Solve for d c=(2d+1)/3

$c = \frac{2 d + 1}{3}$

Step 1

Rewrite the equation as $\frac{2 d + 1}{3} = c$ .

$\frac{2 d + 1}{3} = c$

Step 2

Multiply both sides by $3$ .

$\frac{2 d + 1}{3} \cdot 3 = c \cdot 3$

Step 3

Simplify.

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{2 d + 1}{3} \cdot 3 = c \cdot 3$

Rewrite the expression.

$2 d + 1 = c \cdot 3$

$2 d + 1 = c \cdot 3$

$2 d + 1 = c \cdot 3$

Simplify the right side.

Move $3$ to the left of $c$ .

$2 d + 1 = 3 c$

$2 d + 1 = 3 c$

$2 d + 1 = 3 c$

Step 4

Solve for $d$ .

Subtract $1$ from both sides of the equation.

$2 d = 3 c - 1$

Divide each term in $2 d = 3 c - 1$ by $2$ and simplify.

Divide each term in $2 d = 3 c - 1$ by $2$ .

$\frac{2 d}{2} = \frac{3 c}{2} + \frac{- 1}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 d}{2} = \frac{3 c}{2} + \frac{- 1}{2}$

Divide $d$ by $1$ .

$d = \frac{3 c}{2} + \frac{- 1}{2}$

$d = \frac{3 c}{2} + \frac{- 1}{2}$

$d = \frac{3 c}{2} + \frac{- 1}{2}$

Simplify the right side.

Move the negative in front of the fraction.

$d = \frac{3 c}{2} - \frac{1}{2}$

$d = \frac{3 c}{2} - \frac{1}{2}$

$d = \frac{3 c}{2} - \frac{1}{2}$

$d = \frac{3 c}{2} - \frac{1}{2}$

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