A3: Fractions and reciprocals

You’ll be able to:

• combine fractions over a common denominator,
• solve an equation with reciprocals, like the thin-lens equation, for any of its variables,
• say which values an equation excludes, and what that means physically,
• spot a common misconception: the reciprocal of a sum is not the sum of the reciprocals, $\frac{1}{a+b} \neq \frac{1}{a} + \frac{1}{b}$.

Plan on about 45 minutes, or less if the initial check goes well.

Check yourself first

Answer these two questions without notes. If you’re not sure, it’s ok to answer “I don’t know yet” rather than guessing.

1. Solve for $x$, assuming $a \neq b$:

\[\frac{1}{x} = \frac{1}{a} - \frac{1}{b}\]

2. The thin-lens equation relates object distance $s$, image distance $s^\prime$, and focal length $f$:

\[\frac{1}{s} + \frac{1}{s'} = \frac{1}{f}\]

Solve for $s^\prime$, assuming $s \neq f$.

Answers to the entry check

1. $x = \dfrac{ab}{b - a}$
2. $s^\prime = \dfrac{sf}{s - f}$

Both right? Skip ahead and take the A3 mastery check in Canvas. Either one wrong, or unsure? Keep reading. The idea solves question 1 step by step, and worked example 1 solves question 2.

Where this shows up in Physics 5

• 5B, optics. The thin lens and mirror equation, $\frac{1}{s} + \frac{1}{s^\prime} = \frac{1}{f}$, and lens power in diopters, $P = \frac{1}{f}$ with $f$ in meters.
• 5B, sound. The Doppler shift for a moving source, $f = f_0\,\frac{v}{v - v_s}$, when used to solve for the source’s speed.
• 5C, circuits. Resistors in parallel, $\frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2}$, and capacitors in series, $\frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2}$.
• 5A and 5B, springs. Springs in series, $\frac{1}{k} = \frac{1}{k_1} + \frac{1}{k_2}$.

All of these share the same pattern, a sum of reciprocals, so the same steps can be used to solve all of them.

The idea

Equations with reciprocals all hinge on one rule:

Flipping term by term is not allowed, because the reciprocal of a sum is not the sum of the reciprocals. Check this with numbers: $\frac{1}{1} + \frac{1}{1} = 2$, but $\frac{1}{1 + 1} = \frac{1}{2}$.

So solving any of these equations takes three steps:

1. Isolate the reciprocal of the unknown on one side.
2. Combine the other side into a single fraction over a common denominator.
3. Flip both sides.

Here are the steps on the first equation from the entry check:

\[\frac{1}{x} = \frac{1}{a} - \frac{1}{b} = \frac{b}{ab} - \frac{a}{ab} = \frac{b - a}{ab}\] \[x = \frac{ab}{b - a}\]

Step 1 was already done, since $\frac{1}{x}$ was already isolated. Step 2 found the common denominator $ab$. Step 3 flipped both sides.

Excluded values. A denominator can never be zero. The answer above requires $a \neq b$. If $a = b$, the right side is zero, and $\frac{1}{x} = 0$ has no solution. In physics, an excluded value usually means something. For a lens, $s = f$ is excluded because an object at the focal point produces parallel rays and no finite image.

Check every answer three ways.

• Units. If $a$ and $b$ are lengths, as $s$ and $f$ are in the thin-lens version, then $\frac{ab}{b - a}$ has units of length squared over length, which is a length.
• Size. Combinations like resistors in parallel must come out smaller than either part.
• Limits. Think about what happens as one variable becomes very large or very small and see whether the result makes sense physically.

Worked examples

Worked example 1: solve the thin-lens equation

Problem. Solve $\frac{1}{s} + \frac{1}{s^\prime} = \frac{1}{f}$ for $s^\prime$. Then find $s^\prime$ for an object at $s = 30\ \mathrm{cm}$ in front of a converging lens with $f = 10\ \mathrm{cm}$.

Isolate $\frac{1}{s^\prime}$ by subtracting $\frac{1}{s}$ from both sides:

\[\frac{1}{s'} = \frac{1}{f} - \frac{1}{s}\]

Combine the right side over the common denominator $sf$. Pay attention to the order. $\frac{1}{f}$ becomes $\frac{s}{sf}$ and $\frac{1}{s}$ becomes $\frac{f}{sf}$.

\[\frac{1}{s'} = \frac{s - f}{sf}\]

Flip both sides, which is allowed now that each side is a single fraction:

\[s' = \frac{sf}{s - f}\]

Substitute last: $s^\prime = \frac{(30\ \mathrm{cm})(10\ \mathrm{cm})}{30\ \mathrm{cm} - 10\ \mathrm{cm}} = 15\ \mathrm{cm}$.

Check.

• Substitution: $\frac{1}{30} + \frac{1}{15} = \frac{1}{30} + \frac{2}{30} = \frac{1}{10}$.
• Units: $\mathrm{cm}^2 / \mathrm{cm} = \mathrm{cm}$.
• Distant object: as $s \to \infty$, $s - f \approx s$, so $s^\prime \to f$. Light from far away focuses at the focal point.
• Object at the focal point: at $s = f$, the denominator is zero, so there is no finite image.
• Object inside the focal point: for a converging lens ($f \gt 0$) with $s \lt f$, $s^\prime$ is negative, which means a virtual image.

Worked example 2: resistors in parallel

Problem. Resistors $R_1 = 6.0\ \Omega$ and $R_2 = 3.0\ \Omega$ are connected in parallel, so $\frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2}$. Find $R_{\text{eq}}$.

The first step is already done, because the reciprocal of the unknown is already isolated. Combine the right side over $R_1 R_2$:

\[\frac{1}{R_{\text{eq}}} = \frac{R_2}{R_1 R_2} + \frac{R_1}{R_1 R_2}\]

Your turn. Add the numerators, flip both sides, then substitute the numbers. Before you check your answer, explain why writing $R_{\text{eq}} = R_1 + R_2$ would be wrong.

Solution to worked example 2

\[\frac{1}{R_{\text{eq}}} = \frac{R_1 + R_2}{R_1 R_2}\] \[R_{\text{eq}} = \frac{R_1 R_2}{R_1 + R_2} = \frac{(6.0\ \Omega)(3.0\ \Omega)}{9.0\ \Omega} = 2.0\ \Omega\]

Writing $R_{\text{eq}} = R_1 + R_2$ flips each term separately, which is never allowed.

Check: $2.0\ \Omega$ is smaller than both resistors, which is always the case for resistors combined in parallel. Adding a second path makes it easier for current to flow.

Worked example 3: capacitors in series

Problem. Two capacitors in series satisfy $\frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2}$. Find $C_{\text{eq}}$ in terms of $C_1$ and $C_2$. Then find what your answer reduces to in the limit where $C_2$ is much larger than $C_1$.

Try it yourself before opening the solution.

Solution to worked example 3

The algebra matches worked example 2:

\[C_{\text{eq}} = \frac{C_1 C_2}{C_1 + C_2}\]

When $C_2 \gg C_1$, the sum $C_1 + C_2 \approx C_2$, so $C_{\text{eq}} \approx \frac{C_1 C_2}{C_2} = C_1$. The small capacitor controls the combination.

Check: the units are $\mathrm{F}^2 / \mathrm{F} = \mathrm{F}$, and the result is smaller than $C_1$ for any positive values.

Common issues

1. Flipping term by term. $\frac{1}{a + b} \neq \frac{1}{a} + \frac{1}{b}$. Check this by plugging in $a = b = 1$. The left side is $\frac{1}{2}$ and the right side is $2$.
2. Flipping before combining. Flip only a side that is a single fraction.
3. Forgetting the last flip. If your answer looks like $\frac{s - f}{sf}$, you have found $\frac{1}{s^\prime}$, not $s^\prime$.
4. Reversing a subtraction. $\frac{1}{f} - \frac{1}{s} = \frac{s - f}{sf}$, not $\frac{f - s}{sf}$. Each numerator picks up the other term’s denominator.
5. Cancelling across a plus sign. $\frac{a + b}{a} = 1 + \frac{b}{a}$, not $b$. You can cancel factors, never terms.
6. Treating division by zero as zero. A zero denominator means “no finite answer,” not “the answer is zero.”

Practice and the mastery check

Open the Math Toolkit module in your course’s Canvas site. There are four quizzes for A3, and you can retake any of them as often as you like:

• Entry check. The two questions above with feedback.
• Practice set. Thirteen questions ranging from plain symbols to physics, with feedback on every answer.
• Drill set. Quick repetitions of the algebra alone. Each attempt draws new questions. Use it until the three steps feel automatic.
• Mastery check. Five questions. A passing score is 6 of 7 points. The interpretation and application questions are worth 2 points each, so you need both. If you miss twice in a row, stop and work through the lesson section linked in the feedback, or bring the question to office hours.

Go deeper

These OpenStax sections cover the same algebra with more practice:

• College Algebra 2e, 1.6 Rational Expressions: adding, dividing, and simplifying fractions.
• College Algebra 2e, 2.2 Linear Equations in One Variable: the “Solving a Rational Equation” subsection covers excluded values.