SchoolFreak
INEQUAILTY !!!!
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SchoolFreak
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Please explain how to get it
Mimi_x3
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well..let's do the first first, alright?
Mimi_x3
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first one*
SchoolFreak
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okay please explain how to get answer i have no idea where to start on any of them ")
Mimi_x3
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\[ x- 3 > -9\]
do you know how to do it when its \((x-3) = -9\)
SchoolFreak
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well i would make the x into -3
Mimi_x3
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you are solving for \(x\)
SchoolFreak
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is that what we have to ??
SchoolFreak
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okay i read it
Mimi_x3
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if you have any problems then ask me..
SchoolFreak
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okay for the first one what do we do??
Mimi_x3
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\[x-3 >-9\]
we are solving for \(x\) why not you try it first?
SchoolFreak
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okay-3 is the answer because -3 times -3 is 9 and 9 is greater than -9 right
SchoolFreak
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@Mimi_x3
Mimi_x3
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what? \(x\) is variabe..not times..
Mimi_x3
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a variable*
SchoolFreak
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i thought were suppose to replace x with number??
Mimi_x3
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looks like you have to solve for \(x\) not substituting..
SchoolFreak
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ohhh k so we add 3 to both sides right?
saifoo.khan
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@SchoolFreak , mini isn't already helping you!
saifoo.khan
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is*
SchoolFreak
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i know
SchoolFreak
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@Mimi_x3 so then if we add 3 both sides we have x > -6
Mimi_x3
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Yes!
SchoolFreak
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thats the answer?
Mimi_x3
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yeah lol
Mimi_x3
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you are basically solving for \(x\) thats all
SchoolFreak
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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
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correct
SchoolFreak
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okay lets try the next one
SchoolFreak
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–2x + 1 ≤ –11
Mimi_x3
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why dont you try and tell me what you get
SchoolFreak
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okay i think the first step is to add 2 to both sides?
Mimi_x3
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think first; why add to 2? since its +1?
Mimi_x3
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just isolate the -2x in the first hand so you wont get confused
SchoolFreak
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okay so then you add 1 to both sides?
SchoolFreak
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@Mimi_x3 would substract 1 from both sides right?
Mimi_x3
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yes you subtract because it was a positive in the rhs
SchoolFreak
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–2x + 1 ≤ –11 Okay so now we have -2x ≤ 12
Mimi_x3
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not quite.. whats -11 - 1?
SchoolFreak
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ohhh -12
Mimi_x3
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yes, now you know what to do next
SchoolFreak
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-2x ≤ -12
now we divide both sides by -2 and we have x ≥ 6
Mimi_x3
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yes!
SchoolFreak
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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
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yes
SchoolFreak
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okay let me try this next one all by myself
SchoolFreak
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10 < –3x + 1
Mimi_x3
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alright, if you need help ask me :)
SchoolFreak
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kk
SchoolFreak
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-9 > x @Mimi_x3 is that the answer??
Mimi_x3
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nope..show me what you did
SchoolFreak
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10 < -3x + 1
9 < -3x
-9 > x
Mimi_x3
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you made a small mistake
whats 9/-3?
SchoolFreak
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3
Mimi_x3
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nope 9/ - 3
Mimi_x3
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like 9 divide by negative 3
SchoolFreak
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-3 is the answer??
Mimi_x3
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yes but dont forget the inequality signs
SchoolFreak
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-3 > x Is that the answer @Mimi_x3
Mimi_x3
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not quite..the sign is wrong
SchoolFreak
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No because we divide by a negative so we flip the sign
Mimi_x3
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10 < -3x + 1
9< -3x
-3x > 9
Mimi_x3
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understand so far?
SchoolFreak
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why did u turn the equation on the last part
Mimi_x3
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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
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you have to change*
SchoolFreak
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oh i didnt know that
Mimi_x3
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lets not turn the equation then
lets start from 9< -3x
SchoolFreak
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okay then then we dived and we -3 > x
Mimi_x3
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then you still have to change the x for it to be in the front so x> -3
Mimi_x3
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i mean x<-3
SchoolFreak
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ohh k
SchoolFreak
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@Mimi_x3 2(x + 5) > 8x – 8 this is the next one it looks hard lol
Mimi_x3
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not really..expand it
Mimi_x3
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the first side first
SchoolFreak
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if we do the first side we will have 2x + 10 right on the left side??
Mimi_x3
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yeah.
i gotta go now. sorry.
SchoolFreak
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noo :(
Mimi_x3
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hopefully someone else can help you..if not i'll be back later :)
SchoolFreak
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okay