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We now turn our attention to rational expressions - that is, algebraic fractions - and equations which contain them. The reader
is encouraged to keep in mind the properties of fractions listed on page ?? because we will need them along the way.
Before we launch into reviewing the basic arithmetic operations of rational expressions, we take a moment to
review how to simplify them properly. As with numeric fractions, we ‘cancel common factors,’ not common terms.
That is, in order to simplify rational expressions, we first factor the numerator and denominator. For example:
This equivalence holds provided the factors being canceled aren’t \(0\). Since a factor of \(x\) and a factor of \(x+5\) were
canceled, \(x \neq 0\) and \(x+5 \neq 0\), so \(x \neq -5\). We usually stipulate this as:
Just like their numeric counterparts, you don’t add algebraic fractions by adding denominators of fractions with common
numerators - it’s the other way around: (One of the most common errors students make on college Mathematics
placement tests is that they forget how to add algebraic fractions correctly. This places many students into remedial classes
even though they are probably ready for college-level Math. We urge you to really study this section with great care so that
you don’t fall into that trap.)
As with numeric fractions, we divide rational expressions by ‘inverting and multiplying’. Before we get too
carried away however, we factor to see what, if any, factors cancel.
The ‘\(x \neq 3\)’ is mentioned since a factor of \((x-3)\) was canceled as we reduced the expression. We also canceled a factor of
\((x^2+2)\). Why is there no stipulation as a result of canceling this factor? Because \(x^2 + 2 \neq 0\) for all real \(x\). (See Section AppCmpNums for details.)
At this point, we could go ahead and multiply out the numerator and denominator to get
but for most of the
applications where this kind of algebra is needed (solving equations, for instance), it is best to leave things
factored. Your instructor will let you know whether to leave your answer in factored form or not. (Speaking
of factoring, do you remember why \(x^2-2\) can’t be factored over the integers?)
As with numeric fractions we need common denominators in order to subtract. This is already the case here so
we proceed by subtracting the numerators.
At this point, we need to see if we can reduce this expression so
we proceed to factor. It first appears as if we have no common factors among the numerator and denominator
until we recall the property of ‘factoring negatives’ from Page ??: \(3-w = -(w-3)\). This yields:
The stipulation \(w \neq 3\) comes from the
cancellation of the \((w-3)\) factor.
In this next example, we are asked to add two rational expressions with different denominators. As with numeric
fractions, we must first find a common denominator. To do so, we start by factoring each of the denominators.
To find the common denominator, we examine the factors in the first denominator and note that we need a
factor of \((y-4)^2\). We now look at the second denominator to see what other factors we need. We need a factor of \(y\)
and \((4+y) = (y+4)\). What about \((4-y)\)? As mentioned in the last example, we can factor this as: \((4-y) = -(y-4)\). Using properties of negatives,
we ‘migrate’ this negative out to the front of the fraction, turning the addition into subtraction. We find the (least)
common denominator to be \((y-4)^2 y (y+4)\). We can now proceed to multiply the numerator and denominator of each fraction
by whatever factors are missing from their respective denominators to produce equivalent expressions with
common denominators.
At this stage, we can subtract numerators and simplify. We’ll keep the denominator
factored (in case we can reduce down later), but in the numerator, since there are no common factors, we
proceed to perform the indicated multiplication and combine like terms.
We would like to factor the numerator
and cancel factors it has in common with the denominator. After a few attempts, it appears as if the numerator
doesn’t factor, at least over the integers. As a check, we compute the discriminant of \(2y^2 + 15y + 4\) and get \(15^2 - 4(2)(4) = 193\). This isn’t a
perfect square so we know that the quadratic equation \(2y^2 + 15y + 4=0\) has irrational solutions. This means \(2y^2 + 15y + 4\) can’t factor over
the integers (See the remarks following Theorem discriminanttheoremrealversion.) so we are done.
In this example, we have a compound fraction, and we proceed to simplify it as we did its numeric counterparts
in Example fractionreview. Specifically, we start by multiplying the numerator and denominator of the ‘big’ fraction by the
least common denominator of the ‘little’ fractions inside of it - in this case we need to use \((4-(x+h))(4-x)\) - to remove the
compound nature of the ‘big’ fraction. Once we have a more normal looking fraction, we can proceed as we
have in the previous examples.
Your instructor will let you know if you are to expand the denominator or not. (We’ll keep it factored
because in Calculus it needs to be factored.)
At first glance, it doesn’t seem as if there is anything that can be done with \(2t^{-3} - (3t)^{-2}\) because the exponents on the
variables are different. However, since the exponents are negative, these are actually rational expressions. In
the first term, the \(-3\) exponent applies to the \(t\)only but in the second term, the exponent \(-2\) applies to both the \(3\) and
the \(t\), as indicated by the parentheses. One way to proceed is as follows:
We see that we are being asked to subtract two rational expressions with different denominators, so we need to
find a common denominator. The first fraction contributes a \(t^3\) to the denominator, while the second contributes
a factor of \(9\). Thus our common denominator is \(9t^3\), so we are missing a factor of ‘\(9\)’ in the first denominator and a
factor of ‘\(t\)’ in the second.
We find no common factors among the numerator and denominator so we are done.
A second way to approach this problem is by factoring. We can extend the concept of the ‘Polynomial G.C.F.’
to these types of expressions and we can follow the same guidelines as set forth on page ?? to factor out
the G.C.F. of these two terms. The key ideas to remember are that we take out each factor with the smallest
exponent and that factoring is the same as dividing. We first note that \(2t^{-3} - (3t)^{-2}= 2t^{-3} - 3^{-2} t^{-2}\) and we see that the smallest power on \(t\)
is \(-3\). Thus we want to factor out \(t^{-3}\) from both terms. It’s clear that this will leave \(2\) in the first term, but what about the
second term? Since factoring is the same as dividing, we would be dividing the second term by \(t^{-3}\) which thanks
to the properties of exponents is the same as multiplying by \(\frac {1}{t^{-3}} = t^3\). The same holds for \(3^{-2}\). Even though there are no
factors of \(3\) in the first term, we can factor out \(3^{-2}\) by multiplying it by \(\frac {1}{3^{-2}} = 3^2 = 9\). We put these ideas together below.
While both ways are valid, one may be more of a natural fit than the other depending on the circumstances and
temperament of the student.
As with the previous example, we show two different yet equivalent ways to approach simplifying \(10x(x-3)^{-1} + 5x^2(-1)(x-3)^{-2}\). First up is
what we’ll call the ‘common denominator approach’ where we rewrite the negative exponents as fractions and
proceed from there.
As expected, we got the same reduced fraction as before. □
Next, we review the solving of equations which involve rational expressions. As with equations involving numeric fractions, our
first step in solving equations with algebraic fractions is to clear denominators. In doing so, we run the risk of introducing what
are known as extraneous solutions - ‘answers’ which don’t satisfy the original equation. As we illustrate the
techniques used to solve these basic equations, see if you can find the step which creates the problem for
us.
Our first step is to clear the fractions by multiplying both sides of the equation by \(x\). In doing so, we are implicitly
assuming \(x \neq 0\); otherwise, we would have no guarantee that the resulting equation is equivalent to our original
equation. (See page ??.)
We obtain two answers, \(x = \frac {1 \pm \sqrt {5}}{2}\). Neither of these are \(0\) thus neither contradicts our assumption that \(x \neq 0\). The reader is
invited to check both of these solutions. (The check relies on being able to ‘rationalize’ the denominator -
a skill we haven’t reviewed yet. (Come back after you’ve read Section rationalizingdenomandnumer if you want to!) Additionally, the positive
solution to this equation is the famous Golden Ratio.)
To solve the equation, we clear denominators. Here, we need to assume \(t-1 \neq 0\), or \(t \neq 1\).
We assumed that \(t \neq 1\) in order
to clear denominators. Sure enough, the candidate \(t = 1\) doesn’t check in the original equation since it causes
division by \(0\). In this case, we call \(t = 1\) an extraneous solution. Note that \(t=1\)does work in every equation after we clear
denominators. In general, multiplying by variable expressions can produce these ‘extra’ solutions, which is why
checking our answers is always encouraged. (Contrast this with what happened in Example solveeqnbyfactoring when we
divided by a variable and ‘lost’ a solution.) The other two candidates, \(t = 0\) and \(t = -\frac {1}{2}\), are solutions.
As before, we begin by clearing denominators. Here, we assume \(1 - w\sqrt {2} \neq 0\) (so \(w \neq \frac {1}{\sqrt {2}}\)) and \(2w+5 \neq 0\) (so \(w \neq -\frac {5}{2}\)).
The result is a linear equation in \(w\) so we gather the terms with \(w\) on one side of the equation and put everything
else on the other. We factor out \(w\) and divide by its coefficient.
This solution is different than our excluded values,
\(\frac {1}{\sqrt {2}}\) and \(-\frac {5}{2}\), so we keep \(w = -\frac {14}{6 + \sqrt {2}}\) as our final answer. The reader is invited to check this in the original equation.
To solve our next equation, we have two approaches to choose from: we could rewrite the quantities with
negative exponents as fractions and clear denominators, or we can factor. We showcase each technique below.
Clearing Denominators Approach: We rewrite the negative exponents as fractions and clear
denominators. In this case, we multiply both sides of the equation by \((x^2+4)^2\), which is never \(0\). (Think about that
for a moment.) As a result, we need not exclude any \(x\) values from our solution set.
We leave it to the reader to show that both \(x = -2\) and \(x = 2\) satisfy the original equation.
Factoring Approach: Since the equation is already set equal to \(0\), we’re ready to factor. Following the
guidelines presented in Example rationalexpressionreviewex, we factor out \(3(x^2+4)^{-2}\) from both terms and look to see if more factoring can be
done:
The first equation yields no solutions (Think about this for a moment.) while the second gives us \(x = \pm \sqrt {4} = \pm 2\) as
before.
We are asked to solve this equation for \(y\) so we begin by clearing fractions with the stipulation that \(y-3 \neq 0\) or
\(y \neq 3\). We are left with a linear equation in the variable \(y\). To solve this, we gather the terms containing \(y\) on
one side of the equation and everything else on the other. Next, we factor out the \(y\) and divide by its
coefficient, which in this case turns out to be \(x-2\). In order to divide by \(x-2\), we stipulate \(x - 2 \neq 0\) or, said differently, \(x \neq 2\).
We highly encourage the reader to check the answer algebraically to see where the restrictions on \(x\) and \(y\) come into play.
(It involves simplifying a compound fraction!)
Our last example comes from physics and the world of photography. (See this article on focal length.) We take a
moment here to note that while superscripts in Mathematics indicate exponents (powers), subscripts are used primarily
to distinguish one or more variables. In this case, \(S_1\) and \(S_2\) are two different variables (much like \(x\) and \(y\)) and we
treat them as such. Our first step is to clear denominators by multiplying both sides by \(f S_1 S_2\) - provided each
is nonzero. We end up with an equation which is linear in \(S_1\) so we proceed as in the previous example.
PreCalculus