If $a \equiv 4 (mod 13)$ and $b \equiv 9 (mod 13)$, then how can we find $0 \leq c \leq 12$ such that $c \equiv 9a (mod 13)$ and $ c \equiv 2a+3b (mod 13)$?
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By congruence sum & product rules $!\bmod 13!:\ \color{#c00}{a\equiv 4},, \color{#0a0}{b\equiv 9}\Rightarrow c\equiv 9(\color{#c00}a) \equiv 9(\color{#c00}4)\equiv 10, $ contra $,c\equiv 2(\color{#c00}a)+3(\color{#0a0}b)\equiv 2(\color{#c00}4)+3(\color{#0a0}9)\equiv 9\ \ $ – Bill Dubuque Mar 27 '22 at 17:14
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We need to find the residue of $c$ modulo $13$ so because $c\equiv9a(\bmod13)$, $c\equiv9\cdot4\equiv36\equiv10(\bmod13)$ $c\equiv2a+3b\equiv2\cdot4+3\cdot9\equiv8+27\equiv35\equiv9(\bmod13)$ So we got that $c\equiv9\equiv10(\bmod13)$ which is obviously false, so we have no solutions.
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Please strive not to add more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Mar 27 '22 at 17:15