## anonymous one year ago What is the compound inequality 8 < or = to 2x-4 < 2

1. DanJS

hmm

2. anonymous

Hmm what?

3. DanJS

$8 \le (2x-4) \lt 2$ that right?

4. anonymous

Yes

5. DanJS

that is a compound inequality

6. DanJS

a<b<c

7. anonymous

Except there is no parentheses

8. anonymous

Ya but i have to find the solution to it

9. DanJS

so you have 8 is less than or equal to 2x-4 AND 2x-4 is less than 2

10. DanJS

oh, do you want to just isolate x to get the values for x, solve it

11. anonymous

My options are: 1. -6 < or = to X < -4 2. -2 < or = to X < -1 3. 0 < or = to X < 5 4. -2 < or = to X < 3

12. DanJS

$8 \le (2x-4) \lt 2$ $8 +4\le (2x) \lt 2 + 4$ $\frac{ 12 }{ 2 } \le x \lt \frac{ 6 }{ 2 }$ this says $6 \le x \lt 3$ which is not true, x can not be bigger than 6 AND less than 3 at the same time

13. DanJS

It ca be either OR

14. DanJS

you mistyped something in the prob i think

15. anonymous

I'm confused

16. DanJS

the way you put the problem, you cant have solutions for that, you can break it up into $x \ge 6~~~~OR~~~~x \lt 3$

17. DanJS

a value can't be both those at the same time,

18. anonymous

I just put it the way that the school has it

19. DanJS

you missing a negative sign, or something,

20. DanJS

put the question down again using the draw tool

21. DanJS

the inequality part

22. anonymous

|dw:1443228817681:dw|

23. anonymous

Is that what you mean @DanJS ?

24. anonymous

i say its 1

25. anonymous

can you explain why? I am so confused

26. anonymous

was it really $-8 \le x-4\lt 2$

27. anonymous

or $-8 \le2 x-4\lt 2$

28. anonymous

the second one

29. anonymous

add 4 all the way across$-4\leq 2x<6$ then divide all by 2

30. anonymous

so it is $-2lex <3$

31. anonymous

yeah

32. anonymous

$-2 \le x < 3$

33. anonymous

$-2\leq x<3$

34. anonymous

Thank you so much!