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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:
– 28x + 6x +15 + 8 = – 22x +
23
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.
Using and recognizing –1:
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
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