## anonymous 4 years ago Solve compound inequality. Write solution in interval notation. x+9<0 or 2x> -12 (these kill me)

1. y2o2

it has no solution

2. y2o2

x cant be grater than -6 and smaller than -9

3. ash2326

Jmallis it's easy, if we proceed step by step Let's solve the first equation x+9<0 subtract 9 from both sides x<-9 so x is $$(-\infty,-9)$$ now from the second equation 2x>-12 or x>-6 so from this equation x is $$(-6, \infty)$$ now it's given x+9<0 or x>-12 so we have to take union of the two ie x belongs to $$(-\infty,-9)$$ U $$(-6, \infty)$$

4. anonymous

Thank-you, you think you can help me with another one ash2326

5. ash2326

yeah jmallis, post your question :)

6. ash2326

Jmallis , post your question as a new thread, that way you can get more people to respond:)

7. anonymous

$-6\le2x-12$$\le18$

8. anonymous

should be all together

9. ash2326

we have -6<=2x-12<=18 add 12 to all sides 6<=2x<=30 divide be 2 the whole inequation 3<=x<=10 As we can see x belongs to [3, 10] do notice that it's closed interval

10. anonymous

says to solve compund inequality written in compact form, get x alone in the middle part. Since a compound inequality is really two inequalities in one statement, perform same operations on all 3 parts of inequality

11. ash2326

Yeah , that's what I've done, notice I have added 12 to all three simultaneously, and divided by 2 all together.

12. anonymous

When I put in [3,10] my homework page had a spaz attack

13. ash2326

sorry I made a mistake see this we have -6<=2x-12<=18 add 12 to all sides 6<=2x<=30 divide be 2 the whole inequation 3<=x<=15 As we can see x belongs to [3, 15] do notice that it's closed interval It won't get a attack now:D

14. anonymous

It's fine now haha