Michael Bowen's VC Course Pages

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Math V46 Section A.4

Introduction

Please try these problems on your own before looking at my solutions. There may be other ways to solve these; I make no claim that the methods shown are the quickest, prettiest, or most efficient.

Problem 27

As the fractions have different denominators and we are adding/subtracting them, we need to find a least common denominator (LCD). The most reliable way to do this is to factor all the denominators, and then play the "who has the most?" (of each prime factor) game. This problem has a slight twist in the that factorization of one of the denominators contains a $-1$. What is the easiest way to deal with this peculiarity?

After factoring each denominator, the original problem becomes $$\begin{equation} \label{eq:orig_prob_27} {\frac{{c + 2}}{{5\left( {c - 1} \right)}} - \frac{{c - 2}}{{3\left( {c - 1} \right)}} + \frac{c}{{ - 1\left( {c - 1} \right)}}} \end{equation}$$

Having the $-1$ in the last denominator of Equation $\eqref{eq:orig_prob_27}$ is a bit of an inconvenience. Fortunately, there is a property of fractions which states that a factor of $-1$ may be freely moved from the denominator to the numerator (or vice-versa). A simple example containing numbers only (so it is easier to understand) might be $$\begin{equation} \label{eq:neg_fraction_ex} {\frac{3}{{ - 5}} = \frac{3}{{ - 1\left( 5 \right)}} = \frac{{ - 1\left( 3 \right)}}{5} = \frac{{ - 3}}{5} = - \frac{3}{5}} \end{equation}$$ Any property which works for numbers also works for variables (letters), so we apply this to the $-1$ in the denominator of the last fraction to move it up into the numerator (so it doesn't mess up our LCD). We obtain \begin{align} \frac{{c + 2}}{{5\left( {c - 1} \right)}} - \frac{{c - 2}}{{3\left( {c - 1} \right)}} + \frac{{ - 1\left( c \right)}}{{\left( {c - 1} \right)}} & \Longrightarrow \frac{{c + 2}}{{5\left( {c - 1} \right)}} - \frac{{c - 2}}{{3\left( {c - 1} \right)}} + \frac{{ - c}}{{\left( {c - 1} \right)}} \nonumber \\ \label{eq:move_negative_up} {} & \Longrightarrow \frac{{c + 2}}{{5\cdot\left( {c - 1} \right)}} - \frac{{c - 2}}{{3\cdot\left( {c - 1} \right)}} - \frac{c}{{\left( {c - 1} \right)}} \end{align} Where did the negative sign in the numerator of the last fraction go in the last step of Equation $\eqref{eq:move_negative_up}$? Take a close look at how the sign in front of the entire fraction changed from plus to minus. Also note that the last step exhibits visible multiplication dots to more clearly delineate the various prime factors in each denominator.

From the "who has the most?" game, we quickly determine the LCD: $$\begin{equation} \label{eq:lcd_prob_27} {\text{LCD} = 5 \cdot 3 \cdot \left( {c - 1} \right)} \end{equation}$$

To obtain matching denominators, we now multiply both the numerator and denominator of each fraction by whatever is "missing" in its denominator when compared to the LCD from Equation $\eqref{eq:lcd_prob_27}$. For example, the prime factors in the first denominator are missing a $3$ when compared to the LCD, so we multipy both the top and bottom of this fraction by $3$. We make corresponding adjustments to the other two fractions: $$\begin{equation} \label{eq:make_lcds_prob_27} {\frac{3}{3} \cdot \frac{{c + 2}}{{5\left( {c - 1} \right)}} - \frac{5}{5} \cdot \frac{{c - 2}}{{3\left( {c - 1} \right)}} - \frac{{3 \cdot 5}}{{3 \cdot 5}} \cdot \frac{c}{{\left( {c - 1} \right)}} \Longrightarrow \frac{{3 \cdot \left( {c + 2} \right)}}{{3 \cdot 5 \cdot \left( {c - 1} \right)}} - \frac{{5 \cdot \left( {c - 2} \right)}}{{5 \cdot 3 \cdot \left( {c - 1} \right)}} - \frac{{3 \cdot 5 \cdot c}}{{3 \cdot 5 \cdot \left( {c - 1} \right)}}} \end{equation}$$

We are not quite ready to combine like terms in the numerators, because it is mathematically illegal to combine like terms in an expression containing parentheses or brackets. So to get ready, we first distribute all the parentheses and perform all remaining multiplications in the numerator (because Aunt Sally, or order of operations, wants us to multiply before we add and subtract). Note, however, that we do not distribute the denominators (because it will make cancelling easier later on if we leave them alone). From the multiplications in the numerators, we obtain $$\begin{equation} \label{eq:distrib_numerators} {\frac{{3c + 6}}{{3 \cdot 5\left( {c - 1} \right)}} - \frac{{5c - 10}}{{5 \cdot 3\left( {c - 1} \right)}} - \frac{{15c}}{{3 \cdot 5 \cdot \left( {c - 1} \right)}}} \end{equation}$$ and combining the numerators into a single fraction gives $$\begin{equation} \label{eq:combine_numerators} {\frac{{\left( {3c + 6} \right) - \left( {5c - 10} \right) - \left( {15c} \right)}}{{3 \cdot 5 \cdot \left( {c - 1} \right)}}} \end{equation}$$ Did you notice the return of the pesky parentheses? This is because every fraction in the universe has two pairs of "invisible" parentheses; one pair enclosing the entire numerator, and the other pair enclosing the entire denominator. For example, the fraction $\frac{3}{5}$ is really $\frac{{\left( 3 \right)}}{{\left( 5 \right)}}$ if we make all the implicit parentheses visible. When multiple numerators (that is, two or more) are combined during addition or subtraction, it is required that these parentheses become visible again...even if (as in this case) we just finished expanding several distributives in the immediately preceding steps. If a set of parentheses has a minus sign in front, then we have to treat that minus sign as a $-1$ distributing across the parentheses. Carrying this out leads to the final solution, which is boxed below. Note in particular how the sign in front of the term $10$ changed as a result of distributing the $-1$ in front of its parentheses. $$\begin{equation} \label{eq:final_prob_27} {\frac{{3c + 6 - 5c + 10 - 15c}}{{3 \cdot 5 \cdot \left( {c - 1} \right)}} \Longrightarrow \boxed{\frac{{ - 17c + 16}}{{15\left( {c - 1} \right)}}}} \end{equation}$$

Problem 45

This is a subtraction problem involving a non-fraction as one term and a complex fraction as the other. You should never perform computations with fractions unless every piece of the expression has been transformed into a fraction first. For example, if you were subtracting $7 - \frac{8}{3}$, you would always want to convert the $7$ into a fraction so that you were really subtracting $\frac{7}{1} - \frac{8}{3}$; this would be true even if you were multiplying (rather than subtracting) the $7$ and the $\frac{8}{3}$. As you can see, the general method is to insert a $1$ under the non-fractional part(s) of the expression. After that (if adding or subtracting only), you would use the method of LCDs to force the denominators to match, and then add and simplify the numerators to complete the problem.

There is an additional complication in that the fraction on the right is a complex fraction (meaning it has smaller fractions inside the larger fraction). The approach for these is essentially the same; for each piece of the complex fraction which is not itself a smaller fraction (in this case, the 2 in the "big" numerator and the 1 in the "big" denominator), force it to become a fraction by inserting a 1 beneath it. So, putting these ideas together, the first step of the simplification might look like $$\begin{equation} \label{eq:insert_ones} {\frac{x}{1} - \frac{{\left( {\frac{2}{1}} \right)}}{{\left( {\frac{1}{1} - \frac{1}{x}} \right)}}} \end{equation}$$ if we make visible some of the implicit parentheses associated with the complex fraction to help us determine the order of coming steps.

Thinking now of Aunt Sally (and her order of operations, because she's pretty good about telling us what to do next if we're smart enough to remember her), we realize that the most important operation in Equation $\eqref{eq:insert_ones}$ is the subtraction in the denominator of the complex fraction on the right (because it's parenthesized). However, we need to find the LCD; in this case it should be clear that available primes in the "small" denominators are just $1$ and $x$, so $$\begin{equation} \label{lcd_prob_45} {\text{LCD} = x} \end{equation}$$

The $\frac{1}{1}$ needs to be fixed, LCD-wise, but the $\frac{1}{x}$ already has the "right" denominator, so we will only adjust the $\frac{1}{1}$ by multiplying it by $\frac{x}{x}$. After that, we'll be able to consolidate the "big" denominator into a single fraction. $$\begin{equation} \label{eq:consolidate_big_denom} {\frac{x}{1} - \frac{{\left( {\frac{2}{1}} \right)}}{{\left( {\frac{x}{x} \cdot \frac{1}{1} - \frac{1}{x}} \right)}} \Longrightarrow \frac{x}{1} - \frac{{\left( {\frac{2}{1}} \right)}}{{\left( {\frac{x}{x} - \frac{1}{x}} \right)}} \Longrightarrow \frac{x}{1} - \frac{{\left( {\frac{2}{1}} \right)}}{{\left( {\frac{{\left( x \right) - \left( 1 \right)}}{x}} \right)}} \Longrightarrow \frac{x}{1} - \frac{{\left( {\frac{2}{1}} \right)}}{{\left( {\frac{{x - 1}}{x}} \right)}}} \end{equation}$$ Note that the "pesky parentheses" returned in the third expression in Equation $\eqref{eq:consolidate_big_denom}$, where we combined the numerators, but they had no effect this time (no distributives) so we dropped them.

Thinking again of Aunt Sally, we realize that we've done everything we can inside the parentheses of the complex fraction; namely, both the numerator and denominator have been consolidated into single fractions, and both these fractions are fully reduced. So we now move beyond the parentheses (which only keep their order-of-operations priority for as long as there is a "doable" operation inside them) and look at what needs to be done next. Looking outside the parentheses only, can you spot the two division operations (the fraction lines) and the one subtraction operation remaining to be performed?

Normally we would carry out the division $\displaystyle \frac{x}{1}$ first, using the left-to-right tie-breaker rule. However, doing this would restore the expression to its initial value of just $x$, nullifying the advantage of having the $1$ underneath the $x$ for future LCD purposes. So we move to the other division, which is the "big" fraction line in the complex fraction. For complex fractions having a single "small" fraction in both numerator and denominator, we simply use the "$\div$" in place of the fraction line, giving $$\begin{equation} \label{eq:divide_complex_frac} {\frac{x}{1} - \left( {\frac{2}{1}} \right) \div \left( {\frac{{x - 1}}{x}} \right) \Longrightarrow \frac{x}{1} - \left( {\frac{2}{1}} \right) \cdot \left( {\frac{x}{{x - 1}}} \right) \Longrightarrow \frac{x}{1} - \frac{{2x}}{{x - 1}}}\end{equation}$$

Bravo! The complex fraction is gone! Note that since we were dividing and multiplying, we did not touch the minus sign following the $\displaystyle \frac{x}{1}$ (for order-of-operations reasons). We still need to subtract the remaining fractions, so we undertake another cycle of LCD work. With denominators of $1$ and $x - 1$ for this final subtraction, the LCD is clearly $x - 1$, so we adjust the $\displaystyle \frac{x}{1}$ by multiplying the numerator and denominator by the missing factor $\left( x - 1 \right)$: $$\begin{equation} \label{eq:lcd_and_subtract} {\frac{{\left( {x - 1} \right)}}{{\left( {x - 1} \right)}} \cdot \frac{x}{1} - \frac{{2x}}{{x - 1}} \Longrightarrow \frac{{{x^2} - x}}{{\left( {x - 1} \right)}} - \frac{{2x}}{{\left( {x - 1} \right)}} \Longrightarrow \frac{{\left( {{x^2} - x} \right) - \left( {2x} \right)}}{{\left( {x - 1} \right)}}} \end{equation}$$ We examine the "pesky parentheses" and decide (in this case) that they are safe to drop, giving $$\begin{equation} \label{eq:final_prob_45} {\frac{{{x^2} - x - 2x}}{{\left( {x - 1} \right)}} \Longrightarrow \boxed{\frac{{{x^2} - 3x}}{{\left( {x - 1} \right)}} \Longrightarrow \frac{{x\left( {x - 3} \right)}}{{\left( {x - 1} \right)}}}} \end{equation}$$ Two forms of the solution are boxed to indicate that either version is acceptable for homework and exam purposes. However, it is always wise to factor the final numerator to see whether anything will cross-cancel with the denominator.