Answer
The solution is $\Big[-\dfrac{8}{7},2\Big]$
Work Step by Step
$|7x-3|-1\le10$
Take $1$ to the right side:
$|7x-3|\le10+1$
$|7x-3|\le11$
According to an absolute value inequality property, $|x|\le a$ is equivalent to $-a\le x\le a$. Then, in this particular case, solving $|7x-3|\le11$ is equivalent to solving $-11\le7x-3\le11$
$-11\le7x-3\le11$
Add $3$ on all parts of the inequality:
$-11+3\le7x-3+3\le11+3$
$-8\le7x\le14$
Divide all parts of the inequality by $7$:
$-\dfrac{8}{7}\le\dfrac{7x}{7}\le\dfrac{14}{7}$
$-\dfrac{8}{7}\le x\le2$
Expressing the solution in interval notation:
$\Big[-\dfrac{8}{7},2\Big]$
