## Intermediate Algebra (6th Edition)

The solution is $\Big[-\dfrac{8}{7},2\Big]$
$|7x-3|-1\le10$ 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: $\Big[-\dfrac{8}{7},2\Big]$