Question:solve

1. / 2/3x-4 /+20<10

2. 3 /2x+4/ -24=12

rewrite as single inequality involving absolute value

1. x<negative2 or x>10

PROBLEM 1:

|(2/3)x-4| + 20 < 10

subtract "20" from both sides... like this...

[ |(2/3)x-4| + 20] - 20 < 10 - 20

combine like terms. and you get...

|(2/3)x-4| < -10

There is no solution to this problem since the absolute value of (2/3)x-4 . or anything for that matter cannot be LESS THAN a negative number. Absolute value of ANYTHING always equals a positive number. and Positive numbers are ALWAYS larger than negative numbers, not smaller.

PROBLEM 2:

3 |2x+4| -24=12

( 3 |2x+4| - 24 ) + 24 = 12 + 24

combine like terms...

3 |2x+4| = 36

divide both sides by 3

|2x+4| = 12

2x + 4 = 12 . 2x = 8. x = 4

-(2x+4) = 12 . 2x+4 = -12 ... 2x = -16 . x = -8

So... the solution is "x = 4"... and "x = -8"

LAST PROBLEM:

x< -2 or x > 10 has to make the inequality true.

The inequality can be either.

|4-x| + 4 > 10

OR

|4-x| > 6
if you want to get help on your homeworks, do it properly

you cannot put 2 - 3 problem and ask for solutions

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