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Mathematics 16 Online
OpenStudy (anonymous):

cx over d+f=b

OpenStudy (anonymous):

\[\frac{cx}{d+f}=b\] right?

OpenStudy (anonymous):

\[\frac{ cx }{ d+f }=b\]

OpenStudy (anonymous):

Yes

OpenStudy (anonymous):

and what is the question?

OpenStudy (anonymous):

How to solve it

OpenStudy (anonymous):

solve for \(x\) ?

OpenStudy (anonymous):

No solve for b

OpenStudy (anonymous):

\(b\) is already solved for as it is by itself on one side of the equal sign

OpenStudy (anonymous):

b is already solved for haha

OpenStudy (anonymous):

So how do I answer

OpenStudy (anonymous):

I don't understand what you're asking? Are you solving for x because b is already solved for...?

OpenStudy (anonymous):

Sorry yes solve for x

OpenStudy (anonymous):

\[x=b(d+f)/c\]

OpenStudy (anonymous):

Don't let letters scare you.

OpenStudy (anonymous):

Ok so that's the answer but how do come up with answer?

OpenStudy (anonymous):

It's the same algebra as it would be with numbers.

OpenStudy (anonymous):

Please show how to work it

OpenStudy (anonymous):

\[\frac{ cx }{ d+f}=b\] \[cx = b(d+f)\] because you multiply each side to cancel d+f on the left. \[x=\frac{ b(d+f) }{ c }\] divide by c because you want to get rid of it from the left side.

OpenStudy (anonymous):

Thanks now x^2+y^2=16 Solve

OpenStudy (anonymous):

Hepl

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