INEQUAILTY !!!!

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- anonymous

INEQUAILTY !!!!

- schrodinger

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- anonymous

Please explain how to get it

##### 1 Attachment

- Mimi_x3

well..let's do the first first, alright?

- Mimi_x3

first one*

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## More answers

- anonymous

okay please explain how to get answer i have no idea where to start on any of them ")

- Mimi_x3

\[ x- 3 > -9\]
do you know how to do it when its \((x-3) = -9\)

- anonymous

well i would make the x into -3

- Mimi_x3

you are solving for \(x\)

- anonymous

is that what we have to ??

- Mimi_x3

yes..
why not read this first to gain a better understanding..
http://www.sosmath.com/algebra/inequalities/ineq01/ineq01.html

- anonymous

okay i read it

- Mimi_x3

if you have any problems then ask me..

- anonymous

okay for the first one what do we do??

- Mimi_x3

\[x-3 >-9\]
we are solving for \(x\) why not you try it first?

- anonymous

okay-3 is the answer because -3 times -3 is 9 and 9 is greater than -9 right

- anonymous

- Mimi_x3

what? \(x\) is variabe..not times..

- Mimi_x3

a variable*

- anonymous

i thought were suppose to replace x with number??

- Mimi_x3

looks like you have to solve for \(x\) not substituting..

- anonymous

ohhh k so we add 3 to both sides right?

- saifoo.khan

@SchoolFreak , mini isn't already helping you!

- saifoo.khan

is*

- anonymous

i know

- anonymous

@Mimi_x3 so then if we add 3 both sides we have x > -6

- Mimi_x3

Yes!

- anonymous

thats the answer?

- Mimi_x3

yeah lol

- Mimi_x3

you are basically solving for \(x\) thats all

- anonymous

oh haha x is greater than -6 , the graph would have open circle on -6 shading to the right is that right to?

- Mimi_x3

correct

- anonymous

okay lets try the next one

- anonymous

–2x + 1 ≤ –11

- Mimi_x3

why dont you try and tell me what you get

- anonymous

okay i think the first step is to add 2 to both sides?

- Mimi_x3

think first; why add to 2? since its +1?

- Mimi_x3

just isolate the -2x in the first hand so you wont get confused

- anonymous

okay so then you add 1 to both sides?

- anonymous

@Mimi_x3 would substract 1 from both sides right?

- Mimi_x3

yes you subtract because it was a positive in the rhs

- anonymous

–2x + 1 ≤ –11 Okay so now we have -2x ≤ 12

- Mimi_x3

not quite.. whats -11 - 1?

- anonymous

ohhh -12

- Mimi_x3

yes, now you know what to do next

- anonymous

-2x ≤ -12
now we divide both sides by -2 and we have x ≥ 6

- Mimi_x3

yes!

- anonymous

woo hooo also x is greater than or equal to 6 the graph would have a closed circle on 6 shading to the right

- Mimi_x3

yes

- anonymous

okay let me try this next one all by myself

- anonymous

10 < –3x + 1

- Mimi_x3

alright, if you need help ask me :)

- anonymous

kk

- anonymous

-9 > x @Mimi_x3 is that the answer??

- Mimi_x3

nope..show me what you did

- anonymous

10 < -3x + 1
9 < -3x
-9 > x

- Mimi_x3

you made a small mistake
whats 9/-3?

- anonymous

3

- Mimi_x3

nope 9/ - 3

- Mimi_x3

like 9 divide by negative 3

- anonymous

-3 is the answer??

- Mimi_x3

yes but dont forget the inequality signs

- anonymous

-3 > x Is that the answer @Mimi_x3

- Mimi_x3

not quite..the sign is wrong

- anonymous

No because we divide by a negative so we flip the sign

- Mimi_x3

10 < -3x + 1
9< -3x
-3x > 9

- Mimi_x3

understand so far?

- anonymous

why did u turn the equation on the last part

- Mimi_x3

i dont know how to explain..but for inequalities to have to change the sign when you swap the equation..
you dont need to turn it; i just turned to prevent confusion

- Mimi_x3

you have to change*

- anonymous

oh i didnt know that

- Mimi_x3

lets not turn the equation then
lets start from 9< -3x

- anonymous

okay then then we dived and we -3 > x

- Mimi_x3

then you still have to change the x for it to be in the front so x> -3

- Mimi_x3

i mean x<-3

- anonymous

ohh k

- anonymous

@Mimi_x3 2(x + 5) > 8x – 8 this is the next one it looks hard lol

- Mimi_x3

not really..expand it

- Mimi_x3

the first side first

- anonymous

if we do the first side we will have 2x + 10 right on the left side??

- Mimi_x3

yeah.
i gotta go now. sorry.

- anonymous

noo :(

- Mimi_x3

hopefully someone else can help you..if not i'll be back later :)

- anonymous

okay

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