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

Step 1
Given:
${\displaystyle 25x\equiv 7\left(\text{mod}17\right)}$
and
${\displaystyle 26x\equiv 8\left(\text{mod}6\right)}$
To find:
Solutions of the linear congruences.
Step 2
Consider,
${\displaystyle 25x\equiv 7\left(\text{mod}17\right)}$
It is in the form ${\displaystyle ax\equiv b\left(\text{mod}c\right)}$
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
$a=25,b=7andc=17$
$(25,17)=1$ and ${\displaystyle 1\mid 7}$
Therefore the congruence ${\displaystyle 25x\equiv 7\left(\text{mod}17\right)}$ has a solution.
Let
${\displaystyle 25x\equiv 7\left(\text{mod}17\right)}$
Multiply by 2
${\displaystyle \Rightarrow 50x\equiv 14\left(\text{mod}17\right)}$
${\displaystyle \Rightarrow -x\equiv -3\left(\text{mod}17\right)}$
${\displaystyle \Rightarrow x\equiv 3\left(\text{mod}17\right)}$
Therefore the solution set is
$3,3+17,3+2\cdot 17,...$
${\displaystyle \Rightarrow \{3,20,37,\dots \}}$
Step 3
Now
${\displaystyle 26x\equiv 8\left(\text{mod}6\right)}$
${\displaystyle \Rightarrow 2x\equiv 2\left(\text{mod}6\right)}$
${\displaystyle \Rightarrow x\equiv 1\left(\text{mod}6\right)}$
Therefore the solution set is
$1,1+6,1+2\cdot 6,...$
$1,7,13,..$