Packet Help

Dakota120x

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Joined
Jul 24, 2005
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Here 2 problems in which my friend and I have had very different answers. Thanks for your help.

29. Solve the inequality -4x-7<-6 (lesser or equal to sign)

My answer was x < 5 (lesser or equal to sign)

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30. Solve the absolute value inequality |2x-4|<2

My answer was 0<x<3
 
29) Solve -4x - 7 < -6

You say the solution is "x < 5". I'll check, say, x = -1:

. . . . .-4(-1) - 7 = 4 - 7 = -3 > -6

Check your work. If you get the same answer again, please reply showing your steps.

30) Solve |2x - 4| < 2.

You say the solution is "0 < x < 3". I'll check at x = 1/2:

. . . . .|2(1/2) - 4| = |1 - 4| = |-3| = 3 > 2

Check your work. If you get the same answer again, please reply showing your steps.

Thank you.

Eliz.
 
I did number 29 again and i plugged it in and go it wrong but I did the whole problem again and got -1/4 and thats right!! Now I am off to try the other one. :)
 
Dakota120x said:
29. Solve the inequality -4x-7 <= -6 My answer was x <= 5
Well, let's see...

-4x - 7 <= -6 -- Add 7
-4x <= 1 -- Divide by -4 (Reverse the inequality)
x >= -1/4

I hate to ask, but what was your friend's answer?

30. Solve the absolute value inequality |2x-4|<2 My answer was 0<x<3
2x-4 >= 0
2x >= 4
x >= 2

For x >= 2, |2x-4| = 2x-4
2x - 4 < 2
2x < 6
x < 3 -- For x >= 2, x < 3. Solutions. [2,3)

For x < 2, |2x-4| = -(2x-4)
-(2x-4) < 2
4 - 2x < 2
4 < 2 + 2x
2 < 2x
1 < x -- For x < 2, x > 1. Solutions. (1,2)

Total Solution 1 < x < 3, or (1,3)

You were close. Why did you include (0,1]?
 
For number 30 I did this lol (i did it like a week ago)

-2 < |2x-4| > 2 *then i added 4 to both sides and middle*
-2 + 4 < |2x-4+4| > 2 + 4
2 < | 2x | > 6 *then i divided by 2*

0 < x < 3

so thats how i did it LOL!!
 
Dakota120x said:
-2 < |2x-4| > 2
No; it works like this:

. . . . .-2 < 2x - 4 < 2

The point of splitting it up like this is to get rid of the bars. And the inequality signs have to make sense.

Dakota120x said:
2 < | 2x | > 6
*then i divided by 2*
0 < x < 3
Check your work. What is 2 ÷ 2?

Eliz.
 
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