Solving for a Variable: Free Response

Every letter except the target counts as a fixed number, so read the formula as $V = (\pi r^{2}) \cdot h$: one known quantity multiplying the target. A single division undoes it.

$\frac{V}{\pi r^{2}} = \frac{\pi r^{2} h}{\pi r^{2}} = h, \qquad\text{so}\qquad h = \frac{V}{\pi r^{2}}.$

The divisor $\pi r^{2}$ is built from a letter, so it is not automatically a quantity you are allowed to divide by. Since $\pi$ is not zero, the product $\pi r^{2}$ is zero exactly when $r$ is zero, and the rearranged formula therefore holds provided $r \neq 0$. For an actual cylinder that costs nothing, because a radius is positive, but the restriction belongs to the algebra whether or not the geometry happens to supply it.

Two operations are wrapped around the target: $2$ is subtracted from it, and the whole bracket is then multiplied by $180$. Undo them in reverse, outermost first, so divide both sides by $180$:

$\frac{S}{180} = n - 2.$

Then add $2$ to both sides to free the target:

$n = \frac{S}{180} + 2.$

The only division was by $180$, a number known not to be zero, so nothing has to be assumed about the letters: this rearrangement is as unconditional as the formula it came from.

Now check it on a hexagon, where $n = 6$ and $S = 720$:

$\frac{720}{180} + 2 = 4 + 2 = 6,$

which is the number of sides the hexagon started with. A check like this catches a lost sign or a term left behind. Passing it does not by itself prove the rearrangement right, since one pair of numbers can agree by accident, but failing it proves something is wrong.

Line the two final divisions up beside each other, since that is the only place a condition can enter:

$\pi r^{2} \qquad\text{against}\qquad 180.$

The second is a number. It is not zero, it will never become zero, and no reader needs to be told anything about it. The first contains a letter, and a letter is a stand-in for a number nobody has named yet, so the value $r = 0$ has not been ruled out by anything written down. Dividing by it is a move made on credit, and the restriction $r \neq 0$ is what settles the account.

What decides the difference, then, is not how complicated the formula looks, nor how many steps the rearrangement took. It is one question about the final divisor: is it a quantity whose value is known, or one still standing for something unnamed?

The last question matters because a formula is read by people who did not watch it being derived. Written with nothing attached, it presents itself as holding for every value of its letters, which is more than the algebra ever established. Leaving the restriction off does not make the formula tidier; it makes it claim more than it earned, and it misleads at exactly the value where the formula has nothing to say.

Only two of the three terms contain the target, so move the one that does not. Subtract $2lw$ from both sides:

$A - 2lw = 2lh + 2wh.$

The right-hand side now holds both copies of the target. Read the distributive law from right to left to gather them into one:

$2lh + 2wh = 2h(l + w), \qquad\text{so}\qquad A - 2lw = 2h(l + w).$

The target appears once now, multiplied by the single quantity $2(l + w)$, so one division finishes the job:

$h = \frac{A - 2lw}{2(l + w)}.$

Keep the whole numerator over the bar. Dividing only the $A$ and leaving $2lw$ outside loses a term, exactly as it would when dividing any sum.

The divisor is built from letters, so it carries a restriction: this holds provided $l + w \neq 0$. For a real box that is automatic, since the two base measurements are positive, but as a piece of algebra it is a condition like any other.

Substitute $A = 94$, $l = 5$ and $w = 3$, evaluating the numerator and the denominator separately before dividing:

$h = \frac{94 - 2(5)(3)}{2(5 + 3)} = \frac{94 - 30}{16} = \frac{64}{16} = 4.$

The height is $4$ centimeters, a length, not an area, so it carries centimeters and not square centimeters.

Confirm it against the original formula rather than the one you derived, since that is the version that was given:

$2(5)(3) + 2(5)(4) + 2(3)(4) = 30 + 40 + 24 = 94,$

which is the surface area in the question. Notice also that $l + w = 8$ here, so the restriction from part A is satisfied and the substitution was legitimate.

Carry the division through every term on each side, since a division applied to one term only is not a property of equality at all:

$\frac{A - 2lw}{2l} = \frac{2lh}{2l} + \frac{2wh}{2l} = h + \frac{wh}{l}.$

Nothing about that equation is wrong. It follows from the previous one by the division property of equality, legal here provided $l \neq 0$, and every choice of letters satisfying one satisfies the other. It simply is not an answer to the question that was asked. The target still stands in two places, one of them inside a fraction, so it has not been isolated, and dividing again does not help, because the same thing happens again.

Why the same move succeeds after the factoring is a matter of counting appearances. Division undoes a multiplication, so it can strip a quantity away from the target only when that quantity multiplies the WHOLE side. In $2lh + 2wh$ the factor $2l$ multiplies one term, not the side, because the side is a sum. Factoring rewrites that sum as a product:

$2lh + 2wh = 2h(l + w),$

and in a product the target appears exactly once. One appearance is what one division is able to remove, which is why the factoring is not a tidying step that could be skipped by someone in a hurry. It is the step that makes the division possible.

Take the three questions in order, because they have different answers and it is easy to let one decide the others.

Is it true? Yes. Dividing both sides of $gx + h = kx$ by $k$ is the division property of equality, needing only $k \neq 0$, and it gives

$\frac{gx + h}{k} = \frac{kx}{k} = x.$

So her equation follows from the one she was given, and it is satisfied by exactly the same values of $g$, $h$, $k$ and $x$, at every $k$ that is not zero.

Could her check have failed? No, and this is the part worth slowing down for. Because her equation is equivalent to the original, any values satisfying the original satisfy hers, so the test was decided before she ran it. Trying different numbers changes nothing: at $g = 1$, $h = 12$, $k = 5$ the equation is satisfied by $x = 3$, and her right-hand side gives $\dfrac{1(3) + 12}{5} = 3$ again. A numerical check can catch a step that changed the meaning of an equation; it cannot detect a step that changed nothing.

What does it fail to do? It fails to answer the question. To solve for a letter is to produce an equation with that letter alone on one side and absent from the other, so that its value can be computed from the quantities that are known. Hers has the target on both sides, so it computes the unknown from the unknown. Substituting the numbers makes this concrete: at $g = 2$, $h = 6$, $k = 5$ her result reads $x = \dfrac{2x + 6}{5}$, which is an equation still to be solved, not a formula waiting to be evaluated.

The target stands on both sides, so the first job is to collect every term containing it on one side. Subtract $gx$ from both sides:

$h = kx - gx.$

Both terms on the right carry the target, so factor it out, which is the whole point of the move:

$h = x(k - g).$

The target now appears once, multiplied by the single quantity $k - g$, so one division isolates it:

$x = \frac{h}{k - g}, \qquad\text{provided } k \neq g.$

Unlike Dana's version, this one computes the unknown from quantities that are known. Test it twice, at values that give different answers, so that agreement means something. At $g = 2$, $h = 6$, $k = 5$ it gives $\dfrac{6}{3} = 2$, and the original reads $2(2) + 6 = 5(2)$, that is $10 = 10$. At $g = 1$, $h = 12$, $k = 5$ it gives $\dfrac{12}{4} = 3$, and the original reads $1(3) + 12 = 5(3)$, that is $15 = 15$.

Truth is not the property being asked for, and that is the whole lesson of this question. Every equation obtained from a true equation by properties of equality is true, so a derivation full of legal moves can be beyond reproach and still end somewhere useless. Dana's equation is true, yours is true, and they are equivalent to each other wherever both are defined; what separates them is shape.

The test is therefore a test on the form of the result, and it can be applied by looking rather than by computing. An offered rearrangement is solved for a chosen letter when that letter stands alone on one side of the equals sign, and does not appear anywhere on the other side.

Apply it. Dana's result

$x = \frac{gx + h}{k}$

has the target alone on the left, which is why it can look finished at a glance, but the target also appears in the numerator on the right, so it fails the second half of the test. Yours,

$x = \frac{h}{k - g},$

has the target alone on the left and only $h$, $k$ and $g$ on the right, so it passes. Every letter on the right is one of the quantities you were handed, which is exactly what makes the formula usable: substitute their values and a number comes out, with no equation left to solve.

The test also explains why the factoring in part B was unavoidable. Collecting the terms was what removed the target from one side, and factoring was what reduced it to a single appearance on the other, so that the last division could clear it away entirely.

The target sits in both terms on the right, so it has to be gathered before it can be divided away. Each term carries a factor of $L$, and the lone term is $L \cdot 1$, so a $1$ stays behind inside the bracket:

$S = L - Ld = L(1 - d).$

Check that by expanding it back: $L(1 - d) = L - Ld$, which is what we started from. Dropping the $1$ and writing $L(-d)$ would expand to $-Ld$ alone, losing a whole term.

The target now appears once, multiplied by the single quantity $1 - d$, so one division isolates it:

$L = \frac{S}{1 - d}.$

The divisor contains a letter, so the rearrangement is not unconditional. Since $1 - d$ is zero exactly when $d = 1$, the formula holds provided $d \neq 1$.

The discount rate is $d = 0.15$, so the quantity to divide by is $1 - d = 0.85$, not $0.15$. With $S = 68$:

$L = \frac{68}{0.85} = 80.$

The list price was $80$ dollars, which is above the sale price, as it has to be.

Confirm it forward through the formula as given:

$L - Ld = 80 - 80(0.15) = 80 - 12 = 68,$

which is the sale price stated, and the $12$ dollars taken off is indeed $0.15$ of $80$.

The formula divides by $1 - d$, which is zero exactly when $d = 1$, so $d = 1$ is the rate it cannot serve. The interesting question is what that excluded value means once the letters are read back as prices.

A discount rate of $1$ takes off the entire list price. Whatever an item was listed at, its sale price is

$S = L - L(1) = 0,$

so a coat listed at $200$ dollars and a scarf listed at $12$ dollars both leave a record reading: sale price $0$, rate $1$. The records are identical, and they were identical before any algebra was attempted.

That settles why no arithmetic could help. Recovering the list price means recovering it FROM the record, and here two different list prices produce the very same record. No formula, however clever, can send one input to two different outputs, so no formula can undo this. The restriction $d \neq 1$ is not a technicality about fractions; it is the algebra reporting, faithfully, that the information the accountant wants was destroyed at the moment the discount was applied.

This is worth carrying to every rearrangement whose divisor holds a letter. The excluded value is not a flaw in the rearranging, and it is not always harmless bookkeeping either. It is the one place where the question being asked has no answer, and the formula's refusal to produce one there is the correct behaviour.

The claim says that whatever satisfies one form satisfies the other, which is a statement about every choice of numbers at once. A single choice that breaks it brings it down, so the work is to find one, and there is only one place the second form can be in trouble: its denominator.

Take $l = 0$, and any width at all, say $w = 7$. The first form then forces

$A = lw = 0 \times 7 = 0,$

so the triple $A = 0$, $l = 0$, $w = 7$ satisfies $A = lw$. Put the same triple into the second form:

$w = \frac{A}{l} = \frac{0}{0},$

which names no number, since division by zero is undefined. The second form does not disagree here; it says nothing here, which is worse. A triple satisfies the first and not the second, so the claim is false.

Be careful what has been shown, because a counterexample is a precise instrument. It kills the claim as stated, which was a claim about every triple. It does not show that the rearrangement is wrong, that it is useless, or that the two forms differ anywhere else, and in fact for every nonzero $l$ they agree exactly. What it shows is that the rearranged form covers less ground than the original, and that the difference has to be written down.

Collecting the target on one side turns $ax = bx + c$ into $x(a - b) = c$, so a triple works for a chosen value of $x$ exactly when $c$ equals that value times $a - b$. Pick the two coefficients freely and let that equation decide the constant.

For the first triple take $a = 3$ and $b = 1$, so $a - b = 2$ and $c = (-4)(2) = -8$. Verify it in the equation as given:

$3(-4) = -12, \qquad 1(-4) + (-8) = -4 - 8 = -12.$

The two sides agree, so $x = -4$ satisfies it.

For the second take $a = 1$ and $b = 5$, so $a - b = -4$ and $c = (-4)(-4) = 16$. Verify:

$1(-4) = -4, \qquad 5(-4) + 16 = -20 + 16 = -4.$

Again the two sides agree.

The two triples are genuinely different, sharing neither $a$, nor $b$, nor $c$, and both are answered by the same number. So the solution does not pin the equation down: it names one number, while the equation carries three, and the construction above manufactures as many further triples as you like by choosing any $a$ and $b$ with $a \neq b$ and reading $c$ off. Information travels one way here. The equation determines the solution, and the solution does not determine the equation.

There are two things to prove here, and they are different claims: that this number works, and that nothing else does. A substitution settles the first and cannot touch the second.

It works. Substitute $x = \dfrac{c}{a - b}$, which names a number because $a \neq b$, and evaluate each side. The left side is

$a \cdot \frac{c}{a - b} = \frac{ac}{a - b}.$

The right side needs the constant written over the same denominator before the two can be added:

$b \cdot \frac{c}{a - b} + c = \frac{bc}{a - b} + \frac{c(a - b)}{a - b} = \frac{bc + ca - cb}{a - b} = \frac{ac}{a - b}.$

The two sides agree, so the number satisfies the equation.

Nothing else does. Let $x$ be any number satisfying $ax = bx + c$. Subtract $bx$ from both sides, by the subtraction property of equality:

$ax - bx = c.$

The left side is a sum of two terms both containing the target, so the distributive law, read from right to left, gathers them into one:

$x(a - b) = c.$

Now divide both sides by $a - b$. This is the division property of equality, which permits dividing by a quantity only when that quantity is not zero, and $a - b$ is nonzero precisely because $a \neq b$:

$x = \frac{c}{a - b}.$

So every solution is forced to be that one number. Together with the first half, the equation has that number as a solution and has no other, which is the claim.

Where it is spent. Go along the argument line by line and ask of each whether it would still be legal if the two coefficients happened to be equal. Subtracting $bx$ from both sides is legal for any numbers at all. Gathering $ax - bx$ into $x(a - b)$ is the distributive law, which holds whatever $a$, $b$ and $x$ stand for and never asks anything of them. The one line that asks for something is

$x(a - b) = c \quad\longrightarrow\quad x = \frac{c}{a - b}.$

The division property of equality permits dividing both sides by a quantity only when that quantity is not zero, and $a - b$ is zero exactly when $a = b$. Without the hypothesis, that line asserts a division by something that might be zero, which is not a licensed move, and the same hypothesis is what lets $\dfrac{c}{a - b}$ name a number in the first half.

What follows about the equal case. Nothing at all, and saying so plainly is the point. When $a = b$ the argument does not reach a false conclusion, and it does not reach a true one: it stops, because the step it needs is unavailable. A proof that assumes something and then uses it has established a statement about the cases its assumption covers and has said nothing about the rest. It is tempting to read a proof of "if $a \neq b$ then the solution is that number" as though it also announced what happens when $a = b$, and it does not. That case needs its own argument, run on the equation the substitution $b = a$ actually produces.

What the two have in common. They are one phenomenon seen from two sides. In part A the division by $l$ was carried out and then a triple was found that it could not serve; here the division by $a - b$ is fenced off in advance by a hypothesis, so no such triple can arise. Both are the same transaction: a division by a lettered quantity is not free, and the price is a value, or a whole case, that the result no longer covers. Whether you pay in advance by stating a condition or in arrears by discovering a triple that breaks, the bill is the same size.