For this case we must resolve each of the inequalities and find the solution set.
Inequality 1:
[tex]12x + 7 <-11[/tex]
We subtract 7 from both sides of the inequality:
[tex]12x <-11-7\\12x <-18[/tex]
We divide between 12 on both sides of the inequality:
[tex]x <- \frac {18} {12}\\x <- \frac {9} {6}\\x <- \frac {3} {2}[/tex]
Thus, the solution is given by all values of x less than[tex]- \frac {3} {2}.[/tex]
Inequality 2:
[tex]5x-8> 40[/tex]
We add 8 to both sides of the inequality:
[tex]5x> 40 + 8\\5x> 48[/tex]
We divide between 5 on both sides of the inequality:
[tex]x> \frac {48} {5}[/tex]
Thus, the solution is given by all values of x greater than[tex]\frac {48} {5}.[/tex]
The solution set is given by:
(-∞, [tex]- \frac {3} {2}[/tex]) U ([tex]\frac {48} {5}[/tex],∞)
Answer:
(-∞, [tex]- \frac {3} {2}[/tex]) U ([tex]\frac {48} {5}[/tex],∞)
