Can someone PLEASE help me out with Inequality? SOS

- anonymous

Can someone PLEASE help me out with Inequality? SOS

- schrodinger

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- anonymous

##### 1 Attachment

- anonymous

\[X>7 or -2

- anonymous

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## 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

- 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..

Looking for something else?

Not the answer you are looking for? Search for more explanations.