Originally Posted by

**Devi09** a) $\displaystyle \frac{\2x+3}{3} \leq x-5$

**according to your next line, **

that should be 2x+3 in the numerator which will give the answer provided

I did this:

$\displaystyle \frac{\3(2x+3)}{3} \leq 3(x-5)$

**why didn't you multiply BOTH sides by 3 ?**

$\displaystyle 6x+9 \leq 3x-15$

**Now you multiplied only the left side by 9.**

If you multiply only one side by something other than 1,

you will have a different inequality.

$\displaystyle 6x-3x \leq -15-9$

$\displaystyle 3x \leq -24$

$\displaystyle x \leq -8$

$\displaystyle (inf , -8]$

however the answer should be : $\displaystyle [18,inf)$

b) another one

$\displaystyle \frac{\ x+4}{2} \geq \frac{\ x-2}{4}$

$\displaystyle \frac{\2(x+4)}{2} \geq \frac{\2(x-2)}{2(4)} $

**You halved the right and did nothing to the other side,**

so you now have a different inequality altogether.

You could have multiplied both sides by 4.

$\displaystyle 2x+8 \geq \frac{\2x-4}{8}$

**You multiplied the left by 4**

and changed the denominator of the other side.

Any reason ?

$\displaystyle 8(2x+8) \geq \frac{\8(2x-4)}{8}$

**You multiplied the left by 8 and made a mysterious modification**

to the numerator on the right again.

The rest is also haphazard.

$\displaystyle 16x+64 \geq 16x-32$

$\displaystyle 16x-16x \geq -64 -32$

$\displaystyle 0 \geq -96$

however the answer should be $\displaystyle [-10, inf)$

c) last one

$\displaystyle 3x \leq x+1 \leq x-1$

$\displaystyle 3x-1 \leq x \leq x-1-1$

don't know what to do after this step