Solve C_{0}=h+(21+3)/2

$$C _ { 0 } = \frac { h + ( 21 + 3 ) } { 2 }$$

Solve for C_0

$$C_{0}=\frac{h+24}{2}$$

Solution Steps

Add 21 and 3 to get 24.

$$C_{0}=\frac{h+24}{2}$$

Divide each term of h+24 by 2 to get \frac{1}{2}h+12.

$$C_{0}=\frac{1}{2}h+12$$

Solve for h

$$h=2\left(C_{0}-12\right)$$

Steps for Solving Linear Equation

Add 21 and 3 to get 24.

$$C_{0}=\frac{h+24}{2}$$

Divide each term of h+24 by 2 to get \frac{1}{2}h+12.

$$C_{0}=\frac{1}{2}h+12$$

Swap sides so that all variable terms are on the left hand side.

$$\frac{1}{2}h+12=C_{0}$$

Subtract 12 from both sides.

$$\frac{1}{2}h=C_{0}-12$$

Multiply both sides by 2.

$$\frac{\frac{1}{2}h}{\frac{1}{2}}=\frac{C_{0}-12}{\frac{1}{2}}$$

Dividing by \frac{1}{2} undoes the multiplication by \frac{1}{2}.

$$h=\frac{C_{0}-12}{\frac{1}{2}}$$

Divide C_{0}-12 by \frac{1}{2} by multiplying C_{0}-12 by the reciprocal of \frac{1}{2}.

$$h=2C_{0}-24$$