Can someone PLEASE help me out with Inequality? SOS

- anonymous

Can someone PLEASE help me out with Inequality? SOS

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- anonymous

##### 1 Attachment

- anonymous

\[X>7 or -2

- anonymous

that's the answer, and I don't understand how to do that!

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## More answers

- anonymous

PLEASE @Emeyluv99 ?

- anonymous

factorize the top and the bottom separately first

- anonymous

you mean =0?

- anonymous

yes

- anonymous

so we get (x+3)((x-6) for the nominator

- anonymous

and for the denominator we get (x-7)(x+2)

- anonymous

ok, now, what values of x cannot work

- anonymous

\[\frac{ (x-6)(x+3) }{ (x-7)(x+2) }\]

- anonymous

7 and 2

- anonymous

right?

- anonymous

but how from here?

- anonymous

can you please show how to calculate from here?

- anonymous

@Emeyluv99

- anonymous

sorru. computers being mean.. and be careful, its 7 and -2

- anonymous

now the way I would do it, graph the quadratic you get in the numerator. See what values of x give a positive y and that's you're answer. Be careful to exclude 7 and -2 though

- anonymous

that's what I did but I didn't get the asnswer

- anonymous

from graphing you should get x<-3 and x>6... but, since we can't have 7 you should get X<-3, 67

- anonymous

but that's not the answer... thanks though

- anonymous

@ganeshie8 will you be savior?

- anonymous

my

- ganeshie8

|dw:1440157290238:dw|

- ganeshie8

pick any number to the left of -3
say -4
plug x = -4 in the given inequality

- ganeshie8

\[\frac{ (x-6)(x+3) }{ (x-7)(x+2)}\]
plugging x=-4 gives
\[\frac{ (-4-6)(-4+3) }{ (-4-7)(-4+2) }\]
looks it is positive, yes ?

- anonymous

yes

- anonymous

but why would you put -4 from the first place?

- anonymous

?

- ganeshie8

you can put any number to the left of -3 for testing
since -4 is easy to work..

- anonymous

oh ok, then what am I doing from here?

- ganeshie8

since a number to the left of -3 satisfies the inequality, \(x\lt 3\) is part of the solution :
|dw:1440158449273:dw|

- ganeshie8

lets check the next interval \((-3, -2)\)
pick a number between -3 and -2

- ganeshie8

thr ? @Hipocampus

- anonymous

-2.5

- anonymous

sorry there's a slightly problem with the internet

- anonymous

sorry I'm not following thanks though

- anonymous

it's supposed to be a really easy question

- ganeshie8

\[\frac{ (x-6)(x+3) }{ (x-7)(x+2)}\]
plugging x=-2.5 gives
\[\frac{ (-2.5-6)(-2.5+3) }{ (-2.5-7)(-2.5+2) }\]
looks it is negative, yes ?

- anonymous

yes

- ganeshie8

so the interval (-3, -2) is not a solution :
|dw:1440159462256:dw|

- ganeshie8

to test the next interval \((-2,6)\) pick some easy number between them and plug it in the inequality

- anonymous

so I need to plug in 4 times?

- anonymous

can we try -1.5 together please? I think I've got it!

- ganeshie8

try some easy number like x=0

- ganeshie8

0 is between -2 and 6

- ganeshie8

\[\frac{ (x-6)(x+3) }{ (x-7)(x+2)}\]
plugging x=0 gives
\[\frac{ (0-6)(0+3) }{ (0-7)(0+2) }\]
looks it is positive, yes ?

- anonymous

YES!!

- anonymous

so between -2 and 6 it's ok

- anonymous

8 is negative then no

- anonymous

and 7.5 is fine likewise

- anonymous

so what we get is that -3>x, -27

- anonymous

you're brilliant! thank you so so much!!!!!!

- anonymous

7.5 isn't good right

- ganeshie8

|dw:1440159894339:dw|

- ganeshie8

double check with wolfram
http://www.wolframalpha.com/input/?i=solve+%28x%5E2-3x-18%29%2F%28x%5E2-5x-14%29%3E%3D0

- anonymous

thanks a lot!!!!!

- ganeshie8

np, the basic procedure for solving rational inequalities is :
1) factor both numerator and denomiantor
2) find the x values where the numerator or denominator equal 0
3) plot them on number line
4) pick a number in each interval and test the inequality

- anonymous

@ganeshie8 is there any easier way to solve it?

- ganeshie8

hmm i can't think of any other easy way to solve it.. post it again and see what others have to say..

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