Algebraic Fractions: Free Response

The greatest common factor of $8x^3$ and $12x$ has two pieces: the coefficients $8$ and $12$ share $4$, and the smallest power of $x$ present in both is $x$ itself. So the shared factor is $4x$.

Write each part as that factor times what is left:

$\frac{8x^3}{12x} = \frac{4x \cdot 2x^2}{4x \cdot 3} = \frac{2x^2}{3}.$

The cancellation is legitimate because the fraction contains a hidden $\dfrac{4x}{4x} = 1$, and that is available whenever $4x$ is not zero.

For the restriction, look at the denominator of the ORIGINAL, which is $12x$. It is zero exactly when $x = 0$, so $x = 0$ is excluded and nothing else is.

Neither the top nor the bottom is a product yet, so factor each one before looking for anything to cancel:

$10x + 15 = 5(2x + 3), \qquad 4x + 6 = 2(2x + 3).$

The bracket $2x + 3$ is now a factor of the whole numerator and of the whole denominator, so it cancels and the two numbers in front are all that survive:

$\frac{5(2x + 3)}{2(2x + 3)} = \frac{5}{2}.$

For the restriction, set the original denominator to zero. From $4x + 6 = 0$ comes $4x = -6$, so

$x = -\frac{3}{2}.$

That value is excluded. Notice that the simplified fraction $\dfrac{5}{2}$ has no variable left in it and so could not possibly object to any value of $x$; the restriction belongs to the expression the question asked about.

Cancelling here is dividing the top and the bottom by $2x$, and that move is only available when $2x$ is not zero, since it is the hidden $\dfrac{2x}{2x} = 1$ that licenses it. So the factoring reads

$\frac{6x^2 - 2x}{2x} = \frac{2x(3x - 1)}{2x} = 3x - 1, \qquad x \neq 0.$

Equality between two expressions is a claim tested value by value: substitute an allowed number and both must return the same result. At $x = 4$, for instance, the original is $\dfrac{96 - 8}{8} = 11$, and $3(4) - 1 = 11$ as well.

At $x = 0$ that test cannot even be run. The original becomes $\dfrac{0}{0}$, which names no number at all, while $3x - 1$ cheerfully returns $-1$. The two do not agree there; the second is simply defined where the first is not.

So dropping the restriction would upgrade a statement about every $x$ but one into a statement about every $x$, and the extra case is false. Carrying $x \neq 0$ is what keeps $3x - 1$ a rewriting of the original rather than a different expression that happens to agree with it almost everywhere.

Multiplying fractions means multiplying the tops together and the bottoms together, with no common denominator anywhere in sight:

$\frac{4x}{7} \cdot \frac{21}{8x^3} = \frac{84x}{56x^3}.$

Now reduce. The coefficients $84$ and $56$ share the factor $28$, and the single $x$ on top matches one of the three below:

$\frac{84x}{56x^3} = \frac{28x \cdot 3}{28x \cdot 2x^2} = \frac{3}{2x^2}.$

Only one denominator in the original carries a variable, namely $8x^3$, and it is zero exactly when $x = 0$. So $x = 0$ is the single excluded value.

Turn the division into a multiplication by the reciprocal of the divisor, and factor the first numerator on the way in, since $9x + 12 = 3(3x + 4)$:

$\frac{3(3x + 4)}{2x} \cdot \frac{8x^2}{3x + 4}.$

The bracket $3x + 4$ is now a factor of a numerator and of a denominator, so it cancels, and $\dfrac{8x^2}{2x} = 4x$ takes care of the rest:

$\frac{3 \cdot 8x^2}{2x} = \frac{24x^2}{2x} = 12x.$

Three separate conditions produce the restrictions, and they have to be collected before any cancelling happens. The denominator $2x$ needs $x \neq 0$. The denominator $8x^2$ needs $x \neq 0$ again. And dividing by the second fraction is only meaningful when that fraction is not itself zero, which fails when its numerator vanishes:

$3x + 4 = 0 \;\Rightarrow\; x = -\frac{4}{3}.$

So the work is valid for $x \neq 0$ and $x \neq -\dfrac{4}{3}$. Notice where the second condition came from: $3x + 4$ is a NUMERATOR in the problem as it was posed, and only becomes a denominator once the divisor has been flipped. Reading restrictions off the denominators of the original alone would have missed it entirely.

Start from what division asks for. Writing $A \div B$ asks for the number that multiplies $B$ to give $A$, and the answer is well defined exactly when one such number exists and no other does. Set $B$ to zero and the request collapses: every candidate multiplies $0$ back to $0$, so there is no answer at all when $A$ is not zero, and no single answer when $A$ is zero. Either way $A \div 0$ names nothing. That is a fact about division itself, not about how fractions are written.

So a divisor must be nonzero, and when the divisor is an algebraic fraction, being nonzero becomes a condition on the variable. A fraction is zero exactly when its numerator is zero and its denominator is not, so the NUMERATOR of the divisor is the thing to inspect. That is what makes this restriction easy to miss: numerators are the one place nobody looks for a forbidden value.

Now the example. Its first fraction has $x$ underneath, so $x \neq 0$. Its divisor is $\dfrac{x - 5}{3}$, which is zero when $x - 5 = 0$, so $x \neq 5$. Its remaining denominator, $3$, rules out nothing. Rewriting confirms both:

$\frac{1}{x} \cdot \frac{3}{x - 5} = \frac{3}{x(x - 5)}.$

Both surviving conditions now show up as denominators, which is a useful check, though the reason for the second one was settled before any flipping took place.

Each time is pages divided by rate. The slower printer takes $\dfrac{300}{x}$ minutes. The faster one prints $2x$ pages a minute, so it takes $\dfrac{300}{2x}$ minutes. Run one after the other and the two times add.

The denominators $x$ and $2x$ are not the same, so build the common denominator $2x$ by multiplying the first fraction top and bottom by $2$:

$\frac{600}{2x} + \frac{300}{2x} = \frac{900}{2x}.$

The top and the bottom now share the factor $2$, which cancels:

$\frac{900}{2x} = \frac{2 \cdot 450}{2 \cdot x} = \frac{450}{x}.$

The expression is undefined at $x = 0$, so that value is excluded. The situation is stricter than the algebra, and part C returns to that.

The two times are $\dfrac{300}{x}$ and $\dfrac{300}{2x}$, so the comparison is one algebraic fraction divided by another. Multiply by the reciprocal of the divisor:

$\frac{300}{x} \cdot \frac{2x}{300} = \frac{600x}{300x}.$

Both $300$ and $x$ are factors of the whole top and of the whole bottom, so both cancel:

$\frac{600x}{300x} = 2.$

No value of $x$ makes the divisor $\dfrac{300}{2x}$ equal to zero, since its numerator is the constant $300$, so the divisor contributes no restriction of its own here and the only excluded value is $x = 0$.

The ratio came out as a pure number: the variable cancelled, so the comparison holds whatever the slower printer's rate happens to be. Doubling a rate halves the time, and that is true at $5$ pages a minute and at $50$ alike.

Check it on two values. At $x = 5$ the slower printer takes $\dfrac{300}{5} = 60$ minutes and the faster one takes $\dfrac{300}{10} = 30$ minutes, and

$\frac{60}{30} = 2.$

At $x = 20$ the two times are $15$ minutes and $7.5$ minutes, and $\dfrac{15}{7.5} = 2$ again.

Now the values of $x$. The algebra objects at $x = 0$ and nowhere else, since that is the only value making a denominator zero. The situation objects to more than that. A rate of $0$ pages a minute means the job never finishes, and a negative rate describes nothing at all, so only $x > 0$ corresponds to a real printer.

That gap is worth naming. An expression can be perfectly well defined at a value the problem it models still forbids, so the excluded values and the sensible values are two different lists, and an honest answer reports both.

Take the two removals one at a time, and of each ask whether it multiplies the ENTIRE expression above the bar and the entire expression below it.

Removing the $3$. Writing the numerator as a product makes the answer visible:

$\frac{9x + 6}{3x} = \frac{3(3x + 2)}{3 \cdot x} = \frac{3x + 2}{x}.$

The $3$ multiplies the whole numerator, since $9x + 6$ is $3$ times $3x + 2$, and it multiplies the whole denominator. So the fraction holds a hidden $\dfrac{3}{3} = 1$, and dropping it is legitimate. Since $3x + 2$ and $x$ share no factor beyond $1$, form (i) is also as far as this goes.

Removing an $x$. Here $x$ divides the term $9x$, but it does not divide the term $6$, so it is not a factor of the whole numerator and there is no hidden $\dfrac{x}{x}$ to drop. Carrying the division out honestly instead:

$\frac{3x + 2}{x} = \frac{3x}{x} + \frac{2}{x} = 3 + \frac{2}{x}.$

Be exact about what form (ii) does to that result, because it is not what it first looks like. It does not throw the leftover term away: throwing it away would give $3$. It writes the leftover $\dfrac{2}{x}$ as a bare $2$, giving $3 + 2 = 5$. So it is too big rather than too small, and part B's numbers show exactly that.

The order does not matter, incidentally. Removing the $x$ first, against the $x$ of $3x$, gives $\dfrac{9 + 6}{3} = 5$ by the same unlicensed step.

Every value except $x = 0$ is allowed. It is worth seeing first what happens at the most tempting choice.

At $x = 1$ the printed fraction is $\dfrac{9 + 6}{3} = 5$, form (i) is $\dfrac{3 + 2}{1} = 5$, and form (ii) is $5$. All three agree, and nothing has been settled. That is not bad luck: form (ii) is $3 + 2$ while the true value is $3 + \dfrac{2}{x}$, and those coincide exactly when $\dfrac{2}{x} = 2$, which is to say at $x = 1$ and nowhere else. A wrong answer is entitled to be right somewhere.

So take another value, $x = 2$. The printed fraction:

$\frac{9(2) + 6}{3(2)} = \frac{24}{6} = 4.$

Form (i):

$\frac{3(2) + 2}{2} = \frac{8}{2} = 4.$

Form (ii) is the constant $5$, here as everywhere.

Now the two directions, which are not symmetric. Form (ii) disagrees with the printed fraction at a value both are defined at, and equal expressions agree at EVERY allowed value, so one disagreement refutes it outright and no more testing is needed. Form (i) agrees here, which is evidence and not proof: one value out of infinitely many has been tried, and $x = 1$ is the standing reminder of what that is worth. What actually establishes form (i) is the factoring in part A.

The rule rests on one fact: a fraction whose top and bottom share a factor has a copy of $1$ hidden inside it. If the numerator is $a \cdot c$ and the denominator is $b \cdot c$, then

$\frac{a \cdot c}{b \cdot c} = \frac{a}{b} \cdot \frac{c}{c} = \frac{a}{b},$

since $\dfrac{c}{c} = 1$ for any $c$ that is not zero. Cancelling $c$ is nothing more than dropping that factor of $1$, which is why it leaves the value untouched.

So the condition has two requirements. The quantity must be a factor of the WHOLE numerator and of the WHOLE denominator, and it must not be zero. The second is easy to forget because it is usually satisfied without comment, but it is what the first line above actually needs: $\dfrac{c}{c}$ is $1$ only when $c$ is not zero.

Now what fails when a quantity is merely a term. If it appears in one term of the numerator and not in another, the numerator cannot be written as that quantity times anything, so the fraction cannot be split into $\dfrac{a}{b} \cdot \dfrac{c}{c}$ at all. There is no copy of $1$ available, so there is nothing to drop, and deleting the quantity anyway is not a cancellation but a change of value. Part B is what makes that concrete.

The test is always the same. Write the top and the bottom as products first; whatever appears in both products may go, and whatever does not appear in both stays.

The denominators already agree, so the difference goes over that denominator with the second numerator kept whole inside brackets:

$\frac{(5x + 3) - (x - 9)}{2x + 6}.$

The minus sign reaches both terms inside those brackets, so the $-9$ becomes $+9$:

$(5x + 3) - (x - 9) = 5x + 3 - x + 9 = 4x + 12.$

Now factor the top and the bottom and cancel the shared bracket, using $4x + 12 = 4(x + 3)$ and $2x + 6 = 2(x + 3)$:

$\frac{4(x + 3)}{2(x + 3)} = 2.$

The original denominator $2x + 6$ is zero when $x = -3$, so that value is excluded, and it stays excluded even though the answer is a plain number with no denominator left to object.

The denominators differ, so build a common one before subtracting anything. Both $2x$ and $x^2$ divide $2x^2$, so take that as the common denominator.

Multiply the first fraction top and bottom by $x$, and the second top and bottom by $2$:

$\frac{5}{2x} = \frac{5x}{2x^2}, \qquad \frac{3}{x^2} = \frac{6}{2x^2}.$

Now the numerators can be subtracted over the shared denominator:

$\frac{5x}{2x^2} - \frac{6}{2x^2} = \frac{5x - 6}{2x^2}.$

The terms $5x$ and $6$ are unlike, so the numerator cannot be shortened, and $5x - 6$ has no factor in common with $2x^2$, so this is lowest terms. Both original denominators vanish at $x = 0$ and at no other value, so $x = 0$ is the only exclusion.

A claim like this is settled by one explicit example, so choose numerators where the second one has a sign inside it, since that is the only place the two readings can come apart.

Take $A = 5x$, $B = 2x - 4$ and $D = x + 4$. Keeping $B$ whole:

$\frac{5x - (2x - 4)}{x + 4} = \frac{3x + 4}{x + 4}.$

Ignoring the brackets and writing the terms of $B$ down as they stand:

$\frac{5x - 2x - 4}{x + 4} = \frac{3x - 4}{x + 4}.$

The two differ, so at most one of them can be the original. Test at $x = 1$, which the expression allows. The original is

$\frac{5}{5} - \frac{-2}{5} = 1 + \frac{2}{5} = \frac{7}{5},$

while the two candidates give $\dfrac{3 + 4}{5} = \dfrac{7}{5}$ and $\dfrac{3 - 4}{5} = -\dfrac{1}{5}$. The bracketed reading matches the original and the other does not, so the brackets are not decoration.

What they record is the SCOPE of the minus sign. Subtracting a fraction subtracts the whole of its numerator, so the sign has to reach every term of $B$, and $-(2x - 4)$ is $-2x + 4$, not $-2x - 4$. When $B$ happens to have a single term there is nothing for the brackets to protect, which is why the careless habit survives so long before anything catches it.