College Algebra, Chapter 1, 1.5, Section 1.5, Problem 78

The equation $\displaystyle \frac{1}{F} = \frac{1}{x} + \frac{1}{y}$ represents the lens equation where $F$ is the focal length of a convex lens with an object that is placed at a distance $x$ from the lens. Then its image will be at a distance $y$ from the lens. If the lens has a focal length of $4.8 cm$ and that the image of an object is $4cm$ closer to the lens than the object itself, determine how far from the lens is the object.

$\displaystyle \frac{1}{F} = \frac{1}{x} + \frac{1}{y} \qquad$ Model

If the image of an object is $4cm$ closer to the lens, then $y = x - 4$


$\begin{equation} \begin{aligned} \frac{1}{4.8} =& \frac{1}{x} + \frac{1}{x - 4} && \text{Substitute the given} \\ \\ \frac{x(x - 4)}{4.8} =& x - 4 + x && \text{Multiply both sides by the LCD } x(x - 4) \\ \\ x^2 - 4x =& \frac{48}{5} x - \frac{96}{5} && \text{Multiply $4.8$ or } \frac{24}{5} \text{ then simplify} \\ \\ 5x^2 - 20x =& 48x - 96 && \text{Multiply by } 5 \\ \\ 5x^2 - 68x + 96 =& 0 && \text{Subtract $48x$ and add} \\ \\ (5x - 8)(x - 12) =& 0 && \text{Factor} \\ \\ 5x - 8 =& 0 \text{ and } x- 12 = 0 && \text{Zero Product Property} \\ \\ x =& \frac{8}{5} cm \text{ and } x = 12 cm && \text{Solve for } x \end{aligned} \end{equation}$