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Adding, Subtracting, and Common Denominator Finding with Rational Expressions

Precalculus Algebra Arithmetic with Rational Expressions Adding, Subtracting, and Common Denominator Finding with Rational Expressions

Generalizing How Fractions of Numbers work to Rational Expressions

Suppose we want to add two fractions, such as

$$\frac{2}{5}+\frac{3}{4}$$

Recall that, in order to add or subtract fractions, one first must find a common denominator for the first and second fractions before one can simply add or subtract across the numerators.

To find a common denominator, we can multiply the numerator and denominator of the first fraction by the denominator of the second, and multiply the numerator and denominator of the second fraction by the denominator of the first. Viz:

$$\begin{align}\frac{2}{5}+\frac{3}{4}=\frac{4\cdot 2}{4\cdot 5}+\frac{3\cdot 5}{4\cdot 5}=\frac{8}{20}+\frac{15}{20}\end{align}$$

This method always yields a common denominator. Now since we have a common denominator, we can now add (or subtract) straight across the numerators (because the numerators tell us how many “1/20’s” we have, and adding is basically counting how many “1/20’s” we have total). So, in summary, we end up with:

$$\frac{2}{5}+\frac{3}{4}=\frac{8}{20}+\frac{15}{20}=\frac{23}{20}$$

which cannot be reduced (because 23 shares no divisors with 20 aside from 1).

Adding and subtracting rational expressions works exactly the same way as adding and subtracting ordinary fractions of numbers! Also, in most cases, it is easier to find a common divisor with rational expressions than with ordinary fractions of numbers.

Adding and Subtracting Two Rational Expressions

Suppose we want to add two rational expressions $r$ and $s$

1.) Factor the numerators and denominators of both fractions.

2.) First, find a common denominator by multiplying the numerator and denominator of $s$ by the denominator of $r$, and vice versa,
OR
2.) Find what factors in the denominator of $r$ that $s$ does not have, and multiply those to the numerator and denominator of $s$. Also find the factors in the denominator of $s$ that $r$ does not have, and multiply the numerator and denominator of $r$ by those.

3.) Add (or subtract) straight across the numerators, distributing or factoring to simplify if needed, and put the common denominator in the denominator of your resulting fraction.

It is worth noting that the second method for step 2 above tends to be a LOT more convenient and leads to much simpler final answers.

Find and simplify the result of

$$\frac{2x+4}{x^2+2x+1}-\frac{3x+5}{x^2+3x+2}$$

Start by factoring the numerator and denominator of both fractions.

$$\frac{2x+4}{x^2+2x+1}-\frac{3x+5}{x^2+3x+2}=\frac{2(x+2)}{(x+1)^2}-\frac{3x+5}{(x+1)(x+2)}$$

Now that we have everything factored, it is easy to see what factors the denominators of each fraction have, and which they already have in common. Notice that the denominator of the second fraction has only one factor of $(x+1)$ whereas the first fraction has two (as indicated by the power $2$). So we will multiply a factor of $(x+1)$ to the top and bottom of the second fraction so that both the first and second fractions have the same number of that factor. Viz:

$$\begin{align}\frac{2(x+2)}{(x+1)^2}-\frac{3x+5}{(x+1)(x+2)}&=\frac{2(x+2)}{(x+1)^2}-\frac{(3x+5)(x+1)}{(x+1)(x+2)(x+1)}\\&= \frac{2(x+2)}{(x+1)^2}-\frac{(3x+5)(x+1)}{(x+1)^2(x+2)} \end{align}$$

Similarly, we notice that the denominator of the second fraction has a factor of $(x+2)$ whereas the first fractions’s denominator does not. So, we multiply the top and bottom of the first fraction by this factor so that the denominator matches the second fraction’s denominator. Viz:

$$\begin{align}\frac{2(x+2)}{(x+1)^2}-\frac{(3x+5)(x+1)}{(x+1)^2(x+2)}&=\frac{2(x+2)(x+2)}{(x+1)^2(x+2)}-\frac{(3x+5)(x+1)}{(x+1)^2(x+2)}\\&=\frac{2(x+2)^2}{(x+1)^2(x+2)}-\frac{(3x+5)(x+1)}{(x+1)^2(x+2)} \end{align}$$

Now that we have a common denominator, we can add (or subtract, rather) across the numerators, carrying the common denominator along for the ride. Also, we can distribute our multiplied quantities in the numerator to see if it simplifies at all, or to see if we can factor it. Viz:

$$\begin{align}\frac{2(x+2)^2}{(x+1)^2(x+2)}-\frac{(3x+5)(x+1)}{(x+1)^2(x+2)} &=\frac{2(x+2)^2-(3x+5)(x+1)}{(x+1)^2(x+2)}\\&=\frac{2x^2+8x+8-3x^2+2x+5}{(x+1)^2(x+2)}\\&=\frac{-x^2+10x+13}{(x+1)^2(x+2)}\end{align}$$

Note that the numerator is not (easily) factorable. Sometimes it is in some problems, and better still, we might be able to cancel something in the denominator once we factor the numerator.

Note also that it is almost never practical or useful to distribute/expand the factored denominator. So, unless directly asked to do so in a problem, AVOID EXPANDING THE FINAL ANSWER’S DENOMINATOR.

All this to say: we’re done with the problem!

Generalizing this Method to More Than One Fraction

One can certainly apply the above method to sums and differences of more than one fraction. All you need to do is work with one pair at a time using the above method. Once you find the sum/difference of one pair, you will have a single new fraction from your work that can be added/subtracted with another fraction.

Alternatively, if may be more fruitful to try to adapt the method above for finding a common denominator to help you find a common denominator of ALL fractions in your problem before adding and subtracting all the numerators at once.

$\frac{23}{10x}$

$\frac{3}{5y}$

$\frac{3-4x}{3x^2}$

$\frac{-2y}{x^2-y^2}$

$\frac{2(r^2+1)}{r^2-1}$

$\frac{-1\cdot (x+3)}{(x-1)(x+1)^2}$

$\frac{-x(x+8)}{(x-5)(2x+1)(3x-2)}$

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