c+d over 3 = 2c Solve for c
I interpreted the problem as: \[\dfrac{ c+d }{ 3 }=2c\] ?
The c+d is suppose to up above the 3
Yes. That's right.
Which would be the same general idea... isolate the c. So multiply both sides by 3, then move the c terms together on one side, the non-c term (which will be the d) to the other side. Then divide by the coefficient of c.
if thats so, just multiply both sides by 3 :)
Okay. Sorry. I'm trying to figure this out. Lol.
Is c=5c-d right?
\[\frac{ c + d }{ 3 } = 2c\] multiply both side by 3 \[(3)\frac{ c+d }{ 3 } = 2c(3)\] 3 will be cancelled out so \[c+d=6c\] transpose c to the other side d = 6c - c d = 5c then divide both sides by 5 :) d/5 = c
I'll do a DIFFERENT one for you as an example: \(\dfrac{ 2m+n }{ 3 }=-7m\) solve for m \(\cdot 3\dfrac{ 2m+n }{ 3 }=-7m\cdot 3\) multiply by 3 to clear the den'r \(2m+n=-21m\) \(n=-23m\) subtract 2m from both sides \(-\dfrac{n}{23}=m\) divide by coefficient of m So \(m=-\dfrac{n}{23}\)
But why did you combine the m's if that's what you're solving for?
Join our real-time social learning platform and learn together with your friends!