can you help me with this?

**princess_21** · March 25th 2009, 06:37 AM

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

**skeeter** · March 25th 2009, 07:01 AM

$ 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 $

**princess_21** · March 25th 2009, 07:04 AM

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

**skeeter** · March 25th 2009, 09:26 AM

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

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