Ask your own question, for FREE!
Mathematics 24 Online
OpenStudy (hicklyfe123):

Solve for B. A = 3(B + C) B = (A - C)/3 B = (3A - C)/3 B = (A - 3C)/3 ill give a medal!!

OpenStudy (benlindquist):

@serenity74

OpenStudy (aaronandyson):

Divide the whole equation by (B+C)

OpenStudy (aaronandyson):

So we get : \[\frac{ A }{ B+C } = \frac{ 3(B+C) }{ B+C }\]

OpenStudy (aaronandyson):

The RHS is 3 , right?

OpenStudy (hicklyfe123):

yeah

OpenStudy (hicklyfe123):

whats RHS..?

OpenStudy (aaronandyson):

So we get: \[\frac{ A }{ B+C } = 3\]

OpenStudy (aaronandyson):

We'll start again,okay?

OpenStudy (hicklyfe123):

ok

OpenStudy (aaronandyson):

This time we divide the whole equation by 3. \[\frac{ A }{ 3 } = \frac{ 3(B+C) }{ 3}\]

OpenStudy (aaronandyson):

That gives \[\frac{ A }{ 3 } = B+C\]

OpenStudy (aaronandyson):

Move C to the other side.

OpenStudy (hicklyfe123):

okay

OpenStudy (aaronandyson):

\[\frac{ A }{ 3 } - C = B\]

OpenStudy (aaronandyson):

Solve the above equation.

OpenStudy (aaronandyson):

\[\frac{ A - 3C }{ 3 } = B\]

OpenStudy (hicklyfe123):

thankyouu

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!