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Question Number 110281 by 675480065 last updated on 28/Aug/20

Answered by Rio Michael last updated on 28/Aug/20

$$3x\equiv 1\left(\mathrm{mod}5\right)\left(i\right)$$$$5x\equiv 2\left(\mathrm{mod}7\right)\left(ii\right)$$$$\mathrm{from}\left(i\right)x\equiv 2\left(\mathrm{mod}5\right)\left(iii\right)$$$$\mathrm{from}\left(ii\right)x\equiv 6\left(\mathrm{mod}7\right)\left(iv\right)$$$$\mathrm{from}\left(iii\right)x=5k+2,k\in \mathbb{N}\left(v\right)$$$$\mathrm{into}\left(iv\right)\Rightarrow 5k+2\equiv 6\left(\mathrm{mod}7\right)$$$$\Rightarrow 5k\equiv 4\left(\mathrm{mod}7\right)\iff k\equiv 5\left(\mathrm{mod}7\right)\Rightarrow k=7t+5,t\in \mathbb{N}$$$$\mathrm{hence}x=5(7t+5)+2=35t+27$$$$x=35t+27,$$$$$$$$$$

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