Solve $d = \frac { 4 c b } { b + c }$ for b.
A) $b = \frac { c d } { 4 c - 1 }$
B) $b = \frac { 1 } { 4 }$
C) $b = \frac { d ( b + c ) } { 4 c }$
D) $b = \frac { c d } { 4 c - d }$

Question

Quizplus · Accepted Answer

To solve for $b$, we start with the given equation $d = \frac { 4 c b } { b + c }$. Multiplying both sides by $b + c$ gives $d(b + c) = 4cb$. Expanding the left side gives $db + dc = 4cb$. Rearranging to solve for $b$, we get $db = 4cb - dc$. Factoring out $b$ on the left side and $c$ on the right side, we have $b(d) = c(4 - d)$. Finally, solving for $b$ gives $b = \frac { c d } { 4 c - d }$.

Solve d=4cbb+cd = \frac { 4 c b } { b + c }d=b+c4cb​

Solve $d = \frac { 4 c b } { b + c }$