Solve 3x + 3 = x + 5 mod 11 (= is congruence not equality...)

(3x - x) = (5 - 3) mod 11

2x = 2 mod 11

2x = 13 mod 11

x = 13/2 mod 11

Is this correct?

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- Feb 28th 2010, 07:48 PMjzelltIs this correct?
Solve 3x + 3 = x + 5 mod 11 (= is congruence not equality...)

(3x - x) = (5 - 3) mod 11

2x = 2 mod 11

2x = 13 mod 11

x = 13/2 mod 11

Is this correct? - Feb 28th 2010, 08:27 PMdani
No. To be able to divide by an element $\displaystyle a$ (2 in this example) you need to find $\displaystyle a^{-1}$ such that $\displaystyle a\times a^{-1}=1$. If the inverse does not exist, you cannot "divide" by $\displaystyle a$.

so in your example, $\displaystyle a=2$ and $\displaystyle a^{-1}=6$ ...