Question 348550
Note: I'm assuming that the first inequality should be {{{-4(x+2)>=12}}} and the solution to that inequality is {{{x<=-5}}}



Since x can be either less than or equal to -5 or less than 1, this means that the solution set is really just {{{x<1}}} (since {{{-5}}} is less than 1). We can completely ignore {{{x<=-5}}} as this entire solution set lies within {{{x<1}}}



So the solution to {{{-4(x+2)>=12}}} or {{{3x+8<11}}} is {{{x<1}}} and it is represented in interval notation by *[Tex \LARGE \left(-\infty,1\right)]



If you need more help, email me at <a href="mailto:jim_thompson5910@hotmail.com?Subject=Algebra%20Help">jim_thompson5910@hotmail.com</a>


Also, feel free to check out my <a href="http://www.freewebs.com/jimthompson5910/home.html">tutoring website</a>


Jim