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Mathematics 16 Online
OpenStudy (anonymous):

If x = y/2 + 6, what is y in terms of x?

OpenStudy (anonymous):

the problem looks like this, right? \[x = \frac{ y }{ 2+6 }\]

OpenStudy (anonymous):

No! It's y over 2 plus 6!

OpenStudy (anonymous):

oh, ok. well in that case, you would subtract the 6 on both sides, then multiply by 2 to get your answer

OpenStudy (anonymous):

So i would get y by itself?

OpenStudy (anonymous):

yep

OpenStudy (anonymous):

so the answer would be 2x - 12 = y ?

OpenStudy (anonymous):

hm.. I got x - 12 = y..

OpenStudy (anonymous):

How? :( If you subtract 6 on both sides, you'd get x - 6 = y/2 , then you would multiply 2 to get y by itself, which would give me 2x - 12 = y

OpenStudy (anonymous):

when you multiply by 2, you can't multiply it by both the x and the 6.

OpenStudy (anonymous):

oh okay!

OpenStudy (anonymous):

does that make sense?

OpenStudy (anonymous):

eh a little

OpenStudy (anonymous):

you just have to remember to only multiply the number by a number because you can only combine like terms

OpenStudy (anonymous):

okay! thanks!

OpenStudy (anonymous):

anytime :)

OpenStudy (anonymous):

the problem isn't set up like that.

OpenStudy (anonymous):

\[x = \frac{ y }{ 2 } + 6\]

OpenStudy (anonymous):

The problem is set up like that

OpenStudy (anonymous):

where did the 3 come from?

OpenStudy (anonymous):

it should look like this\[x= \frac{ y }{ 2 } + 6\]\[x - 6 = \frac{ y }{ 2 }\]\[x - 12 = y\]

OpenStudy (anonymous):

you're doing the wrong kind of problem. They asked what y was in terms of x.

OpenStudy (anonymous):

@SlightlyConfused just ignore this guy

OpenStudy (anonymous):

yeah i took the question in other way it would be like \[x=\frac{ y }{ 2 } +6\] \[x-6=\frac{ y }{ 2 }\] \[2(x-6)=y\] \[2x-12=y\] \[y=2x-12\] this is the correct answer

OpenStudy (anonymous):

now is it correct ??? @samanthablank

OpenStudy (anonymous):

@samanthablank I think @Ibad is correct on this one

OpenStudy (anonymous):

Simply because, 2x-12 is one of the multiple choices listed for this question

OpenStudy (anonymous):

ok.. maybe I was wrong.

OpenStudy (anonymous):

@samanthablank No it's fine!! Thanks for helping me

OpenStudy (anonymous):

Your too @Ibad

OpenStudy (anonymous):

*you

OpenStudy (anonymous):

ok, yeah I was wrong. This site ( http://www.mathway.com/default.aspx?subject=trigonometry)e says @Ibad is correct.

OpenStudy (anonymous):

no problem @samanthablank , it happens, same i took question in a wrong manner 3 times, in mathematics even a scientist can do a mistake :)

OpenStudy (anonymous):

haha yeah, true :)

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