When fractions are multiplied, they are multiplied by multiplying the numerators by each other and the denominators by each otehr. No cross-multiplication in involved! Always remember that variables stand for numbers! Because of that fact, nothing changes when dealing with variables in a multiplication problem with frations that have rational expressions in them.
Example:
_ _
1. Expand: x^2| x^2 3y^3 |
---| --- - ---- |
y^2|_ y m _|
Solution: Two multiplications are supposed to be done.
x^2 x^2
You have to first multiply --- by --- and then
y^2 y
x^2 -3y^3
you have to multiply --- by -----.
y^2 m
That gives you the following:
(x^2)(x^2) (x^2)(3y^3)
---------- - -----------.
(y^2)y (y^2)m
Lastly, you simplify both expressions since answers
are not considered correct in Algebra unless they are
simplified. After simplification (multiply terms
together and cancel things out if possible), you get
the answer --
x^4 3yx^2
--- - -----.
y^3 m
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Specifically, we will help you understand how to find LCMs of algebraic expressions.
Example:
1. Find the LCM of 15(a^2)b and 10ab^3.
Solution: Write the expressions as products of prime and
literal factors.
15(a^2)b 10ab^3
3 * 5 * a * a * b 2 * 5 * a * b * b * b
From the listing of prime and literal factors, take
the groups of factors with the most instances of that
factor. For example, there are three bs in
10ab^3's factors, therefore, three bs are
listed in the LCM.
The LCM is the following:
2 * 3 * 5 * a * a * b * b * b = 30(a^2)(b^3)
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Example:
1. Add: a b
- + -------
x (x + y)
Solution: First, find the LCM of the denominators, which will
become the new denominator.
-------- + --------
x(x + y) x(x + y)
So that the problem does not change, the numerator of
each term has to be multipled by the same quantity
that its respective denominator was. The original
denominator of the first term was x, and it has
been multiplied by (x + y), so the original
numerator, a, must be multiplied by (x + y),
too.
a(x + y)
-------- + --------
x(x + y) x(x + y)
The original denominator of the second term was
(x + y), but it was multiplied by x, so the
original numerator must also be multiplied by x.
Now, the fractions can be added together.
a(x + y) xb a(x + y) + xb
-------- + -------- = -------------
x(x + y) x(x + y) x(x + y)
This is one of the rare times in Algebra that there
are multiple forms of the correct answer. In this
case, you can multiply out the numerator and/or
denominator if you want, and since doing that does
not help you simplify the answer any further, they
are also correct forms of th answer.
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a
-
b
---
c
-
d
where b, c, and d do not equal 0. This is the same as (a/b)/(c/d). To solve these fractions, you will need to multiply the numerator, or the first term of the problem by the reciprocal of the denominator, or the second term of the problem.
Example:
1. Simplify: a
-
b
---
c (b, c do not equal 0)
Solution: Multiply the numerator (a/b) by the reciprocal
of the denominator, which is (1/c).
a 1 a
- * - = --
b c bc
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Take the quiz on fractions. The quiz is very useful for either review or to see if you've really got the topic down.