How to Divide Whole Numbers by Fractions - A Simple Guide
In this guide to dividing whole numbers by fractions, you'll find kid-friendly explanations, solved examples, and practice questions to test your skills!
Algebra with fractions might seem tricky at first, but once you know the right steps, it’s a piece—or fraction—of a cake!
In this guide, we’ll walk you through the basics with a step-by-step guide, clear examples, and practical tips, so you can approach algebraic fractions with confidence and ease.
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An algebraic fraction is a fraction where either the numerator, denominator, or both contain algebraic expressions. For example:
\(\Large\frac{3x}{4}\), \(\Large\frac{x+2}{y-1}\) and \(\Large\frac{5}{x^2+3x+2}\)
In these examples, the variables x and y make the fractions algebraic. For some students, they also make the fractions scary, because after all, they do represent the unknown.
But, worry not!
Just like with the “regular” fractions, working with algebraic ones follows the same basic rules, with a small twist; The variables add some extra steps when simplifying, adding, multiplying, or dividing them.
Before we get into the details of doing algebra with fractions, let’s refresh some concepts:
Algebraic fractions are undefined if the denominator equals zero.
For instance, in \(\Large\frac{1}{x-3}\), the fraction is undefined if x = 3 because dividing by zero is not allowed in math.
Algebraic fractions can also include more complex expressions like \(\Large\frac{x+5}{x^2-4}\).
These may need factoring to simplify or solve.
Just like reducing a numeric fraction (e.g., \(\Large\frac{6}{8}\) = \(\Large\frac{3}{4}\)), we can simplify algebraic fractions by dividing both the numerator and denominator by their greatest common factor (GCF).
The key to solving algebraic fractions is to follow the same rules as regular fractions while carefully handling variables.
Let’s start with addition and subtraction, which often require finding a common denominator before proceeding.
To add or subtract algebraic fractions, we need them to have a shared denominator.
If the denominators are already the same, great—you can combine the fractions directly!
If not, you’ll need to find their equivalents with their least common denominator (LCD) first.
The steps are:
Identify the denominators of the fractions.
Find the LCD (the smallest expression that both denominators can divide into).
Rewrite each fraction with the LCD as the new denominator.
Combine the numerators by adding or subtracting them.
Simplify the resulting fraction if possible.
Let’s put this into practice with some solved examples!
Example 1: Fractions with the same denominator
Add \(\Large\frac{3x}{4}\) and \(\Large\frac{5x}{4}\)
The denominators are the same, so we can add the numerators and then simplify the fraction:
\(\Large\frac{3x}{4}\) + \(\Large\frac{5x}{4}\) = \(\Large\frac{3x+5x}{4}\) = \(\Large\frac{8x}{4}\) = 2x
Answer: 4x
Example 2: Fractions with different denominators
Add \(\Large\frac{x}{3}\) and \(\Large\frac{2}{5}\)
First, we need to find the LCD of 3 and 5, which is 15. Then we can rewrite each fraction:
\(\Large\frac{x}{3}\) = \(\Large\frac{5×x}{5×3}\) = \(\Large\frac{5x}{15}\)
and
\(\Large\frac{2}{5}\) = \(\Large\frac{2×3}{5×3}\) = \(\Large\frac{6}{15}\)
Now that the denominators are the same, we can add the fractions:
\(\Large\frac{5x}{15}\) + \(\Large\frac{6}{15}\) = \(\Large\frac{5x+16}{15}\)
The result cannot be simplified further, so the answer is:
\(\Large\frac{5x+16}{15}\).
Example 1: Fractions with the same denominator
Subtract \(\Large\frac{7x}{9}\) from \(\Large\frac{10x}{9}\)
Since the denominators are the same, we can just subtract the numerators and then simplify the fraction:
\(\Large\frac{10x}{9}\) - \(\Large\frac{7x}{9}\) = \(\Large\frac{10x-7x}{9}\) = \(\Large\frac{3x}{9}\)
As 9 is divisible by 3, we can simplify \(\Large\frac{3x}{9}\)to \(\Large\frac{1x}{3}\).
\(\Large\frac{3x}{9}\) = \(\Large\frac{x}{3}\)
Answer: \(\Large\frac{x}{3}\).
Example 2: Fractions with different denominators
Subtract \(\Large\frac{2}{x}\) from \(\Large\frac{3}{x+1}\)
The denominators are different, so find the LCD: x(x + 1). Now we can rewrite each fraction:
\(\Large\frac{3}{x+1}\) = \(\Large\frac{3x}{x(x+1)}\)
and
\(\Large\frac{2}{x}\) = \(\Large\frac{2(x+1)}{x(x+1)}\)
The fractions now have the same denominator, so we subtract them:
\(\Large\frac{3x}{x(x+1)}\) - \(\Large\frac{2(x+1)}{x(x+1)}\) = \(\Large\frac{3x-2(x+1)}{x(x+1)}\)
To make things easier, we simplify the numerator:
3x - 2(x + 1) = 3x - 2x - 2 = x - 2
Now, we rewrite the fraction and get the answer: \(\Large\frac{x-2}{x(x+1)}\)
Many of our students find multiplication and division easier than addition and subtraction because these don't require finding a common denominator.
We simply multiply or divide the numerators and denominators directly, but we have to be careful to simplify and watch for restrictions on the variables.
Example 1: Multiplying simple fractions
Multiply \(\Large\frac{3x}{4}\) and \(\Large\frac{5}{2x}\).
First, we multiply the numerators: 3x × 5 = 15x.
Then, we multiply the denominators: 4 × 2x = 8x
Now, we can combine the fraction and simplify it:
\(\Large\frac{15x}{8x}\) = \(\Large\frac{15}{8}\)
Answer: \(\Large\frac{15}{8}\).
Since our numerator is larger than the denominator (improper fraction), we can convert this result into a mixed number: 1\(\Large\frac{7}{8}\)
Example 2: Multiplying with polynomials
Multiply \(\Large\frac{x+2}{x-3}\) and \(\Large\frac{3x}{x+2}\)
Like in the previous example, we first multiply the numerators: (x + 2) × 3x = 3x(x + 2).
Then we multiply the denominators: (x - 3) × (x + 2) = (x - 3)(x + 2).
You can see that we have a common factor (x + 2) so we can cancel it out. What remains is:
\(\Large\frac{3x(x+2)}{(x-3)(x+2)}\) = \(\Large\frac{3x}{(x-3)}\)
We cannot simplify further, so the final answer is \(\Large\frac{3x}{(x-3)}\) (provided x \(\neq\) (-2) or 3).
Example 1: Dividing simple fractions
Divide \(\Large\frac{6x}{5}\) by \(\Large\frac{2x}{3}\)
When dividing fractions, we should ‘flip’ (take the reciprocal of) the second fraction, then multiply as usual. So here we can rewrite as multiplication by the reciprocal:
\(\Large\frac{6x}{5}\) ÷ \(\Large\frac{2x}{3}\) = \(\Large\frac{6x}{5}\) × \(\Large\frac{3}{2x}\)
Now, we take the same steps as with multiplying algebraic fractions.
First, we multiply the numerators: 6x × 3 = 18x. Then we multiply the denominators: 5 × 2x = 10x
Combine the fraction and then simplify:
\(\Large\frac{18x}{10x}\) = \(\Large\frac{18}{10}\) = \(\Large\frac{9}{5}\)
The final answer is \(\Large\frac{9}{5}\).
Since this is an improper fraction, we can write it as a mixed number 1\(\Large\frac{4}{5}\).
Example 2: Dividing with polynomials
Divide \(\Large\frac{x^2-9}{x+1}\) by \(\Large\frac{x-3}{x+1}\)
First, rewrite as multiplication by the reciprocal:
\(\Large\frac{x^2-9}{x+1}\) ÷ \(\Large\frac{x-3}{x+1}\) = \(\Large\frac{x^2-9}{x+1}\) × \(\Large\frac{x+1}{x-3}\)
Simplify first: x2 - 9 as (x - 3)(x + 3):
\(\Large\frac{(x-3)(x+3)}{x+1}\) × \(\Large\frac{x+1}{x-3}\)
We can cancel out the common factors (x + 1) and (x - 3):
\(\Large\frac{(x-3)(x+3)}{x+1}\) × \(\Large\frac{x+1}{x-3}\) = x + 3
Final answer: x + 3 (provided x \(\neq\) (-1) or 3)
Working with algebraic fractions once you master the main steps.
At Mathnasium of Allen, we help students of all skill levels learn and master algebraic fractions, so over the years, we noticed where they tend to stumble.
Let’s explore these mistakes and learn how to avoid them so you can solve algebraic fractions with confidence and accuracy.
When adding or subtracting algebraic fractions, some students skip finding the least common denominator (LCD) and try to combine terms directly. This leads to incorrect results.
Example Mistake: \(\Large\frac{x}{3}\) + \(\Large\frac{2}{5}\) \(\neq\) \(\Large\frac{x+2}{8}\)
When adding or subtracting algebraic fractions, first identify and rewrite both fractions with the LCD before combining numerators. In this case, the correct approach would be:
\(\Large\frac{x}{3}\) + \(\Large\frac{2}{5}\) = \(\Large\frac{5x}{15}\) + \(\Large\frac{6}{15}\) = \(\Large\frac{5x+6}{15}\)
Canceling Terms Incorrectly
Canceling terms is a powerful tool for simplifying fractions, but it must be done properly. Students often cancel terms across addition or subtraction, which is a big no-no.
Example Mistake: \(\Large\frac{x+2}{x+4}\) \(\neq\) 1 + \(\Large\frac{2}{4}\)
Tip: Cancel only factors (entire expressions multiplied together), not individual terms. For example:
\(\Large\frac{x(x+2)}{x(x+4)}\) = \(\Large\frac{x+2}{x+4}\) and cannot be simplified further.
Ignoring Restrictions on Variables
Fractions are undefined if their denominator equals zero, but students sometimes forget to check for values that make the denominator zero.
Example Mistake: Simplifying \(\Large\frac{x^2-9}{x+3}\) to x + 3 without stating x \(\neq\) 3.
To avoid this mistake, always check for restrictions by setting the original denominator equal to zero and solving for x. In this case:
x - 3 = 0 → x \(\neq\) 3
Skipping Factoring
When simplifying or multiplying algebraic fractions, skipping the step of factoring can prevent students from recognizing opportunities to cancel terms.
Example Mistake: \(\Large\frac{x^2-9}{x^2+3x}\) \(\neq\) \(\Large\frac{x-9}{x+3}\)
Factor all expressions fully before simplifying. In this case:
\(\Large\frac{x^2-9}{x^2+3x}\) = \(\Large\frac{(x-3)(x+3)}{x(x+3)}\) = \(\Large\frac{x-3}{x}\) (provided x \(\neq\) 0 or (-3))
\(\Large\frac{x}{3}\) + \(\Large\frac{2x}{5}\)
\(\Large\frac{4}{x}\) + \(\Large\frac{5}{2x}\)
\(\Large\frac{2x}{7}\) - \(\Large\frac{x}{14}\)
\(\Large\frac{x}{x+1}\) - \(\Large\frac{1}{x+1}\)
\(\Large\frac{3x}{4}\) × \(\Large\frac{2}{x}\)
\(\Large\frac{x+1}{x}\) × \(\Large\frac{x}{2}\)
\(\Large\frac{x^2}{5}\) ÷ \(\Large\frac{x}{10}\)
\(\Large\frac{2}{3x}\) ÷ \(\Large\frac{4}{x}\)
Questions about algebraic fractions? We have your answers!
You can only cancel terms when they are factors—entire expressions multiplied together—not when they’re part of an addition or subtraction operation.
Example Mistake: \(\Large\frac{x+1}{x+3}\) \(\neq\) 1 \(\Large\frac{1}{3}\)
In the fraction \(\Large\frac{x+1}{x+3}\), x + 1 and x + 3 are terms that are being added, not multiplied, so they cannot be canceled. This algebraic fraction can’t be simplified further.
However, if the expression were \(\Large\frac{(x+1)(x-1)}{(x+3)(x-1)}\), you could cancel (x - 1) because it’s a factor:
\(\Large\frac{(x+1)(x-1)}{(x+3)(x-1)}\) = \(\Large\frac{x+1}{x+3}\) (provided x \(\neq\) (-3) or 1).
Division by zero is undefined because it breaks the fundamental rules of mathematics. To understand this, think about dividing something into parts:
If you have 10 ÷ 2, you can divide 10 into two equal parts: 5 and 5.
However, if you try 10 ÷ 0, how many groups of zero can you make? It’s impossible to define because you can’t create groups of “nothing.”
In algebraic fractions, always check the denominator to make sure it’s not zero for any value of the variable. For example, in \(\Large\frac{3}{x-5}\), the fraction is undefined if x = 5.
When algebraic fractions have multiple variables, the same rules apply, but you must carefully handle each variable.
When simplifying, it is important to treat each variable independently when factoring and canceling terms. For example:
\(\Large\frac{xy}{x^2y^2}\) = \(\Large\frac{1}{xy}\) (provided x \(\neq\) 0, y \(\neq\) 0)
You should also watch for values of variables that make any denominator zero. For example, in \(\Large\frac{1}{x(y+2)}\), x \(\neq\) 0 and y \(\neq\) (-2).
Handling multiple variables might feel tricky at first, but with practice, it becomes manageable.
Algebraic fractions pop up in real life more often than you might expect! Here are a few examples:
Science and Engineering: Calculations involving rates, such as speed or flow rate, often use algebraic fractions (e.g., distance/time)
Finance: Fractions are used to determine interest rates, tax calculations, and investment growth. For example, interest earned/principle
Cooking: Recipes often use fractions to adjust ingredient quantities based on servings. Algebraic fractions can model how ingredients scale. For example, scaling \(\Large\frac{1}{2}\)x cups of sugar for x = 3 servings gives 1.5 cups.
Mathnasium of Allen’s specially trained math tutors work with students of all skill levels to help them understand and excel in any math class and topic, including algebraic fractions.
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\(\Large\frac{11x}{15}\)
\(\Large\frac{13}{2x}\)
\(\Large\frac{3x}{14}\)
\(\Large\frac{x-1}{x+1}\)
\(\Large\frac{3}{2}\)
\(\Large\frac{x+1}{2}\)
2x
\(\Large\frac{1}{6}\)