Question

help please !

5+3/|2-3x| < 8

Answer 1

\[5 + \frac{3}{|2 - 3x|} < 8\]

Answer 2

Is that what you're working on?

Answer 3

So anyway, I'm just going to post the solution assuming that what I have posted is correct .

Answer 4

yup , sorry for late reply .

Answer 5

\[
\frac{3}{|2 - 3x|} < 8 - 5
\\\frac{3}{|2 - 3x|} < 3
\\\frac{1}{|2 - 3x|} < \frac{3}{3}
\\\frac{1}{|2 - 3x|} < 1
\\1 < |2 - 3x|
\\|2 - 3x| > 1
\\|3x - 2| > 1
\\3x - 2 > \pm 1
\\-1> 3x - 2 > 1
\\-1 + 2 > 3x > 1 + 2
\\1 > 3x > 3
\\1/3 > x > 3/3
\\1/3 > x > 1
\\ x > 1 \text{ or } x < 1/3
\]

Answer 6

okay thank you for the help . Really appreciate it

Answer 7

I hope you understood the steps. You can do it several ways.

Answer 8

There are a couple of rules of use to know like for example

|x - y| = |y - x|

Answer 9

for example if x = 5 and y = 9, then

|5 - 9| = |-4| = 4
|9 - 5| = | 4| = 4

Answer 10

I like that rule because I never have to worry about dividing by negative numbers. I never divide by a negative number.

Answer 11

okay , but how about this question .

Answer 12

1/( x+2 ) > - 1/( x -2)

Answer 13

umm , can you show the calculations ?

Answer 14

\[\frac{1}{x+2} > -\frac{1}{x-2}
\\\space{}\\
\\\text{Multiply the fraction on the left side by (x+2)/(x+2):
}\\\space{}\\
\\\frac{x+2}{(x+2)^2} > -\frac{1}{x-2}
\\\space{}\\
\text{Cross Multiply:}
\\\space{}\\
\\(x+2)(x-2) > -(x+2)^2
\\\space{}\\
\text{Expand:}
\\\space{}\\
\\x^2 - 4 >-(x^2 + 4x + 4)
\\x^2 - 4 > -x^2 - 4x - 4
\\x^2 + x^2+4x + 4 - 4 > 0
\\2x^2 + 4x > 0
\\2x(x+2) > 0
\\2x > 0
\\x + 2 > 0
\]

Obviously when we solve 2x > 0, x = 0, so 0 > 0 makes no sense, however we can use it as a critical point.

Solving
x + 2 >0
x > -2

So we have critical points -2, 0, and 2

Thus our interval is
\[(-2,0), (2,\infty)\]