iBeYoshiii
  • iBeYoshiii
c+d over 3 = 2c Solve for c
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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DebbieG
  • DebbieG
I interpreted the problem as: \[\dfrac{ c+d }{ 3 }=2c\] ?
iBeYoshiii
  • iBeYoshiii
The c+d is suppose to up above the 3
iBeYoshiii
  • iBeYoshiii
Yes. That's right.

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DebbieG
  • DebbieG
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.
petiteme
  • petiteme
if thats so, just multiply both sides by 3 :)
iBeYoshiii
  • iBeYoshiii
Okay. Sorry. I'm trying to figure this out. Lol.
iBeYoshiii
  • iBeYoshiii
Is c=5c-d right?
petiteme
  • petiteme
\[\frac{ c + d }{ 3 } = 2c\] 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
DebbieG
  • DebbieG
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 \(m=-\dfrac{n}{23}\)
iBeYoshiii
  • iBeYoshiii
But why did you combine the m's if that's what you're solving for?

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