sylinan
help please(: 1 over 3 (2x − 8) = 4



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Limitless
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\[\frac{1}{3(2x8)}=4\]
What do you suggest from here?

hba
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\[\frac{ 1 }{ 3(2x8) }=4\]
Use distributive property to evaluate your denominator

sylinan
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@Limitless its 1 over 3 and next to (2x − 8) = 4

Limitless
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\[\frac{1}{3}(2x8)=4\]
What do you suggest from here?

sylinan
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@Limitless distributing?

hba
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\[\frac{ 1 }{ 3 }(2x8)=4\]
1)Multiply both sides by 3

Limitless
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@sylinan Sure. Then what?

sylinan
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@hba doesnt it cancel out?

hba
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Yes.

Limitless
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You get \[\frac{2}{3}x\frac{8}{3}=4.\]

hba
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@Limitless
Stop complicating it.

sylinan
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im getting confused ):

hba
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Multiply your question by 3

Limitless
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@hba Complicating it would be using the Taylor series for \(x\). I apologize for replying when she was talking to you; I misread the post.

hba
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@sylinan \[\frac{ 1 }{ 3 }(2x8)=4\]
Multiply both sides by 3.

hba
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\[3 \times\frac { 1 }{ 3 }(2x8)=4 \times 3\]

hba
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@sylinan
I know you can do it now :)

sylinan
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@hba okay so far i got 3(2x8)=12

hba
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Now see this
\[\ \cancel 3 \times \frac{ 1 }{\ \cancel 3 }(2x8)=4 \times 3\]

sylinan
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so then its 1(2x8)=12?

hba
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Now add both sides by 8

sylinan
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i thought i was suposed to do distribute the one?

hba
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It doesn't make a diff
It will stay the same :)

sylinan
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true okay so i got x=10?

hba
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Yeaahhhh :D

hba
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You are right :P

sylinan
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thank you (: help me with one more?(:

hba
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Sure,Please post your question in a new tab :)

sylinan
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okay so open a new question?

hba
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Yeah.
The CoC says that not me :P

sylinan
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oh okay (: