Exercise 42 Page 198

To solve the inequality, we have to isolate x. By distributing - 2 on the left-hand side and using inverse operations, we can isolate the variable.

- 2(3x+7)≥- 14-6x

Distr

Distribute - 2

- 6x-14≥- 14-6x

AddIneq

LHS+14≥RHS+14

- 6x≥- 6x

AddIneq

LHS+6x≥RHS+6x

0≥ 0

The statement 0≥ 0 is always true. A number will always be equal to itself. Therefore, all real numbers are solutions to this equation. Now we can use the result to form a solution set in set-builder notation. { x | xis a real number }