# Thread: can you help me with this?

1. ## can you help me with this?

am I in the right track??

$\displaystyle 9\ge\frac{8-x}{4}>0$

$\displaystyle 9-2\ge\frac{-x}{4}>-2$

$\displaystyle 7(4)\ge-x>2(4)$

$\displaystyle 28\ge-x>8$

i am confused when changing -x to x

$\displaystyle 28\le x<-8$

is this correct??

another example
-x>-8
x<-8

thanks I'm confused

2. $\displaystyle 0 < \frac{8-x}{4} \leq 9$

multiply every term by 4 ...

$\displaystyle 0 < 8-x \leq 36$

subtract 8 ...

$\displaystyle -8 < -x \leq 28$

one inequality at a time ...

$\displaystyle -8 < -x$ ... $\displaystyle 8 > x$ ... $\displaystyle x < 8$

$\displaystyle -x \leq 28$ ... $\displaystyle x \geq -28$

final solution ...

$\displaystyle -28 \leq x < 8$

3. Originally Posted by skeeter
$\displaystyle 0 < \frac{8-x}{4} \leq 9$

multiply every term by 4 ...

$\displaystyle 0 < 8-x \leq 36$

subtract 8 ...

$\displaystyle -8 < -x \leq 28$

one inequality at a time ...

$\displaystyle -8 < -x$ ... $\displaystyle 8 > x$ ... $\displaystyle x < 8$

$\displaystyle -x \leq 28$ ... $\displaystyle x \geq -28$

final solution ...

$\displaystyle -28 \leq x < 8$

so this means i will change the signs of both terms?/
like this -x>8 then x<-8

4. Originally Posted by princess_21
so this means i will change the signs of both terms?/
like this -x>8 then x<-8
yes ... what you have done is multiply both sides of the inequality by -1 ... you also must change the direction of the sign anytime you multiply or divide by a negative value.

for example ...

5 > 4

multiply both sides by -1 ...

-5 < -4