Solve the linear congruence $3x \equiv 4\left(mod\, \, \, \, 7\right)$, and find the smallest positive integer that is a solution of this congruence

My Approach-:

$3x \equiv 4\left(mod\, \, \, \, 7\right)$

$\Rightarrow x \equiv 3^{-1}\, \,4\left(mod\, \, \, \, 7\right)$

$3^{-1}$ means it is the multiplicative inverse of $3\, \,mod\, \,7$

multiplicative inverse of $3\, \,mod\, \,7$

$\Rightarrow 7=3*2+1$

$\Rightarrow 3=1*3+0$

$\Rightarrow 1=1*7+\left(-2\right)3$

thus $-2$ or $5$ is the inverse.

Thus i am getting

$\Rightarrow x \equiv 3^{-1}\, \,4\left(mod\, \, \, \, 7\right)$

$\Rightarrow x \equiv 20\left(mod\, \, \, \, 7\right)$

But in the solution they are multiplying the inverse $5$ to both sides and get equation as-:

$15 \,x \equiv20 \, \left(mod\,\,7\right) $

and then

$x \equiv 15x\,\equiv\,20\,\equiv\,\,6\,\left(mod\,\,7\right)$

The solution is given here

Please help me out ,where i am wrong!

thanks!

Answer

You're not wrong, when you write $x \equiv 3^{-1} 4$, then you have
also multiplied both sides by the inverse of $3$. You're just using
different notation.