Simplifying
expressions is a skill that requires a significant amount of
practice. Prerequisite is a clear and thorough understanding of the
properties
of integers which
apply to the real numbers as well. These properties may seem
obvious; when applied to variables, somehow they lose their opacity.
These laws should be reviewed and reinforced as they are used (with
numerical examples as examples) to simplify algebraic expressions.
Variables and constants represent real numbers and therefore abide
by these laws.
Example:
simplify 4(x + 2)
using
the distributive property we get 4•x + 4•2 = 4x + 8
We
say that we are “clearing grouping symbols.” In this case
the grouping symbols are ( and ).
Many
times we need to perform the distributive property in reverse( we
call this operation “factoring” ) before we simply the
expression.
For
instance,
4x
+ 8 can be factored. We have two terms here: 4x and 8. We are looking
for the greatest
common divisor of
these two terms. So, 4x is divisible by 4, as it 8.
We
have 4x + 8 = 4x + 4•2 = 4(x + 2).
The
expression 4x + 8 has been factored into 4(x + 2).
The
point of all this is: Watch the patterns when simplifying
expressions, those same patterns are used to factor expressions. In
the above example the pattern is simply the number 4.
Example:
simplify
First
we need to factor both the numerator and denominator into terms, then
cancel like terms if any.
Factoring
first:
We
now have one term above and one term below because of multiplication
and the grouping symbols ( and ).
5
+ 3x = 3x + 5 commutative property of addition
So,
we have:
Canceling the common
factor (3x + 5) in both the numerator and the denominator we have the
answer:
Example:
simplify 3(5 + 2x) – 4(7x – 2)
Using
the distributive property (twice, once for each term) we have
3(5
+ 2x) = 3•5 + 3•2x = 15 + 6x
– 4(7x
– 2) = – 4•7x – (– 4)• 2 = –
28x – – 8 = –28x + 8
now
adding these results together we get:
Example:
simplify
Only
when there is a single term in the denominator can we write this
expression as:
Now,
using the distributive property of multiplication over addition we
get:
which
is
reducing
the fractions we arrive at the answer:
Example:
simplify
Notice the denominator is not a single term as in the previous example.
What
we need to do is factor the denominator into a single term, the
product of factors. If we could not factor the denominator, we would
be done, no simplification would be possible in that case.
15x
+ 45 = 15 (x + 3) and this is the term we are looking for.
Rewriting
the original expression with this term we have:
15
= 5 • 3
so
we have
and
canceling the common factor 3 in both the numerator and denominator
we get:
Clearing
the grouping symbols we get the answer:
Example:
simplify
Since
we have only terms in the numerators and denominators in the
fractions we can cancel common factors in the fractions like so:
We
now have
Now
we rearrange and combine like terms like so:
x
– 3x = -2 x so,
So
we arrive at the answer 15 - x.
Example:
simplify
The
horizontal bar (vinculum) is a grouping symbol for both the numerator
and the denominator of this fraction. Before we can simplify this
fraction, we need to get a single term in the denominator (and a
single term in the numerator will be helpful.)
We
have two terms in the denominator: 4 and – 6y. The gcd of these
two terms is 2,
so
we have 2•2 – 2• 3• y which is 2 (2 – 3y).
Likewise
with the numerator 4 is the gcd of the terms 8x and – 4,
so
we have 4•2• x – 4•1 which is 4(2x – 1).
This
fraction becomes
4
divided by 2 is 2 so the simplified answer is (clearing grouping
symbols)
Now
for a final example
Example:
simplify
As
before we need to combine terms in the numerator to form a single
term and also combine terms in the denominator to obtain a single
term there as well. We'll use the technique of cross multiplying to
arrive at this term in the numerator.
Cross
multiplying in the numerator we get:
In
the denominator notice that 9•9 = 81, so writing
and
subtracting we get in the denominator:
Now,
rewriting the original expression with these two results we get:
Inverting
the second fraction and multiplying:
Take
a close look at this last result. Remember I said earlier that the
vinculum acts as a grouping symbol for both the numerator and
denominator of a fraction? Well what we have here is a fraction with
3 factors in the numerator and also in the denominator. Here's what
we have:
The
factors (9x-1) can be canceled; 9*3 = 27 and 27 * 3 = 81
so
the factors 81, 9, and 3 reduce to 3 in the numerator.
The
answer is 3.
Example:
factoring out –1 from an expression:
consider
(3 – x)
note
that: –1 • –3 = 3 and –1 • x =
–x
substituting
in and factoring out the -1 we get
(3
– x) = ( (–1 • –3) + ( –1 •
x) )
=
(–1) • ( –3 + x) = – ( x – 3)
(3
– x) = – ( x – 3) (notice that prefixing the
expression with
"-”
changes each sign in the expression.)
and
Example:
simplify
factor
out a -1 from the numerator and we have:
and
rearranging the numerator canceling like factors:
= -1
This is another FREE ALGEBRA PRINTABLE presented to you from the
Algebra section of
K12math.com