Simplifying Fractions 2
Simplifying Fractions
Examples with solutions
Example 1:
Simplify 
.
solution:
Since the numerator is the sum of two terms, we must begin by
factoring it as much as possible:
(both terms in the original expression share a
factor of 3)
(both terms in the square brackets of the preceding
line share a factor of x )
(both terms in the square brackets of the preceding
line share a factor of y )
Thus
as the final answer.
The initial factoring step is absolutely essential. If instead
you decided to begin by cancelling between the denominator and
the first term of the numerator, along the lines of
you would end up with a completely wrong result. This kind of
thing is not wrong so much because we say so, or because you
didn’t use “our” method, the method “we”
told you to use. Rather, it is wrong because the result does not
have the same mathematical meaning, nor will it give the same
mathematical results as the original expression, and so it’s
about the same sort of thing as deciding the change the meaning
of words when you speak (but pretending you haven’t!). If
you substitute specific pair of values for x and y into this
proposed simplified expression, you will get different value from
it than you would get when you substitute the same values of x
and y into the original expression. Simplification is intended to
give a mathematical expression which is easier to write down and
work with than the original expression, but which gives exactly
the same results as the original expression in all circumstances.
If our apparent simplified expression does not do this, then we
have made a mistake.
Example 2:
Simplify 
.
solution:
It is so tempting just to cancel the ‘5x 2’
terms and be done:
or even
By now you’re screaming back, “No! No! Both of these
results are wrong! Wrong! Wrong!” In both cases, the that is
being cancelled is not a factor in the denominator (it is a
“term” in the denominator) and hence cannot be validly
cancelled. In the second version above, another error is made as
well. When factors are cancelled, they always leave a result of
“1” behind. If it is a multiplication by 1, we need not
write down the 1, as in
But, if it is an “added” 1, you cannot drop it:
(1)(20x) = 20x but (1) + (20x) 
20x
This latter is as incorrect as writing
This is supposed to look like a fraction with nothing written
in the numerator or in the denominator (that is, the ultimate in
cancellation – perhaps we should consider this to be
annihilation!). Of course, you can easily see this is nonsense.
Anyway, back to the original example. Before any cancelling can
be contemplated, both the numerator and denominator must be
factored completely. The numerator is already in factored form.
However, for the denominator we have
5x 2 + 20x = 5[x 2 + 4x] = 5x(x + 4)
Thus
as the final correct answer.
Example 3:
Simplify 
.
solution:
The denominator is obviously fully factored. For the
numerator, only slight factoring is possible:
3t 2 + 5ts - 7s 2 t = t (3t + 5s - 7s
2 + 4st )
Therefore
as the final simplified form.
Example 4:
Simplify 
.
solution:
First, factoring a 2 + 4a = a (a + 4)
and ax + 4x = x (a + 4)
Thus
as the final simplified result.
