Adding and Subtracting Rational Expressions: Free Response

Neither denominator is a product yet, and nothing about a common denominator can be settled while they are sums. The first is a trinomial whose two numbers multiply to $10$ and add to $7$, and the second is a perfect square:

$x^2+7x+10 = (x+2)(x+5), \qquad x^2+10x+25 = (x+5)^2$

The distinct factors across the pair are $x+2$ and $x+5$. The highest power of $x+2$ reached inside a single denominator is the first, and the highest power of $x+5$ is the second, so

$\text{LCD} = (x+2)(x+5)^2$

The excluded values come from the denominators as they were given, and those vanish at $x=-2$ and at $x=-5$. You can also read them straight off the factored least common denominator, which is zero at exactly those two values and nowhere else.

Each fraction is missing exactly the factors of the least common denominator that its own denominator lacks. The first is short one copy of $x+5$, and the second is short one copy of $x+2$. Multiply the numerator and the denominator of each by what it lacks, and by nothing else:

$\frac{3(x+5)}{(x+2)(x+5)^2} + \frac{4(x+2)}{(x+2)(x+5)^2}$

Now the numerators sit over one denominator and may be added:

$3x+15+4x+8 = 7x+23$

so the sum is

$\frac{7x+23}{(x+2)(x+5)^2}$

Nothing cancels. A factor shared with the denominator would have to be $x+2$ or $x+5$, and those would need the numerator to vanish at $x=-2$ or at $x=-5$. It gives $9$ and $-12$ there, so the fraction is already in lowest terms.

Multiplying the denominators gives $(x+2)(x+5) \cdot (x+5)^2$, which is $(x+2)(x+5)^3$. Set beside the least common denominator $(x+2)(x+5)^2$, it carries one extra copy of $x+5$.

Run the addition over it and watch what that copy costs. The first fraction now lacks $(x+5)^2$ and the second lacks $(x+2)(x+5)$:

$3(x+5)^2 + 4(x+2)(x+5)$

Expanding the two products gives

$3x^2+30x+75+4x^2+28x+40$

which collects to $7x^2+58x+115$, so the route arrives at

$\frac{7x^2+58x+115}{(x+2)(x+5)^3}$

That is not in lowest terms, and finishing means factoring a quadratic you created yourself, $7x^2+58x+115 = (x+5)(7x+23)$, then cancelling the surplus $x+5$. The route through the least common denominator never built it, so there was nothing to take back out.

Now the condition. A factor common to both denominators appears twice in their product but only once in the least common denominator, so the product runs surplus exactly when a common factor exists. The product is therefore the least common denominator precisely when the two denominators share no factor other than $1$.

Whole numbers count as factors here, which is what the pair $2x$ and $4$ is for. They share $2$. Their product is $8x$, while the least common denominator takes one $x$ and $\operatorname{lcm}(2,4)=4$:

$\text{LCD}(2x, 4) = 4x$

A criterion that inspected only the polynomial factors would have called $8x$ least, and it is not.

Factor first: $x^2-25 = (x-5)(x+5)$ and $x^2-5x = x(x-5)$. The distinct factors are $x$, $x-5$ and $x+5$, each reaching only the first power, so the least common denominator is $x(x-5)(x+5)$ and the excluded values are $x=0$, $x=5$ and $x=-5$.

The first fraction lacks a factor of $x$ and the second lacks $x+5$. Rewrite both, and put brackets round the numerator being subtracted before anything is expanded:

$\frac{x(2x+7) - (x+2)(x+5)}{x(x-5)(x+5)}$

Expand inside the brackets first:

$2x^2+7x-(x^2+7x+10)$

Then distribute the minus across all of what is there:

$2x^2+7x-x^2-7x-10 = x^2-10$

So the difference is

$\frac{x^2-10}{x(x-5)(x+5)}$

A factor shared with the denominator would have to be $x$, $x-5$ or $x+5$, so the numerator would have to vanish at $0$, at $5$ or at $-5$. It gives $-10$, $15$ and $15$ there, so nothing cancels and the fraction is in lowest terms.

Line 1 is right. The second denominator is a perfect square, $x^2-16x+64 = (x-8)^2$, and the first denominator $x-8$ divides it, so $(x-8)^2$ is the least common denominator.

Line 2 is where the work departs. Building the first fraction up over $(x-8)^2$ gives it the numerator $6(x-8) = 6x-48$, and the second fraction's numerator is the whole package $3x-4$, so the subtraction is

$6x-48-(3x-4) = 6x-48-3x+4 = 3x-44$

The student wrote $6x-48-3x-4$, leaving the $-4$ with the sign it carried while it was still on the far side of the subtraction sign. Subtracting a negative term is the case people lose most often, because the correction runs the unexpected way: the $-4$ has to arrive as $+4$. Lines 3 and 4 are faithful continuations of that one slip, which is why the first wrong line, rather than the wrong answer, is the thing to name.

The corrected difference is

$\frac{3x-44}{(x-8)^2}$

in lowest terms, since the only factor available to cancel is $x-8$ and the numerator gives $-20$ at $x=8$. The expression excludes $x=8$ and nothing else.

One value settles the disagreement. Take $x=10$, which both expressions allow. There $x-8$ is $2$ and $(x-8)^2$ is $4$, while the second numerator is $26$, so the original is

$\frac{6}{2} - \frac{26}{4} = 3 - \frac{13}{2} = -\frac{7}{2}$

The corrected fraction agrees, giving $\dfrac{30-44}{4} = -\dfrac{7}{2}$, while the student's fraction gives $\dfrac{30-52}{4} = -\dfrac{11}{2}$. One disagreement at an allowed value is enough to separate them, because two expressions that are equal agree at every value both are defined at.

The check claims that every correctly worked subtraction ends with a common factor available to cancel. One correct subtraction whose answer is already in lowest terms refutes that, and such subtractions are not hard to write. Take

$\frac{2}{x-1} - \frac{1}{x+4} = \frac{2(x+4)-(x-1)}{(x-1)(x+4)}$

whose numerator collects to $x+9$, so the difference is

$\frac{x+9}{(x-1)(x+4)}$

The numerator was formed with the minus reaching both terms of $x-1$, so nothing was lost, and $x+9$ is zero only at $x=-9$, where neither denominator factor vanishes. Nothing cancels, and the work is right. Part A of this question is a second instance.

Why the check feels convincing is worth naming. When a problem has been designed so that its answer simplifies, losing a sign destroys the common factor, and in that setting the failure to cancel really does track the mistake. What does not follow is the reverse reading. Most differences anyone writes down do not simplify at all, so nothing cancelling is the ordinary outcome and carries no information about signs.

The honest version of the advice runs one way only. If you expected a simplification and did not get one, re-read the sign step, since that is the cheapest thing to check. An answer in lowest terms is not a verdict either way, and the real verification is redoing the numerator with the subtracted package bracketed, or testing the original against your answer at one allowed value.

Each leg takes its distance over its own speed, so the outward leg takes $\dfrac{36}{v}$ hours and the return leg takes $\dfrac{36}{v+6}$ hours. One follows the other, so the two times add.

The denominators $v$ and $v+6$ share no factor, so the least common denominator is their product $v(v+6)$. The first fraction lacks $v+6$ and the second lacks $v$:

$\frac{36(v+6)+36v}{v(v+6)} = \frac{72v+216}{v(v+6)}$

The numerator has $72$ as a common factor, and $72v+216 = 72(v+3)$, so the ride takes

$\frac{72(v+3)}{v(v+6)}$

hours. That is in lowest terms: the numerator is zero only at $v=-3$, while the denominator's factors are zero at $v=0$ and $v=-6$.

The algebra excludes $v=0$ and $v=-6$. The ride excludes more than that. A speed of zero never reaches the ridge, and a negative speed describes nothing at all, so only $v>0$ describes an actual cyclist. Those are two different questions and an honest answer reports both lists.

A distance divided by a fraction of hours is a fraction with a fraction underneath it, and the main bar means divide:

$72 \div \frac{72(v+3)}{v(v+6)} = 72 \cdot \frac{v(v+6)}{72(v+3)}$

The two copies of $72$ cancel, leaving

$\frac{v(v+6)}{v+3}$

kilometres per hour, in lowest terms since neither $v$ nor $v+6$ is $v+3$.

Check it on a number. At $v=12$ the legs take $\dfrac{36}{12}=3$ hours and $\dfrac{36}{18}=2$ hours, so the ride covers $72$ kilometres in $5$ hours, an average of $14.4$ kilometres per hour. The expression gives $\dfrac{12 \cdot 18}{15}$, which is $14.4$ as well.

The distance dropped out entirely, which is worth noticing: the same average would come out on a road of any length, because lengthening both legs lengthens the distance and the time together.

Combine the two over the denominator $v+3$, bracketing the expression being subtracted:

$(v+3) - \frac{v(v+6)}{v+3} = \frac{(v+3)^2 - v(v+6)}{v+3}$

The numerator is $v^2+6v+9-v^2-6v = 9$, so the gap between the app's figure and the ride's average is

$\frac{9}{v+3}$

which is positive for every positive $v$. The app's figure is always too high, and not by a rounding error either: at $v=12$ the gap is $\dfrac{9}{15} = 0.6$, which is exactly the $15$ the app would report against the $14.4$ the ride actually averaged.

The reason lies in the ride rather than in the algebra. The slow leg takes longer than the fast one, so the cyclist spends more of the ride at $v$ than at $v+6$, and an average taken over time leans towards the speed held for longer. The midpoint would be right only if the two speeds were held for equal times, which these two equal-length legs cannot manage.

The size of the error follows the same reasoning. At $v=3$ the gap is $1.5$ kilometres per hour, while at $v=27$ it is $0.3$: the faster the cyclist, the closer in duration the two legs are, and the less the weighting matters.

Combine the top into one fraction and the bottom into one fraction, giving each whole term the denominator $x$:

$x - \frac{9}{x} = \frac{x^2-9}{x}, \qquad 1 + \frac{3}{x} = \frac{x+3}{x}$

The main bar is a division, so the whole thing is one fraction divided by another:

$\frac{x^2-9}{x} \div \frac{x+3}{x} = \frac{(x-3)(x+3)}{x} \cdot \frac{x}{x+3}$

The $x$ cancels and so does the $x+3$, leaving

$x-3$

Now the excluded values, which come from two places. The inner fractions put $x$ underneath, so $x \neq 0$. The bottom of the main bar cannot be zero either, and $\dfrac{x+3}{x}$ is zero exactly when $x=-3$, so $x \neq -3$ as well.

The answer is a polynomial and shows no sign of either restriction, so both have to be carried by hand:

$x-3, \qquad x \neq 0, \; x \neq -3$

The top and the bottom are built from the same two denominators, $x$ and $x+7$, so both combine over $x(x+7)$. On the top the numerator being subtracted is a package, so bracket it:

$\frac{4}{x+7} - \frac{1}{x} = \frac{4x-(x+7)}{x(x+7)} = \frac{3x-7}{x(x+7)}$

On the bottom nothing is subtracted:

$\frac{1}{x} + \frac{1}{x+7} = \frac{(x+7)+x}{x(x+7)} = \frac{2x+7}{x(x+7)}$

Dividing one by the other, the shared $x(x+7)$ cancels:

$\frac{3x-7}{x(x+7)} \cdot \frac{x(x+7)}{2x+7} = \frac{3x-7}{2x+7}$

The restrictions come from three places. The inner denominators give $x \neq 0$ and $x \neq -7$. The bottom of the main bar must not be zero, and $\dfrac{2x+7}{x(x+7)}$ is zero when $2x+7$ is, that is at $x = -\dfrac{7}{2}$, so that value goes on the list too. Only the last of the three is visible in the final form.

Choose the multiplier by asking what would clear every inner denominator at once, which is the least common denominator of all of them. Here the inner denominators are $x$ and $x+7$, twice each, so that is $x(x+7)$.

Multiply the top and the bottom of the whole fraction by it. On the top,

$\left(\frac{4}{x+7} - \frac{1}{x}\right)x(x+7) = 4x-(x+7) = 3x-7$

and on the bottom,

$\left(\frac{1}{x} + \frac{1}{x+7}\right)x(x+7) = (x+7)+x = 2x+7$

so the fraction is $\dfrac{3x-7}{2x+7}$, exactly as before, and no inner fraction survived past the first line.

The multiplier has two jobs, and they are worth keeping apart, because each requirement answers a different question.

Clearing is the first job, and it is what forces a common multiple of every inner denominator. Miss one of them and that inner fraction survives the step, leaving a complex fraction behind.

Preserving the value is the second job, and it asks for much less: only that the multiplier is never zero, since what licenses the whole move is the fact that also licenses building a fraction up, namely that multiplying a numerator and a denominator by the same nonzero quantity leaves the value alone. Nothing about clearing is needed for that. Multiplying above and below by $x^2+1$, for instance, preserves the value everywhere and clears nothing at all, which is exactly why the two requirements cannot be run together into one.

Here $x(x+7)$ meets both. It is a common multiple of every inner denominator, and it is nonzero at every value still in play, since it vanishes only at $x=0$ and $x=-7$, both already excluded.

Both routes are honest, and which is shorter depends on the fraction. This one favours clearing, because its four inner denominators are built from only two distinct factors. When the top and the bottom are already single fractions, combining first is the shorter road.

Factor both denominators. The first has a common factor of $x$, and the second is a trinomial whose two numbers multiply to $-12$ and add to $-4$:

$x^2-6x = x(x-6), \qquad x^2-4x-12 = (x-6)(x+2)$

The distinct factors are $x$, $x-6$ and $x+2$, each reaching the first power only, so the least common denominator is $x(x-6)(x+2)$ and the excluded values are $x=0$, $x=6$ and $x=-2$. Write them down now, because one of them is about to become invisible.

The first fraction lacks $x+2$ and the second lacks $x$, and the second numerator is being subtracted, so bracket it:

$\frac{3(x+2)-4x}{x(x-6)(x+2)} = \frac{6-x}{x(x-6)(x+2)}$

The numerator $6-x$ is $-(x-6)$, which is one of the denominator's own factors wearing a minus sign, so it cancels:

$\frac{-(x-6)}{x(x-6)(x+2)} = \frac{-1}{x(x+2)}$

The denominator left standing is smaller than the one that was built. All three restrictions survive: the answer raises no objection at $x=6$, but the expression it came from is undefined there.

Write the two expressions as $\dfrac{A}{P}$ and $\dfrac{B}{Q}$, each in lowest terms, with $P$ and $Q$ sharing no common factor. Because they share nothing, the recipe for the least common denominator takes every factor of each, so it is the product $PQ$, and combining gives

$\frac{A}{P} \pm \frac{B}{Q} = \frac{AQ \pm BP}{PQ}$

Call that numerator $N$. Being in lowest terms means no irreducible factor divides both $N$ and $PQ$, so suppose some irreducible $f$ divides $PQ$. A polynomial factors into irreducible pieces in essentially one way, and the pieces of $PQ$ are the pieces of $P$ together with the pieces of $Q$, so $f$ is a factor of $P$ or a factor of $Q$. Take the first case; the second is the same argument with the letters swapped.

So let $f$ be a factor of $P$. Then $f$ divides $BP$. Suppose it divided $N$ as well. It would then divide $N \mp BP$, which is $AQ$, and by the same uniqueness of factorization it would have to be a factor of $A$ or of $Q$. It is not a factor of $Q$, since $P$ and $Q$ share no factor and $f$ divides $P$. So $f$ is a factor of $A$, and then $A$ and $P$ share the factor $f$, contradicting $\dfrac{A}{P}$ being in lowest terms.

No irreducible factor of $PQ$ therefore divides $N$, so the combined fraction cannot be reduced at all. The argument never used which sign was taken, so a sum and a difference are covered alike.

The hypothesis is doing real work, and part A is where you can watch it fail: those two denominators shared the factor $x-6$, the least common denominator was smaller than their product, and the numerator turned out to be divisible by the shared factor.

Take

$\frac{1}{x(x+4)} + \frac{1}{x(x-3)}$

Both fractions are in lowest terms, and their denominators share the factor $x$. The distinct factors are $x$, $x+4$ and $x-3$, so the least common denominator is $x(x+4)(x-3)$, one copy of $x$ smaller than the product of the two denominators. The first fraction lacks $x-3$ and the second lacks $x+4$:

$\frac{(x-3)+(x+4)}{x(x+4)(x-3)} = \frac{2x+1}{x(x+4)(x-3)}$

For anything to cancel, one of $x$, $x+4$ or $x-3$ would have to divide $2x+1$, which would mean $2x+1$ vanishing at $0$, at $-4$ or at $3$. It gives $1$, $-7$ and $7$ there, so nothing cancels and the combination is already in lowest terms.

Set that beside part A, where the denominators also shared a factor and the fraction did reduce. Two cases with the same hypothesis and opposite outcomes show that a shared factor cannot settle the matter by itself: it makes a reduction possible and no more. Part B supplies the other half of the picture, and its hypothesis has to be quoted in full: for two fractions ALREADY IN LOWEST TERMS whose denominators share no factor, a reduction is impossible, and there the built denominator is guaranteed to be the answer's. Drop the lowest-terms half and the guarantee goes with it, since $\dfrac{x^2+x}{x^2}$ and $\dfrac{1}{x+2}$ have denominators sharing no factor yet combine to $\dfrac{x^2+4x+2}{x(x+2)}$, over less than the denominator built for them.

What that means in practice is that the check has to be carried out rather than predicted. Combine first, then test each factor of the denominator against the numerator by substituting the value that kills it.