Anonymous

# inequalities and abs. value?

could someone please walk me through how to do a problem like this:

3<|3x-4|<12

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• Anonymous

3<|3x-4|<12

|3x -4| = 3x -4 when x > 4/3

|3x -4| = -(3x -4) when x < 4/3

3<|3x-4|<12

a) x > 4/3 , we solve

3< 3x-4 <12

let's add 4 to all term of the inequalities

3 +4 < 3x < 12 +4

7< 3x < 16

divide all term by 3 to get

7/3 < x < 16/3

b) x <4/3

we solve

3< -3x+4 <12

let's add -4 to all term of the inequalities

3 -4 < -3x < 12 -4

-1 < -3x < 8

let's divide by (-3) all terms (dividing by a negative number changes all < to > )

1/3 > x > -8/3

Summary:

a) x > 4/3

7/3 < x < 16/3 ---> solution domain1 =

b) x <4/3

1/3 > x > -8/3

-8/3 < x < 1/3 = solution domain2

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• Break off the absolute symbol by breaking the inequality into its two parts. For example, |x| > 5 turns into x > 5 and x < -5

So:

3 < (3x -4) < 12

or

-3 > (3x-4) > -12

7 < 3x < 16

or

1 > 3x > -8

Divide by 3 to all sides. If it were -3x, you'd have to flip the inequality symbol again, but here it stays the same

7/3 < x < 16/3

or

1/3 > x > -8/3

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• 3<|3x-4|<12

case 1: 3x-4 >0

|3x-4| = 3x-4

3x-4 < 12

3x < 16

x < 16/3

3x-4 >3

3x > 7

x > 7/3

7/3 < x < 16/3 --- (1)

case 2:

3x-4 < 0

3x-4 =-(3x-4)

=-3x+4

-3x+4 < 12

-3x < -8

3x > 8

x > 8/3

-3x+4 > 3

-3x > -1

3x < 1

x < 1/3

7/3 < x < 16/3 -- (1)

8/3 < x < 1/3 --- (2)

Draw a line and see where x lies based on (1) and (2). That is your solution.

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