## anonymous 4 years ago x-2>-5 or 5x>25

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1. lgbasallote

first solve x - 2 > -5 do this by adding 2 to both sides. what do you get?

2. anonymous

-3 ?

3. lgbasallote

right x > -3

4. lgbasallote

now solve 5x > 25

5. anonymous

btw the said was suppose to be " greater than equal to"

6. lgbasallote

which one?

7. anonymous

so i just wrote greater than

8. ParthKohli

Doesn't matter.

9. ParthKohli

Sorry for interrupting, but you have to solve these exactly like you solve $$\rm x - 2=-5$$ and $$5x = 25$$.

10. lgbasallote

$x - 2 \ge -5 \; \text{or} \; 5x \ge 25$ ??

11. anonymous

Yes that correct except the first one is the the greater than symbol

12. lgbasallote

oh so $x - 2 > -5 ~\text{or} ~ 5x \ge 25$

13. anonymous

Yess

14. lgbasallote

so anyway $5x \ge 25$ do you know how to solve for x?

15. anonymous

x=5 ?

16. lgbasallote

x $$\ge$$ 5 the symbols matter

17. lgbasallote

so since x - 2 > -5 is x > -3 and 5x $$\ge$$ 25 is x$$\ge$$ 5 your answer would be x > -3 or x$$\ge$$ 5

18. lgbasallote

@ParthKohli or is different from and

19. anonymous

but when it comes to graphing x>-3 is not included

20. lgbasallote

if you mean open circle..yes

21. anonymous

yeah i mean open circle , greater than is always included

22. lgbasallote

greater than is always included?

23. anonymous

lol nvm i cant explain it here :P

24. lgbasallote

i notice

25. anonymous

Okay guys one more question

26. lgbasallote

shoot

27. anonymous

excluded....you have to draw a dotted line if it is greater than and you have to draw a dark complete line for greater than or equal to....

28. anonymous

hahah thats exactly what i meant ^

29. anonymous

sorry @lgbasallote if I interrupted you..you can carry on with your client..

30. anonymous

l3xl +8>5

31. lgbasallote

first of all, subtract 8 from both sides

32. anonymous

yeah so its 5-8=-3

33. lgbasallote

right |3x| > -3

34. anonymous

so absolute value of 3x=-3 ?

35. lgbasallote

again. remember that signs matter. > is different from =

36. anonymous

yeah sorry

37. lgbasallote

you'll see the importance in the next step

38. lgbasallote

since |3x| > -3 you turn this into two equations 3x > -3 and 3x < 3 now solve for x in each one

39. anonymous

6 and 1 ?

40. lgbasallote

hmm first solve 3x > -3 divide both sides by 3

41. anonymous

-1

42. anonymous

x=-1 or x=1 ?

43. lgbasallote

...signs...

44. lgbasallote

it's not equal signs...

45. anonymous

sign

46. anonymous

nvm then

47. lgbasallote

it should be x>-1 and x<1 signs are very important

48. anonymous

Alright im done , thank you guys so much . you really helped me

49. ParthKohli

@lgbasallote Try to think! If $$\rm x > 5$$, it still satisfies $$\rm x\ge -3$$

50. lgbasallote

in this case yes it is. but the fact remains that or is different from and