# Simplifying Fractions

Three concepts help explain the process of simplifying fractions:

1. Multiplying a quantity by 1 has no effect
2. A fraction whose numerator is exactly the same as its denominator is equal to 1 (unless the denominator equals zero)
$\dfrac{17a^2b}{17a^2b}~~=~~1~~~~a\ne 0,~~b\ne 0$
3. A product of two fractions can be rewritten as a fraction of two products (and vice versa)
$\dfrac{a}{b} \cdot \dfrac{c}{d}~~=~~\dfrac{ac}{bd}\\*~\\*\dfrac{ac}{bd}~~=~~\dfrac{a}{b} \cdot \dfrac{c}{d}$

To simplify a fraction:

• Rewrite both numerator and denominator as products of factors (if they are not already factored)
• Examine both numerator and denominator to see if they share any factors
• If they do share factors, use concept (3) above to move the shared factors into a separate fraction
• That separate fraction should now have a numerator that is exactly the same as its denominator, which by concept (2) above means that it must equal 1, therefore by concept (1) above we can drop it from the expression

Consider the following fraction… can it be simplified?

$\dfrac{12}{15}$

Following the process described above produces the following steps:

$\dfrac{12}{15}~~=~~\dfrac{3\cdot 4}{3\cdot 5}~~=~~\dfrac{3}{3}\cdot\dfrac{4}{5}~~=~~1\cdot\dfrac{4}{5}~~=~~\dfrac{4}{5}$

This is the only process that allows you to “cancel out” something from both numerator and denominator at the same time. Common factors of the numerator and denominator “cancel each other out” because they correspond to multiplying then dividing by the same number, which does not change a quantity.  Multiplying and dividing an expression by the same number is equivalent to multiplying it by one.

BEWARE! Unless you are positive that an expression is a common factor of both numerator and denominator, take an extra step or two in your work to verify it is indeed have a common factor before cancelling it out.

That’s all there is to it.  Let’s look at some examples to see this approach in action:

### Example 1

Does the five in the numerator cancel out with the five that is a factor of the denominator?

$\dfrac{12+5}{3\cdot 5}~~=~~?$

The big question here is: is five a factor of both the numerator and the denominator? The expression above shows that five is a factor of the denominator, but five cannot be factored out of all terms in the numerator (because 5 does not divide evenly into 12). Therefore, we cannot rewrite the numerator as “five times” some sum, and since five is not a factor of every term in the numerator, the fives do NOT cancel out:

$\dfrac{12+5}{3\cdot 5}~~\ne~~\dfrac{12+1}{3}$

This is something you can also verify using a calculator. The original fraction could be evaluated:

$\dfrac{12+5}{15}~~=~~\dfrac{17}{15}~~=~~1.1333$

then compared to the result we would get if we incorrectly cancelled out the fives:

$\dfrac{12+5}{3\cdot 5}~~=>~~\dfrac{12+1}{3}~~=~~\dfrac{13}{3}~~=~~4.3333$

This produces a very different result, which demonstrates that “cancelling out” the two fives when the one in the numerator was NOT a factor of the entire numerator produces an expression that is NOT equivalent to the starting expression.

### Example 2

What if a fraction includes variables? For example:

$\dfrac{3a^2}{5a}~~=~~\dfrac{3a\cdot a}{5\cdot a}$

Is “a” a factor of every term in the numerator and denominator?

Both numerator and denominator have only one term, and “a” is indeed a factor of both. Therefore, we can proceed to cancel out the common factor of “a”:

$\dfrac{3a\cdot a}{5\cdot a}~~=~~\dfrac{3a}{5}\cdot \dfrac{a}{a}~~=~~\dfrac{3a}{5}\cdot 1~~=~~\dfrac{3a}{5}$

### Example 3

What about fractions that have more than one term in either numerator or denominator:

$\dfrac{3a^2 + 3}{5a}~~=~~\dfrac{3\cdot (a^2 + 1)}{5a}$

What are the factors of the numerator, and what are the factors of the denominator?

The factors of the numerator are 3 and $(a^2+1)$. The factors of the denominator are 5 and “a”. Therefore there are no common factors, and this fraction cannot be further simplified.

Note that the “a” in the numerator will not cancel out with the “a” in the denominator because “a” is not a factor of all terms in the numerator.

Also note that a factor can include more than one term, such as $(a^2+1)$. Therefore, a numerator or denominator that cannot be factored can still be considered “one big factor”, and might still cancel out.

### Example 4

Consider a slight variation of the last fraction:

$\dfrac{3a^2 + a}{5a}$

Can this be simplified? What are the factors of the numerator? What are the factors of the denominator?

Hopefully you found the common factor of “a”:

$\dfrac{3a^2 + a}{5a}~~=~~\dfrac{(3a + 1)\cdot a}{5\cdot a}~~=~~\dfrac{3a + 1}{5}\cdot\dfrac{a}{a}~~=~~\dfrac{3a + 1}{5}$

### Example 5

And finally, let’s turn to an “uglier” fractional expression, one with more terms in both numerator and denominator. What are the factors of the numerators and the denominators here?

$\dfrac{2x^2+6}{5y^3}\cdot \dfrac{10y^4}{3x^2+9}\\*~\\*~\\*=~~\dfrac{2\cdot (x^2+3)\cdot 2 \cdot 5\cdot y^4}{5\cdot y^3\cdot 3\cdot (x^2+3)}$

Now let’s rearrange the factors to move all the common factors to the left side of the numerator and denominator:

$=~~\dfrac{5\cdot y^3\cdot (x^2+3)\cdot 2\cdot 2\cdot y}{5\cdot y^3\cdot (x^2+3)\cdot 3}\\*~\\*~\\* =~~\dfrac{5\cdot y^3\cdot (x^2+3)}{5\cdot y^3\cdot (x^2+3)}\cdot \dfrac{2\cdot 2\cdot y}{3}$

Since the numerator of the fraction on the left is exactly the same as its denominator, it is equal to one and can be dropped from the expression to produce a simplified fraction:

$=~~ \dfrac{2\cdot 2\cdot y}{3}\\*~\\*~\\* =~~\dfrac{4y}{3}$