$10x+7\equiv 2$ mod 25

Question

$10x+7\equiv 2$ mod 25

Solve for x, Correct answer is 2,7,12,17,22

What method can we use to solve this problem?

I tried and got this far:

$10x+7\equiv 2$ mod 25

$10 x + 7 = 2 + 25n$

$10x + 5 = 25n$

$2x = -1$ mod 5

Answer 1

Note that for each integer $x$ the equation $10x \equiv_{25} 10(x+5)$ holds, and that $10x \not \equiv_{25} 10(x+i)$ for each $i=1,2,3,4$.

Thus if $x_0$ is an integer satisfying $10x_0 + 7 \equiv_{25} 2$ then the set of integers $x$ satisfying $10x + 7 \equiv_{25} 2$ is precisely the set $\{x: x = x_0 \pm 5n$ for some nonnegative integer $n\}$. Note that $x_0=2$ satifies the equation $10x_0 + 7 \equiv_{25} 2$. So the set of integers $x$ satisfying $10x + 7 \equiv_{25} 2$ is precisely the set $\{x: x = 2 \pm 5n$ for some nonnegative integer $n\}$.

This is precisely the set of integers $x$ satisfying $2x \equiv_5 -1 \equiv_5 4$ or equivalently $x \equiv_5 2$, relating to your solution.

Answer 2

$10x+7\equiv2\bmod 25\\10x\equiv 20\bmod 25\\2x\equiv 4\bmod 5\\x\equiv 2\bmod 5$