# Solve the following linear congruences using Diophantine equations: 25x -= 7(mod 17) and 26x -= 8(mod 6)

Solve the following linear congruences using Diophantine equations: $25x\equiv 7\left(\text{mod}17\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}26x\equiv 8\left(\text{mod}6\right)$
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Step 1
Given:
$25x\equiv 7\left(\text{mod}17\right)$
and
$26x\equiv 8\left(\text{mod}6\right)$
To find:
Solutions of the linear congruences.
Step 2
Consider,
$25x\equiv 7\left(\text{mod}17\right)$
It is in the form $ax\equiv b\left(\text{mod}c\right)$
We know if greatest common divisor of a and c divides b that is (a, c)|b then the linear congruence has a solution.
Here

$\left(25,17\right)=1$ and $1\mid 7$
Therefore the congruence $25x\equiv 7\left(\text{mod}17\right)$ has a solution.
Let
$25x\equiv 7\left(\text{mod}17\right)$
Multiply by 2
$⇒50x\equiv 14\left(\text{mod}17\right)$
$⇒-x\equiv -3\left(\text{mod}17\right)$
$⇒x\equiv 3\left(\text{mod}17\right)$
Therefore the solution set is
$3,3+17,3+2\cdot 17,...$
$⇒\left\{3,20,37,\dots \right\}$
Step 3
Now
$26x\equiv 8\left(\text{mod}6\right)$
$⇒2x\equiv 2\left(\text{mod}6\right)$
$⇒x\equiv 1\left(\text{mod}6\right)$
Therefore the solution set is
$1,1+6,1+2\cdot 6,...$
$1,7,13,..$