# Math Help - can you help me with this?

1. ## can you help me with this?

am I in the right track??

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

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

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

$28\ge-x>8$

i am confused when changing -x to x

$28\le x<-8$

is this correct??

another example
-x>-8
x<-8

thanks I'm confused

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

multiply every term by 4 ...

$
0 < 8-x \leq 36
$

subtract 8 ...

$
-8 < -x \leq 28
$

one inequality at a time ...

$-8 < -x$ ... $8 > x$ ... $x < 8$

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

final solution ...

$
-28 \leq x < 8
$

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

multiply every term by 4 ...

$
0 < 8-x \leq 36
$

subtract 8 ...

$
-8 < -x \leq 28
$

one inequality at a time ...

$-8 < -x$ ... $8 > x$ ... $x < 8$

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

final solution ...

$
-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