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OpenStudy (mimi_x3):
if you have any problems then ask me..
OpenStudy (anonymous):
okay for the first one what do we do??
OpenStudy (mimi_x3):
\[x-3 >-9\]
we are solving for \(x\) why not you try it first?
OpenStudy (anonymous):
okay-3 is the answer because -3 times -3 is 9 and 9 is greater than -9 right
OpenStudy (anonymous):
@Mimi_x3
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OpenStudy (mimi_x3):
what? \(x\) is variabe..not times..
OpenStudy (mimi_x3):
a variable*
OpenStudy (anonymous):
i thought were suppose to replace x with number??
OpenStudy (mimi_x3):
looks like you have to solve for \(x\) not substituting..
OpenStudy (anonymous):
ohhh k so we add 3 to both sides right?
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OpenStudy (saifoo.khan):
@SchoolFreak , mini isn't already helping you!
OpenStudy (saifoo.khan):
is*
OpenStudy (anonymous):
i know
OpenStudy (anonymous):
@Mimi_x3 so then if we add 3 both sides we have x > -6
OpenStudy (mimi_x3):
Yes!
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OpenStudy (anonymous):
thats the answer?
OpenStudy (mimi_x3):
yeah lol
OpenStudy (mimi_x3):
you are basically solving for \(x\) thats all
OpenStudy (anonymous):
oh haha x is greater than -6 , the graph would have open circle on -6 shading to the right is that right to?
OpenStudy (mimi_x3):
correct
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OpenStudy (anonymous):
okay lets try the next one
OpenStudy (anonymous):
–2x + 1 ≤ –11
OpenStudy (mimi_x3):
why dont you try and tell me what you get
OpenStudy (anonymous):
okay i think the first step is to add 2 to both sides?
OpenStudy (mimi_x3):
think first; why add to 2? since its +1?
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OpenStudy (mimi_x3):
just isolate the -2x in the first hand so you wont get confused
OpenStudy (anonymous):
okay so then you add 1 to both sides?
OpenStudy (anonymous):
@Mimi_x3 would substract 1 from both sides right?
OpenStudy (mimi_x3):
yes you subtract because it was a positive in the rhs
OpenStudy (anonymous):
–2x + 1 ≤ –11 Okay so now we have -2x ≤ 12
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OpenStudy (mimi_x3):
not quite.. whats -11 - 1?
OpenStudy (anonymous):
ohhh -12
OpenStudy (mimi_x3):
yes, now you know what to do next
OpenStudy (anonymous):
-2x ≤ -12
now we divide both sides by -2 and we have x ≥ 6
OpenStudy (mimi_x3):
yes!
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OpenStudy (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
OpenStudy (mimi_x3):
yes
OpenStudy (anonymous):
okay let me try this next one all by myself
OpenStudy (anonymous):
10 < –3x + 1
OpenStudy (mimi_x3):
alright, if you need help ask me :)
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OpenStudy (anonymous):
kk
OpenStudy (anonymous):
-9 > x @Mimi_x3 is that the answer??
OpenStudy (mimi_x3):
nope..show me what you did
OpenStudy (anonymous):
10 < -3x + 1
9 < -3x
-9 > x
OpenStudy (mimi_x3):
you made a small mistake
whats 9/-3?
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OpenStudy (anonymous):
3
OpenStudy (mimi_x3):
nope 9/ - 3
OpenStudy (mimi_x3):
like 9 divide by negative 3
OpenStudy (anonymous):
-3 is the answer??
OpenStudy (mimi_x3):
yes but dont forget the inequality signs
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OpenStudy (anonymous):
-3 > x Is that the answer @Mimi_x3
OpenStudy (mimi_x3):
not quite..the sign is wrong
OpenStudy (anonymous):
No because we divide by a negative so we flip the sign
OpenStudy (mimi_x3):
10 < -3x + 1
9< -3x
-3x > 9
OpenStudy (mimi_x3):
understand so far?
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OpenStudy (anonymous):
why did u turn the equation on the last part
OpenStudy (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
OpenStudy (mimi_x3):
you have to change*
OpenStudy (anonymous):
oh i didnt know that
OpenStudy (mimi_x3):
lets not turn the equation then
lets start from 9< -3x
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OpenStudy (anonymous):
okay then then we dived and we -3 > x
OpenStudy (mimi_x3):
then you still have to change the x for it to be in the front so x> -3
OpenStudy (mimi_x3):
i mean x<-3
OpenStudy (anonymous):
ohh k
OpenStudy (anonymous):
@Mimi_x3 2(x + 5) > 8x – 8 this is the next one it looks hard lol
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OpenStudy (mimi_x3):
not really..expand it
OpenStudy (mimi_x3):
the first side first
OpenStudy (anonymous):
if we do the first side we will have 2x + 10 right on the left side??
OpenStudy (mimi_x3):
yeah.
i gotta go now. sorry.
OpenStudy (anonymous):
noo :(
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OpenStudy (mimi_x3):
hopefully someone else can help you..if not i'll be back later :)