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  1. Suppose $4x\equiv 6 \pmod {18} $ Then $2x\equiv 3 \pmod 9$ Then $6x\equiv 9 \pmod 9$ Then $6x\equiv 0 \pmod 9$ Then $x\equiv 0 \pmod 9$ Then $x=9k$

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  1. Suppose $4x\equiv 6 \pmod {18}$ Then $2x\equiv 3 \pmod 9$ Then $2x\equiv 12 \pmod 9$ Then $x\equiv 6 \pmod 9$ Then $x = 9k+6$
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    $\begingroup$ In (i): $\;6x=0\pmod 9\rlap{\;\;\;\;/}\implies x=0\pmod 9\;$ . For example, $\;6\cdot 3=0\pmod 9\;$ but $\;3\neq 0\pmod 9\;$ $\endgroup$ – DonAntonio Apr 5 '14 at 18:20
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    $\begingroup$ No theorem is ignored, but there is a "non sequitur" in 1. $\endgroup$ – Hagen von Eitzen Apr 5 '14 at 18:25
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$6x\equiv 0\pmod 9$ does not imply $x\equiv 0\pmod 9$. Because $6$ and $9$ are not coprime. What you do get by dividing out one $3$ is that $2x\equiv 0\pmod 3$, though - but you knew that already.

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In $1$ the step from $6x\equiv 0$ (mod $9$) to $x\equiv 9$ (mod $9$) is invalid. The conclusion should be $2x\equiv 0$ (mod $3$), because the modulus is divisible by $3$ and then $x\equiv 0$ (mod $3$), because $(2,3)=1$. Or equivalently go all the way in one step because $(6,9)=3$ and if you want to divide by $6$ you must divide the modulus by $(6,9)$.

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The problem is that you perform a noninvertible transformation on the equation by scaling it by $\,3,\,$ because $\,3\,$ is noninvertible mod $\,9.\,$ Doing that will generally introduce extraneous roots. Indeed $\!\bmod 9\!:\ 2x\equiv 3\,\Rightarrow\, 6x\equiv 9\equiv 0,\,$ and $\,6x\equiv 0\!\iff\! 9\mid 6x\!\iff\! 3\mid 2x\!\iff\! 3\mid x\, $ (not $\,9\mid x).$ Now $\,3\mid x\!\iff\! x\equiv \color{#c00}{0,3},6\pmod{\! 9},\,$ but $\,x\equiv \color{#c00}{0,3}\,$ are not roots of $\,2x\equiv 3\pmod{\! 9},\,$ even though they are roots of $\,6x\equiv 9.$

But if we scale by an invertible $\,c\,$ then $\,ax\equiv b\,\smash[t]{\underset{\color{#c00}{\ \times\ c^{-1}}}{\color{#c00}{\Longleftarrow}}\!\!\color{#0a0}{\overset{\times\ c}\Longrightarrow}} \ acx\equiv bc,\,$ since the direction $\,(\color{#c00}{\Longleftarrow})$ follows by multiplying the RHS by $\,\color{#c00}{c^{-1}}.\,$ So the lhs and rhs congruences have exactly the same set of roots. Hence no extraneous roots are introduced by scaling by an invertible element.

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It's the multiplication law of modulus arithmetic. For all $a_1,a_2,b_1,b_2,n$ integers, $ a_1 a_2 \equiv b_1 b_2 \pmod n.\,$

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