Fractions

A fraction consists of two parts: a numerator and a denominator.

$\frac{numerator}{denominator}$

If the numerator is smaller than the denominator, the fraction is called proper and is less than one. For example: $\frac{1}{2}$, $\frac{4}{5}$ and $\frac{3}{\mathrm{Ï€}}$ are all proper fractions and therefore less than 1.

If the numerator is larger than the denominator, the fraction is called improper and is greater than 1. For example: $\frac{3}{2},\frac{5}{4}$ and $\frac{\mathrm{Ï€}}{3}$ are all improper fractions and therefore greater than 1.

An improper fraction can be converted into a mixed fraction by dividing its denominator into its numerator. For example, since 2 divides into 7 three times with a remainder of 1, we get

To convert a mixed fraction into an improper fraction, multiply the denominator and the integer and then add the numerator. Then, write the result over the denominator. For example,

5 $\frac{2}{3}$= $\frac{3Â·5+2}{3}$= $\frac{17}{3}$

In a negative fraction, the negative symbol can be written on the top, in the middle, or on the bottom; however, when a negative symbol appears on the bottom, it is usually moved to the top or the middle:

$\frac{5}{-3}$= $\frac{-5}{3}$=- $\frac{5}{3}$.

If both terms in the denominator of a fraction are negative, the negative symbol is often factored out and moved to the top or middle of the fraction:

Strategy: To compare two fractions, cross-multiply. The larger number will be on the same side as the larger fraction.

Example:

Column A

$\frac{9}{10}$

Column B

$\frac{10}{11}$

Cross-multiplying gives 9 . 11 versus 10 . 10, which reduces to 99 versus 100. Now, 100 is greater than 99. Hence, 10/11 is greater than 9/10, and the answer is (B).

Strategy: Always reduce a fraction to its lowest terms.

Example:

Column A

$\frac{2{x}^{2}+4x+2}{{\left(x+1\right)}^{2}}$

Column B

2

Factor out the 2 in column A:

$\frac{2\left({x}^{2}+2x+1\right)}{{\left(x+1\right)}^{2}}$

$\frac{2\left(x+1\right)\left(x+1\right)}{\left(x+1\right)\left(x+1\right)}$

Finally, canceling the ( x+ 1)â€™s gives 2. Hence, the columns are equal.

Strategy: To solve a fractional equation, multiply both sides by the LCD (lowest common denominator) to clear fractions.

Example:
If $\frac{x+3}{x-3}$= y, what is the value of x in terms of y?

(A) 3 â€“ y

(B) $\frac{3}{y}$

(C) âˆš y+12

(D) $\frac{-3y-3}{1-y}$

(E) 3 y 2

First, multiply both sides of the equation by x Â–â€“ 3:
$\left(x-3\right)\frac{x+3}{x-3}$ =(x - 3)y

Cancel the (x Â–â€“ 3)'s on the left side of the equation:
x + 3 = (x-3)y

Distribute the y:
x + 3 = xy - 3y

Subtract xy and 3 from both sides:
x - xy = -3y - 3

Factor out the x on the left side of the equation:
x(1 - y) = -3y -3

Finally, divide both sides of the equation by 1 Â–â€“ y:
x = $\frac{-3y-3}{1-y}$

Complex Fractions: When dividing a fraction by a whole number (or vice versa), you must keep track of the main division bar:
$\frac{\mathrm{a}}{\mathrm{b}}{\mathrm{c}}}$ = $aÂ·\frac{c}{b}$ = $\frac{ac}{b}$. But $\frac{\mathrm{a}}{\mathrm{b}}}{c}$ = $\frac{a}{b}Â·\frac{1}{c}$ = $\frac{a}{bc}$

Example:

$\frac{1-\frac{1}{2}}{3}$= ?

(A) 6

(B) 3

(C) 1/3

(D) 1/6

(E) 1/8

Solution: $\frac{1-\frac{1}{2}}{3}$ = $\frac{\frac{2}{2}-\frac{1}{2}}{3}$ = $\frac{\frac{2-1}{2}}{3}$ = $\frac{\frac{1}{2}}{3}$ = $\frac{1}{2}Â·\frac{1}{3}$ = $\frac{1}{6}$.

Example:

If z â‰  0 and yz â‰  1, then $\frac{1}{y-\frac{1}{z}}$

(A) $\frac{yz}{zy-1}$

(B) $\frac{y-z}{z}$

(C) $\frac{yz-z}{z-1}$

(D) $\frac{z}{zy-1}$

(E) $\frac{y-z}{zy-1}$

Solution: $\frac{1}{y-\frac{1}{z}}$ = $\frac{1}{\frac{z}{z}y-\frac{1}{z}}$ = $\frac{1}{\frac{zy-1}{z}}$ = $1Â·\frac{z}{zy-1}$ = $\frac{z}{zy-1}$

Multiplying fractions is routine: merely multiply the numerators and multiply
the denominators: $\frac{a}{b}·\frac{c}{d}$ = $\frac{ac}{bd}$ .

For example,

Two fractions can be added quickly by cross-multiplying: $\frac{a}{b}Â±\frac{c}{d}$ = $\frac{adÂ±bc}{bd}$

Example:
1 2 - 3 4 = ?

(A) â€“5/4

(B) â€“2/3

(C) â€“1/4

(D) 1/2

(E) 2/3

Cross multiplying the expression $\frac{1}{2}$- $\frac{3}{4}$yields = $\frac{4-6}{8}$= $\frac{-2}{8}$ = - $\frac{1}{4}$.

Example:
Which of the following equals the average of x and 1 x ?

(A) $\frac{x+2}{x}$

(B) $\frac{{x}^{2}+1}{2x}$

(C) $\frac{x+1}{{x}^{2}}$

(D) $\frac{2{x}^{2}+1}{x}$

(E) $\frac{x+1}{x}$

The average of x and $\frac{1}{x}$ is$\frac{x+\frac{1}{x}}{2}$ = $\frac{\frac{{x}^{2}+1}{x}}{2}$ =$\frac{{x}^{2}+1}{x}Â·\frac{1}{2}$ = $\frac{{x}^{2}+1}{2x}$. Thus, the answer is (B).

To add three or more fractions with different denominators, you need to form a common denominator of all the fractions.

For example, to add the fractions in the expression $\frac{1}{3}$+ $\frac{1}{4}$ + $\frac{1}{18}$, we have to change the denominator of each fraction into the common denominator 36 (note, 36 is a common denominator because 3, 4, and 18 all divide into it evenly). This is done by multiplying the top and bottom of each fraction by an appropriate number (this does not change the value of the expression because any number divided by itself equals 1):

$\frac{1}{3}$( $\frac{12}{12}$)+ $\frac{1}{4}$($\frac{9}{9}$)+ $\frac{1}{18}$($\frac{2}{2}$)= $\frac{12}{36}$ + $\frac{9}{36}$ + $\frac{2}{36}$ = $\frac{12 +9 +2}{36}$ = $\frac{23}{36}$

You may remember from algebra that to find a common denominator of a set of fractions, you prime factor the denominators and then select each factor the greatest number of times it occurs in any of the factorizations. That is too cumbersome, however. A better way is to simply add the largest denominator to itself until all the other denominators divide into it evenly. In the above example, we just add 18 to itself to get the common denominator 36.

To find a common denominator of a set of fractions, simply add the largest denominator to itself until all the other denominators divide into it evenly.

Fractions often behave in unusual ways: Squaring a fraction makes it smaller, and taking the square root of a fraction makes it larger.
(Caution:This is true only for proper fractions, that is, fractions between 0 and 1.)

Example:
${\left(\frac{1}{3}\right)}^{2}$= $\frac{1}{9}$ and $\frac{1}{9}$ is less than $\frac{1}{3}$

Also and $\frac{1}{2}$is greater than $\frac{1}{4}$

For example, the câ€™s in the expression cannot be canceled. However, the câ€™s in the expression can be canceled as follows: $\frac{cx+c}{c}=\frac{\overline{)c}\left(x+1\right)}{\overline{)c}}=x+1$