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$x\equiv 2\:(\text{mod }6)$ and $x \equiv 3\:(\text{mod }9)$

attempted solution:

$x = 2, 8, 14, 20,$

$x = 2+6m$

$x = 3, 12, 21, 30, 39$

x = $3+9m$

$2+6m = 3+9m$ $-1 = 3m$ $-1/3 = m$

$m $is not an integer, therefore there is no common solutions?

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    Where is the system of equations?2012-11-29
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    Something wrong with the question as stated, since solutions of the form $(x=2,y=3)$ abound.2012-11-29
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    Do you mean $x \equiv 3 (\mod 9)$?2012-11-29
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    yeah i mean that2012-11-29
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    @internetlearning Then please edit your question, and also give the relation between $x$ an $y$2012-11-29

2 Answers 2