## anonymous 5 years ago need help solving this inequality please. x >-7

1. anonymous

...isn't that already solved? You know the values of x. That's about as simple as it gets

2. anonymous

it's a question on an exam prep....

3. anonymous

hm... anything more to it than just "solve..."?

4. anonymous

So what are the values of x that satisfy the relation x > -7

5. anonymous

like choices? Otherwise i think im at a loss

6. myininaya

(-7,infinity)

7. anonymous

how would I graph it....thats the other part of the question...sorry guys

8. anonymous

$x \in (-7, \infty)$ Graph the relation? or the solutions?

9. anonymous

on a number line, draw it with an open circle on -7, with an arrow to the right

10. anonymous

Thanks Bwah!

11. anonymous

on a cartesian plane, a vertical line at x=-7, dotted (non inclusive). shade to the right

12. myininaya

x=-7 is a vertical line so since we have x>-7 and not x>=-7 then you draw a broken vertical line at x=7 and since we have x>-7 shade everything to the right 0f -7

13. anonymous

14. anonymous

15. anonymous

well, the process is the same as any equation u just have an inequality sign at the end

16. anonymous

Depends on which dimensions he has to include in the graph. In $$R^1$$ it's just a number line. In $$R^2$$ it is a field comprised of all (x,y) pairs such that $$y\in (-\infty, \infty) \text{ and } x \in (-7, \infty)$$ . In $$R^3$$ it would be a boxy thing, and so on.

17. anonymous

I think I am lost on the process :)

18. anonymous

How would you solve 3-10x = 33?

19. anonymous

-10x>30 x<-3 (if i recall correctly, u switch the sign if u divide or multiply by a negative number)

20. anonymous

subtract 3 from both sides?

21. anonymous

Indeed. You can do the same thing to inequalities.

22. anonymous

Then what would you do?

23. anonymous

divide by 10......I think I am seeing the light :)

24. anonymous

Yep. The only rule is that if you divide or multiply by a negative you have to change the direction of the inequality. $$-x < 5 \implies x > 5$$

25. anonymous

Ack

26. anonymous

$-x < 5 \implies x > -5$ rather

27. anonymous

Thanks so very much!

28. myininaya

I don't know if you have ever though of like if we have 5>2 then and we multiply both sides by negative one or divide by negative one then it becomes -5<-2 since -2 is bigger than -5

29. myininaya

any any negative number