## Intermediate Algebra (6th Edition)

The solution is $(-\infty,1]\cup\Big(\dfrac{29}{7},\infty\Big)$
$3x+2\le5$ or $7x\gt29$ Solve the first inequality: $3x+2\le5$ Take $2$ to the right side: $3x\le5-2$ $3x\le3$ Take $3$ to divide the right side: $x\le\dfrac{3}{3}$ $x\le1$ Expressing the solution in interval notation: $(-\infty,1]$ Solve the second inequality: $7x\gt29$ Take $7$ to divide the right side: $x\gt\dfrac{29}{7}$ Expressing the solution in interval notation: $\Big(\dfrac{29}{7},\infty\Big)$ Since the compound inequality is formed by the word "or", the solution is composed by the solutions of both inequalities: The solution is $(-\infty,1]\cup\Big(\dfrac{29}{7},\infty\Big)$