# Algebra II: Fractional Expressions

On this page we hope to clear up any problems you might have with fractional expressions and equations.  Fractions are used continually in math, and even with the advent of powerful calculators, are used extensively in advanced math courses.  Click any of the links below or scroll down to start understanding fractions better!

Multiplication.
Division.
Lowest Common Multiple.
Complex Fractions.
Division of Polynomials.
Fractional Equations.
Variation (direct and indirect).

## Multiplication of Fractions

The multiplication of fractions is rather straightforward.  The Fraction Multiplication Theorem says for any fractional expressions, (a/b) and (c/d) where b and d do not equal 0, (a/b)(c/d) = (a * c)/(b * d).  Example:
```1. Problem:  x + 3    x3
---- * -----
y - 4  y + 5
Solution: Using the Fraction Multiplication
Theorem, multiply the numerator
of the first fraction by the numer-
ator of the second fraction and the
denominator of the first fraction
by the denominator of the second
fraction.

(x + 3)x3
--------------
(y - 4)(y + 5)
```

## Divison of Fractions

The division of fractions is also not very complicated.  The Fraction Division Theorem says for any fractional expressions, (a/b) and (c/d) where b, c, and d do not equal 0, (a/b)/(c/d) = (a/b)(d/c).  In other words, you divide by multiplying by a reciprocal.  Example:
```1. Problem:  x - 2   x + 5
----- / -----
x + 1   x - 3
Solution: Utilizing the Fraction Division
Theorem, we take the reciprocal of the
divisor and multiply.

x - 2   x - 3
----- * -----
x + 1   x + 5

Use the Fraction Multiplication
Theorem to multiply the problem out.

x2 - 5x + 6
-----------
x2 + 6x + 5
```

## Lowest Common Multiple

Finding the LCM, or Lowest Common Multiple is a necessary skill if you want to be able to add to and subtract fractions from other numbers.  However, finding the LCM is usually covered in most elementary algebra (Algebra I) courses.  This custom has been followed on this site, so you can click here to get a better understanding of Lowest Common Multiples.

## Addition and Subtraction of Fractions

Adding and subtracting fractions from other numbers is a very useful skill since you encounter fractions so often.  This skill is usually covered in elementary algebra (Algebra I) courses.  We have followed that custom on this site, so click here to get a better understanding of addition and subtraction when fractions are involved.

## Complex Fractions

Complex fractions, or fractions that have a fraction for its numerator, denominator, or both, are encountered less often than normal fractions, but are just as easy to solve because they, like normal fractions, are division problems.  The most common way to simplify complex fractions is to simplify the numerator and denominator, treat it as a division problem, and then simplify as usual.  Example:
```1. Simplify:     1
1 + -
x
-----
1
1 - -
x2
Solution: Combine the numerator into one
fraction and then do the same to the
denominator.  (If you do not know how

x   1   x + 1
- + -   -----
x   x     x
----- = ------
x2 - 1   x2 - 1
- - -    ------
x2  x2      x2

Now you have a normal division problem
with fractional expressions.

x + 1   x2 - 1
----- / ------
x       x2

Using the Fractional Division
Theorem, take the reciprocal of the
divisor and multiply.

x + 1     x2
----- * ------
x     x2 - 1

After multiplication, you have the
following expression:

(x + 1)x2
---------------
x(x + 1)(x - 1)

Now, cancel out any factors that are
in both the denominator and the
numerator to simplify.
(They both have (x + 1) and
x as a factor.  Cancel them
out.)

You are left with the following:

x
-----
x - 1
```

## Divison of Polynomials

Always keep in mind that a fraction bar is the same as a division sign. Division by a monomial can be done by rewriting the problem as a fraction.  Example:
```1. Divide:   12x3 + 8x2 + x + 4 by 4x
Solution: Rewrite the problem as a fraction.
12x3 + 8x2 + x + 4
-----------------
4x

This expression shows that each
term in the numerator is divided
by 4x.  Rewrite the prob-
lem to show that.

12x3    8x2   x    4
---- + --- + -- + --
4x     4x   4x   4x

Now do the four divisions in-
dicated by each fraction.

3x2 + 2x + (1/4) + (1/x)
```
When the divisor is not a monomial, you have to use a procedure that resembles long division as you learned way back in 5th grade arithmetic!  Example:
```2. Divide:   x2 + 5x + 6 by x + 3
Solution: Write the problem as a long
division problem.
___________
x + 3 )x2 + 5x + 6

Divide first term by first term —
(x2/x) = x.

x__________
x + 3 )x2 + 5x + 6
x2 + 3x      Multiply x by divisor.
-------
2x      Subtract.

Bring down the next term and repeat
the process.

x_+_2______
x + 3 )x2 + 5x + 6
x2 + 3x
-----------
2x + 6
2x + 6
------
0

The quotient is x + 2.
```

## Solving Fractional Expressions

A fractional equation is an equation that contains one or more fractional expressions.  To solve a fractional equation you multiply each side of the equation by the LCM of all the denominators.  Example:
```1. Solve:    2   5   1
- - - = -
3   6   x
Solution: The LCM of all denominators is
6x.  Multiply each side of the
equation by 6x.

6x((2/3) - (5/6)) = 6x(1/x)

Use the distributive law of multi-
plication, which says a(b + c) =
ab + ac, to rewrite the
equation.

6x(2/3) - 6x(5/6) = 6x(1/x)

Multiply each group of terms
together.

(12x/3) - 30x/6 = 6x/x

Perform each of the indicated
divisions.

4x - 5x = 6

Solve for x.
Combine like terms.

-x = 6

Multiply each side by -1.

x = -6
```

## Variation

A worker earns \$10.00/hr.  In one hour, \$10.00 is earned.  In two hours, \$20.00 is earned, and so forth.  This gives us a set of ordered pairs which all have the same ratio — (1, 10), (2, 20), (3, 30), . . .  This situation, where there are pairs of numbers in which the ratio is constant, we have direct variation.  Whenever a situation, such as the one described above gives rise to the function f(x) = kx, where k is a positive constant, there is direct variation.  Example:
```1. Problem:  Find the variation constant k
and an equation of variation where y
varies directly as x, and where
y = 34 and x = 7.

Solution: The problem tells us (7, 34)
is a solution of the direct variation
equation, y = kx.
Plug in the givens for y and x
and solve for l.

34 = k(7)
(34/7) = k

The constant of variation is (34/7).
```
The above problem could be used in real life if someone made \$4.857/hr. and they wanted to know how much they would make if they worked for 7 hours. Using 7 for x would tell us they would make \$34.00.

Sometimes, things do not vary directly.  An example of this is a bus that travels 20 kilometers in one hour at a speed of 20 km/hr.  At 40 km/hr. it would only take half of an hour to go 20 kilometers.  This gives us a pair of ordered pairs — (20, 1), (40, .5), . . .  These numbers, whose product is constant, vary indirectly.  Whenever a situation such as the one described above gives rise to the function f(x) = (k/x), where k is a positive constant, there is inverse variation.  Example:

```2. Problem:  Find the variation constant and
then an equation of variation where
y varies inversely as x, and
y = 32 when x = (1/5).
Solution: We know that (.2, 32) is
a solution of the inverse variation
equation.
Plug in the given information and
solve for k.

32 = (k/.2)
6.4 = k

The equation of variation is the
following:

y = (6.4/x)
```
This can be applied to real life, too.  If you were driving 100 km/hr., you would go 6.4 kilometers in 3.84 minutes.

Take the quiz on fractional expressions.  The quiz is very useful for either review or to see if you've really got the topic down.

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