## Intermediate Algebra (6th Edition)

We are given that $x-2\lt6$ or $3x+1\gt1$. Therefore, a solution only needs to satisfy one of the inequalities. We can first solve $x-2\lt6$ for x. Add 2 to both sides. $x\lt8$ Next, can first solve $3x+1\gt1$ for x. Subtract 1 from both sides. $3x\gt0$ Divide both sides by 3. $x\gt0$ Therefore, the solution set consists of all numbers that are either greater than 0 or less than 8, which includes all real numbers.