Exercise 44 Page 198

To solve the inequality, we have to isolate x. By distributing - 3 on the left-hand side and using inverse operations, we can isolate the variable.

- 3(4x+8)+1≥- 23-12x

Distr

Distribute - 3

- 12x-24+1≥- 23-12x

AddTerms

Add terms

- 12x-23≥- 23-12x

AddIneq

LHS+23≥RHS+23

- 12x≥ -12x

AddIneq

LHS+12x≥RHS+12x

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 | x is a real number}