Literal Equations and Formulas: Free Response

Here $r$ is the unknown, and $d$ and $t$ are held as fixed but unstated numbers. The equation reads $d = t \cdot r$: one known quantity multiplying the target, so a single division isolates it.

$r = \frac{d}{t}.$

The divisor $t$ is a letter, so the rearrangement is not unconditional: it holds provided $t \neq 0$, which is also the natural requirement that the trip actually took some nonzero amount of time.

Substituting the given values,

$r = \frac{126}{3} = 42,$

so the drone's rate is $42$ kilometers per hour.

This time $t$ is the letter singled out, and $d$ and $r$ are the ones held fixed. Read the same equation as $d = r \cdot t$, a known quantity multiplying the new target:

$t = \frac{d}{r}.$

The divisor here is $r$, a different letter from part A's divisor, so the restriction is $r \neq 0$, not $t \neq 0$.

At $d = 200$ and $r = 25$,

$t = \frac{200}{25} = 8,$

so the trip takes $8$ hours.

In part A the roles were: unknown $r$, parameters $d$ and $t$. In part B the roles were: unknown $t$, parameters $d$ and $r$. The equation $d = rt$ never changed; only the assignment of roles did.

In both cases the rule for isolating the target is the same: divide by whatever else is multiplying it. In part A that other factor was $t$, so $t$ had to be shown nonzero. In part B the target was $t$ itself, and the other factor doing the multiplying was $r$, so $r$ had to be shown nonzero instead:

$t \neq 0 \ \text{(part A)} \qquad\text{versus}\qquad r \neq 0 \ \text{(part B)}.$

So the two conditions are not two versions of one fact. They are the same rule, that the divisor must be nonzero, applied to two different quantities, because the divisor in a rearrangement is always the OTHER factor multiplying the target, and a different letter held that role in each part.

Collect the $x$ terms on the left and the constants on the right:

$mx - 3x = 2m + 5, \qquad (m-3)x = 2m+5.$

The coefficient is $m-3$, zero exactly at $m=3$. For $m \neq 3$, dividing gives

$x = \frac{2m+5}{m-3},$

exactly one solution. At $m = 3$ the coefficient vanishes and the constant becomes $2(3)+5 = 11$, so the equation reads $0 \cdot x = 11$. No value of $x$ satisfies that, so there is no solution at $m=3$, and since the boundary case already fails to be an identity, no value of $m$ can ever produce infinitely many solutions here.

Collect terms exactly as before:

$mx - 3x = 5m - 15, \qquad (m-3)x = 5(m-3).$

The coefficient is the same $m-3$ as in part A. For $m \neq 3$, the factor $m-3$ cancels:

$x = \frac{5(m-3)}{m-3} = 5,$

the identical value $5$ no matter which $m \neq 3$ is chosen. At $m=3$ the constant is $5(3-3) = 0$ as well, so the equation reads $0 \cdot x = 0$, an identity satisfied by every real number. Since the boundary already produces an identity, no value of $m$ can ever make this family have no solution.

Both families reduce to a coefficient of $m - 3$, so both share the same boundary value $m = 3$ and the same coefficient away from it, even though their one-solution formulas there, $\dfrac{2m+5}{m-3}$ and $5$, are different. The coefficient alone cannot be what separates the two families' behavior AT the boundary, since it is identical in both.

What differs is the constant term at that one value:

$2(3) + 5 = 11 \neq 0 \qquad\text{versus}\qquad 5(3-3) = 0.$

In the first family the constant is $11$ at the boundary, nonzero, so the boundary equation $0 \cdot x = 11$ is a contradiction and no solution results there; the family can never reach infinitely many, because that would need the constant to vanish at the boundary too, and it never will, however the rest of the family is built. In the second family the constant vanishes at the boundary too, so the boundary equation is $0 \cdot x = 0$, an identity; the family can never reach no solution, for the mirror reason.

So the check that predicts the outcome without a full classification is short: evaluate the constant at the parameter value that zeros the coefficient. A nonzero result rules out infinitely many everywhere in the family; a zero result rules out no solution everywhere in the family.

Isolate the term holding the target by subtracting $\dfrac{1}{a}$ from both sides:

$\frac{1}{b} = \frac{1}{t} - \frac{1}{a}.$

Combine the right side over the common denominator $at$:

$\frac{1}{b} = \frac{a - t}{at}.$

Both sides are reciprocals of nonzero quantities exactly when $a \neq t$, so take the reciprocal of each side to reach

$b = \frac{at}{a - t}.$

The divisor $a - t$ is built from two letters, so the rearrangement carries a restriction the original combined-rate equation did not visibly show: it holds only when $a \neq t$.

Substitute $a = 12$ and $t = 4$ into the rearranged formula, evaluating the numerator and the denominator separately:

$b = \frac{(12)(4)}{12 - 4} = \frac{48}{8} = 6.$

Pump B alone would take $6$ hours.

Confirm it in the original equation:

$\frac{1}{12} + \frac{1}{6} = \frac{1}{12} + \frac{2}{12} = \frac{3}{12} = \frac{1}{4},$

which is $\dfrac{1}{t}$ for $t = 4$, the combined time given.

Read the equation as a statement about rates rather than times: $\dfrac{1}{t}$ is the combined rate, and $\dfrac{1}{a}$ is pump A's own rate. Rearranged, pump B's rate is

$\frac{1}{b} = \frac{1}{t} - \frac{1}{a}.$

At $a = t$ this reads

$\frac{1}{b} = \frac{1}{t} - \frac{1}{t} = 0.$

No finite $b$ satisfies $\dfrac{1}{b} = 0$, since a finite time always has a positive rate; the equation is only satisfied in the limit of an infinitely long time, which is another way of saying pump B does no real work at all.

That is also why the situation cannot occur for an actual second pump. Adding a pump that contributes any positive rate can only shorten the combined time below pump A's own time alone, never leave it exactly equal. So $a = t$ describes a pump B that is not really pumping, and the algebra's refusal to name a finite $b$ there is reporting that fact honestly rather than failing to find one.

Collect every term containing $x$ on one side. Subtract $rx$ and $q$ from both sides:

$px - rx = s - q.$

Both terms on the left carry the target, so factor it out:

$x(p - r) = s - q.$

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

$x = \frac{s - q}{p - r}.$

The divisor is built from the two rate letters, so the rearrangement holds only when $p \neq r$: two identical per-gigabyte rates never cross unless the flat fees already agree, in which case every usage costs the same rather than one particular usage.

First case, $p = 6$, $r = 6$: $p - r = 0$, so

$x = \frac{40 - 10}{0}$

is undefined, even though $p = 6 \neq 0$ and $r = 6 \neq 0$. The classmate's rule says nothing should go wrong here, yet the formula fails.

Second case, $p = 0$, $r = 5$: $p - r = -5$, nonzero, so

$x = \frac{25 - 10}{0 - 5} = \frac{15}{-5} = -3$

is perfectly well defined, even though $p = 0$. The classmate's rule says this should have broken down, yet the formula works.

Both cases test the claim against a fact it did not anticipate, and both defeat it: satisfying $p \neq 0$ and $r \neq 0$ is neither SUFFICIENT for the formula to work, since both rates nonzero can still fail, nor NECESSARY for it to work, since the formula can succeed even when one rate is zero.

The division that actually happens in the rearrangement is by $p - r$, a single quantity built from a subtraction, not by $p$ and then separately by $r$. So the honest restriction is the one statement $p \neq r$.

A restriction stated as two separate conditions on the individual letters is not a safer version of that; it answers a different question, whether $p$ and $r$ are each individually nonzero, which is unrelated to whether their difference is nonzero. Part B's two checks make the disagreement concrete:

$6 - 6 = 0 \qquad\text{but}\qquad 0 - 5 = -5.$

The two-condition version under-restricts at the first pair: it would wave through $p=6$, $r=6$ as safe, when the true formula divides by zero there. It over-restricts at the second pair: it would forbid $p=0$, $r=5$, when the true formula is perfectly well defined there, because the divisor $p-r=-5$ never involved $p$ on its own.

So the two versions do not even agree on which cases are dangerous; they disagree in both directions from the correct answer, and neither error is the safer one to make.

A solution exists. Fix any $k \neq u$, so $k - u \neq 0$. The division property of equality permits dividing both sides of $(k-u)x = vk - w$ by the nonzero quantity $k - u$:

$x = \frac{vk - w}{k - u}.$

Substituting this value back in recovers the original equation, since dividing and then multiplying by the same nonzero number returns what you started from, so this number satisfies the equation.

It is the only one. Suppose $x_1$ and $x_2$ both satisfy $(k-u)x = vk-w$ for this same $k$. Then $(k-u)x_1 = (k-u)x_2$, since both equal $vk-w$. Subtracting gives $(k-u)(x_1 - x_2) = 0$. Since $k \neq u$, the factor $k - u$ is not zero, so the other factor must be:

$x_1 - x_2 = 0, \qquad x_1 = x_2.$

So any two solutions coincide, meaning there is exactly one.

The left side $0 \cdot x$ equals $0$ for every real number $x$, since multiplying anything by zero gives zero:

$0 \cdot x = vu - w \quad\Longrightarrow\quad 0 = vu - w.$

If $vu \neq w$, the right side $vu - w$ is a fixed nonzero number, so this reads $0 = vu-w$ with $vu-w\neq0$, a false statement that no choice of $x$ can affect, since $x$ never appears in it. So no $x$ satisfies the equation: the solution set is empty.

If $vu = w$, the right side is $0$, so the equation reads $0 = 0$, a true statement holding regardless of $x$. Every real number then satisfies it, so the solution set is all of $\mathbb{R}$.

Parts A and B together classify every value of $k$: part A covers every $k \neq u$, giving exactly one solution each time, and part B covers the single remaining value $k = u$, splitting it by whether $vu = w$:

$k \neq u \Rightarrow \text{one solution}; \qquad k = u,\ vu\neq w \Rightarrow \varnothing; \qquad k=u,\ vu=w \Rightarrow \mathbb{R}.$

Since every real number is either equal to $u$ or not, these cases cover every $k$ at all, so they are exhaustive; and no $k$ can be both equal to $u$ and not equal to $u$, nor can the single fixed fact $vu = w$ be both true and false, so the cases do not overlap.

The striking part is that whether "no solution" or "infinitely many" is reachable at all is decided by $u$, $v$, $w$ alone, before any particular $k$ is examined. If $vu \neq w$, then at the one place where the coefficient vanishes the constant does not, so this family can produce no solution (at $k=u$) but can never produce infinitely many, no matter how many other values of $k$ are tried, since every $k \neq u$ gives one solution and $k=u$ gives none. If $vu = w$, the mirror holds: this family can produce infinitely many (at $k=u$) but can never produce no solution. Exactly one of these two possibilities is true for any fixed $u$, $v$, $w$, because $vu = w$ is a plain equation between two fixed numbers, either true or false, not something that varies with $k$.

This is why the two worked families in the lesson behaved as they did: each is one instance of this shape, and which of "no solution" or "infinitely many" it could ever reach was already settled the moment its constants were chosen, not something that needed checking separately at each value of the parameter.