Question

How do i solve question 25

Answer 1

I never learned to do operations with absolute values, so I don't have
the standard tools to use.  The only thing I have to work with when I run
into one of these is my brain.  So there are probably better ways to do it,
but here's how I would approach this one:

The fraction itself is either positive or negative.  Either way, its absolute
value is less than 4.

If the fraction is positive, then you can just throw away the absolute-value
bars, and work it like a normal inequality ... multiply each side by (x-3), then
work it down so there's nothing but 'x' on the left side, etc.

If the fraction is negative AND its absolute value is less than 4, then the
fraction itself is greater than -4 .  Throw away the absolute-value bars,
change the inequality to say that, and then work it like a normal inequality.

Do BOTH of these, and you'll come out with two inequalities for 'x' ... it's
more than something AND less than something else.  Put these together
into one inequality, and you have the whole range for what 'x' can be.  

Sounds complicated.  One of these days, I should learn the right way
to do these.

Answer 2

$\frac{x+2}{x-3} \ \textgreater \ 4$         or        $\frac{x+2}{x-3} \ \textless \ -4$

$x+2 \ \textgreater \ 4(x-3)$

$x+2 \ \textgreater \ 4x-12$

$x+14 \ \textgreater \ 4x$

$14 \ \textgreater \ 3x$

$4 \frac{2}{3} \ \textgreater \ x$  <-answer



$\frac{x+2}{x-3}\ \textless \ - 4$

$x+2\ \textless \ - 4(x-3)$

$x+2\ \textless \ - 4x+12$

$x-10\ \textless \ - 4x$

$-10\ \textless \ - 5x$

$2\ \textless \ x$  <-answer



2 < x < 14/3