Solve P_{50}=L+[frac{50(n)/100-0F_{b}}{f}]i

$$P _ { 50 } = L + [ \frac { \frac { 50 ( n ) } { 100 } - 0 F _ { b } } { f } ] i$$

Solve for L

$$L=P_{50}-\frac{in}{2f}$$

$$f\neq 0$$

Steps for Solving Linear Equation

Multiply both sides of the equation by f.

$$P_{50}f=fL+\left(\frac{50n}{100}-0F_{b}\right)i$$

Divide 50n by 100 to get \frac{1}{2}n.

$$P_{50}f=fL+\left(\frac{1}{2}n-0F_{b}\right)i$$

Anything times zero gives zero.

$$P_{50}f=fL+\left(\frac{1}{2}n-0\right)i$$

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

$$fL+\left(\frac{1}{2}n-0\right)i=P_{50}f$$

Subtract \left(\frac{1}{2}n-0\right)i from both sides.

$$fL=P_{50}f-\left(\frac{1}{2}n-0\right)i$$

Reorder the terms.

$$Lf=P_{50}f-\frac{1}{2}in$$

The equation is in standard form.

$$fL=P_{50}f-\frac{in}{2}$$

Divide both sides by f.

$$\frac{fL}{f}=\frac{P_{50}f-\frac{in}{2}}{f}$$

Dividing by f undoes the multiplication by f.

$$L=\frac{P_{50}f-\frac{in}{2}}{f}$$

Divide P_{50}f-\frac{in}{2} by f.

$$L=P_{50}-\frac{in}{2f}$$

Solve for F_b

$$F_{b}\in \mathrm{C}$$

$$P_{50}=L+\frac{in}{2f}\text{ and }f\neq 0$$