Linear Equations in Disguise: Free Response

The denominators are $3$, $2$ and $6$, and the smallest number all three divide into is $6$. Multiply every term of both sides by $6$, keeping the numerator that is a difference inside a bracket as the multiplier lands on it:

$6 \cdot \frac{2x}{3} - 6 \cdot \frac{x-4}{2} = 6 \cdot \frac{5}{6}, \qquad 4x - 3(x - 4) = 5.$

Distribute the $-3$ across both terms of the bracket, and watch the $-4$ turn into $+12$:

$4x - 3x + 12 = 5, \qquad x + 12 = 5, \qquad x = -7.$

Check in the original equation rather than in the cleared one, since it is the original that was asked about:

$\frac{2(-7)}{3} - \frac{-7-4}{2} = -\frac{14}{3} + \frac{11}{2} = \frac{-28 + 33}{6} = \frac{5}{6}.$

The coefficient reaching furthest to the right of the point is $0.25$, which runs to the hundredths, so two factors of ten are needed and the multiplier is $100$:

$100(0.25x + 1.5) = 100(0.4x - 0.9), \qquad 25x + 150 = 40x - 90.$

Gather the variable on one side and the constants on the other:

$240 = 15x, \qquad x = 16.$

Check in the original equation: the left side is $0.25(16) + 1.5 = 4 + 1.5 = 5.5$, and the right side is $0.4(16) - 0.9 = 6.4 - 0.9 = 5.5$.

A single factor of ten moves every digit one place only, so multiplying by $10$ leaves

$2.5x + 15 = 4x - 9,$

which is tidier but not yet a whole-number equation. One factor of ten cleared the tenths; the hundredths needed the second.

Clearing multiplies every term by one number, so undoing it divides every term by that same number. Divide $6x - 4 = 9$ through by $12$:

$\frac{6x}{12} - \frac{4}{12} = \frac{9}{12}, \qquad \frac{x}{2} - \frac{1}{3} = \frac{3}{4}.$

Multiplying that back by $12$ returns $6x - 4 = 9$ term for term, so it is a fraction equation with exactly the disguise asked for.

Now the second half. Nothing forced the number $12$. Dividing the same whole-number equation through by $5$ instead gives

$\frac{6x}{5} - \frac{4}{5} = \frac{9}{5},$

and that one clears with a multiplier of $5$. Every choice of divisor produces a different-looking equation, and each clears back to $6x - 4 = 9$, so a whole-number equation has as many disguises as there are numbers to divide it by.

All of them have the same solution, $x = \dfrac{13}{6}$, and that is the point worth carrying away: multiplying both sides by a number that is not zero changes how an equation looks and not which values satisfy it.

Test each line against the one directly above it, in order.

Line 1 is the equation as given. Line 2 is the step to examine. It is a cross-multiplication, the move that turns $\dfrac{a}{b} = \dfrac{c}{d}$ into $ad = bc$, and that move is nothing more exotic than multiplying both sides by $bd$. What it needs is that each side IS one of those fractions. The left side here is $\dfrac{x}{4} + \dfrac{1}{2}$, a sum of two terms, so nothing licenses the step, and the $\dfrac{1}{2}$ simply vanishes from the page.

After that, the work is faithful to its own line 2. Distributing gives

$3x = 4x + 24,$

and subtracting $4x$ from both sides gives $x = -24$. So lines 3 and 4 are correct consequences of a line that should never have been written, which is exactly why nothing further down looks wrong.

The denominators are $4$, $2$ and $3$, so the smallest number all three divide into is $12$. Multiply every term of both sides by it:

$12 \cdot \frac{x}{4} + 12 \cdot \frac{1}{2} = 12 \cdot \frac{x+6}{3}, \qquad 3x + 6 = 4(x + 6).$

Distribute on the right and collect:

$3x + 6 = 4x + 24, \qquad -18 = x.$

Substitute $x = -18$ into the original equation. The left side is

$\frac{-18}{4} + \frac{1}{2} = -4.5 + 0.5 = -4,$

and the right side is $\dfrac{-18 + 6}{3} = \dfrac{-12}{3} = -4$, so the two agree.

Substitute Priya's $x = -24$ into the same equation. The left side is

$\frac{-24}{4} + \frac{1}{2} = -6 + 0.5 = -5.5,$

while the right side is $\dfrac{-24 + 6}{3} = -6$. Those disagree, so $-24$ is not a solution. It solves line 2 faithfully, and line 2 is not the equation she was given.

Give the two terms on the left a common denominator and add them:

$\frac{x}{4} + \frac{1}{2} = \frac{x}{4} + \frac{2}{4} = \frac{x + 2}{4}.$

The equation is now a single fraction equal to a single fraction,

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

which is the shape cross-multiplication was made for. Multiplying both sides by $12$, the product of the two denominators, sends each numerator across to the opposite denominator:

$3(x + 2) = 4(x + 6), \qquad 3x + 6 = 4x + 24, \qquad x = -18.$

That is the value clearing the denominators produced, and the reason it had to be is worth stating. Rewriting $\dfrac{x}{4} + \dfrac{1}{2}$ as $\dfrac{x+2}{4}$ is an identity: the two expressions are worth the same at every value of $x$, so the equation before the rewriting and the equation after it are satisfied by exactly the same numbers. Priya's step was not of that kind. It deleted a term instead of absorbing it, and produced an equation with a different solution.

The general condition is the one her line 2 walked past. Cross-multiplication applies when the equation is one ratio equal to one ratio. If a side is a sum, or carries an extra term, either combine that side into a single fraction first, as here, or clear every denominator with their least common denominator, as in part B. Both routes leave the solutions alone; the shortcut on its own does not.

The scale says that the ratio of paper centimetres to ground kilometres is the same everywhere on the sheet, and the pair $3$ and $8$ names that ratio:

$\frac{c}{d} = \frac{3}{8}.$

Put $c = 10.5$ into it and cross-multiply:

$\frac{10.5}{d} = \frac{3}{8}, \qquad 3d = 8(10.5) = 84, \qquad d = 28.$

So $10.5$ centimetres of paper stands for $28$ kilometres of trail. The size is worth a glance before moving on: $10.5$ centimetres is three and a half times $3$ centimetres, and $28$ kilometres is three and a half times $8$ kilometres, exactly as a fixed ratio demands.

This time the ground distance is known and the map length is not, but the same proportion governs both:

$\frac{c}{20} = \frac{3}{8}.$

Cross-multiply and solve:

$8c = 3(20) = 60, \qquad c = 7.5.$

The trail is $7.5$ centimetres long on the map. Check it against the scale itself: $20$ kilometres is two and a half times $8$ kilometres, and $7.5$ centimetres is two and a half times $3$ centimetres.

Cross-multiply each arrangement and set the resulting equations beside one another. A proportion and the equation it cross-multiplies to are satisfied by the same pairs wherever the denominators are not zero, so two arrangements that produce the same equation are two ways of writing one fact.

The arrangement $\dfrac{c}{d} = \dfrac{3}{8}$ gives

$8c = 3d.$

The arrangement $\dfrac{c}{3} = \dfrac{d}{8}$ gives

$8c = 3d$

as well, so it says exactly what the first one says. Only what sits underneath what has changed, and the cross-products do not notice that swap.

The arrangement $\dfrac{c}{d} = \dfrac{8}{3}$ gives

$3c = 8d,$

which is a different equation altogether. Now test all three on the pair the scale itself hands you, $c = 3$ with $d = 8$. The first two become $8(3) = 3(8)$, that is $24 = 24$, which is true. The last becomes $3(3) = 8(8)$, that is $9 = 64$, which is false, so it does not describe this map. Read as a sentence it puts $8$ centimetres of paper against $3$ kilometres of ground, which is this map's scale the wrong way round.

The denominators are $2$, $3$ and $6$, so the least common denominator is $6$. Multiply every term of both sides by it, including the term that carries only $m$:

$6 \cdot \frac{x}{2} + 6 \cdot \frac{x}{3} = 6 \cdot \frac{kx}{6} + 6m, \qquad 3x + 2x = kx + 6m.$

Combine the two variable terms on the left, then subtract $kx$ from both sides and take the $x$ out as a factor:

$5x = kx + 6m, \qquad 5x - kx = 6m, \qquad (5 - k)x = 6m.$

No value of $k$ or of $m$ was assumed anywhere along the way, so this one line stands for every member of the family at once.

Everything hangs on the number multiplying $x$, which is $5 - k$.

If $k \neq 5$, that number is not zero, so both sides may be divided by it:

$x = \frac{6m}{5 - k}.$

That is one value and no other, whatever $m$ happens to be. Even $m = 0$ belongs here: it gives the single solution $x = 0$, not a family of them.

If $k = 5$, the number multiplying $x$ is zero, and since $0 \cdot x$ is $0$ whatever $x$ is, the variable has left the equation. The line now reads

$0 = 6m,$

with no $x$ in it to adjust, so the constant decides. When $m = 0$ the line says $0 = 0$, which is true whatever $x$ was, so every number is a solution. When $m \neq 0$ the line says that $0$ equals a number that is not $0$, which is false, and no choice of $x$ can rescue a numeric falsehood, so there is no solution.

So the coefficient decides whether there is a verdict to read at all, and the constant decides which verdict it is.

Take the cases from part B and turn them back into equations.

Every number is a solution in the case $k = 5$ with $m = 0$, which is the single member

$\frac{x}{2} + \frac{x}{3} = \frac{5x}{6}.$

It is an identity for a reason you can see without any of the machinery: $\dfrac{x}{2} + \dfrac{x}{3} = \dfrac{3x + 2x}{6} = \dfrac{5x}{6}$, so the two sides are one expression written two ways.

A member with no solution needs $k = 5$ and any $m$ that is not zero. Taking $m = 1$ gives

$\frac{x}{2} + \frac{x}{3} = \frac{5x}{6} + 1.$

Clearing it gives $5x = 5x + 6$, and subtracting $5x$ leaves $0 = 6$, which is false, so nothing satisfies it. Any other nonzero $m$ does the same work.

Now the inference, which runs one way only. Told nothing but that a member has no solution, you may conclude that $k = 5$, because every member with $k \neq 5$ has the solution part B named, and you may conclude that $m \neq 0$, because $m = 0$ would make every number a solution instead. What you may not conclude is which nonzero value $m$ has: $m = 1$, $m = -4$ and $m = 100$ all produce members with no solution, and the verdict cannot tell them apart. So the outcome fixes $k$ completely, and fixes only whether $m$ is zero, not which number it is.

The denominator $x - 5$ is zero at $x = 5$, so the equation says nothing at all there, and that value is set aside before any algebra begins.

Multiply every term of both sides by $x - 5$:

$x = 2 + 3(x - 5), \qquad x = 3x - 13.$

Collect and solve:

$13 = 2x, \qquad x = \frac{13}{2}.$

Now the comparison that the excluded value was written down for: $\dfrac{13}{2}$ is $6.5$, which is not $5$, so nothing rules it out. Substituting confirms it. The left side is

$\frac{6.5}{1.5} = \frac{13}{3},$

and the right side is $\dfrac{2}{1.5} + 3 = \dfrac{4}{3} + 3 = \dfrac{13}{3}$.

The denominator is the same, so $x = 5$ is excluded again. Multiply every term of both sides by $x - 5$:

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

Collect and solve:

$5 = x.$

The candidate is exactly the value set aside at the start. It cannot be a solution, because the original equation is not defined there at all: putting $x = 5$ into it asks for $\dfrac{5}{0}$. So the candidate is thrown out, and since the clearing produced no other, the equation has no solution.

Notice what the answer is not. It is not $x = 5$, and it is not $x = 0$. A rejected candidate leaves nothing behind it.

A claim that says always is finished off by a single instance, so produce one and evaluate both equations at the same value.

Start from

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

and multiply both sides by $x - 5$ to reach $x = 5 + 2(x - 5)$, that is, $x = 2x - 5$. Test $x = 5$ in each of them. In the cleared equation,

$5 = 2(5) - 5 = 5,$

which is true, so $x = 5$ is a solution of the cleared equation. In the original, the left side is $\dfrac{5}{5-5} = \dfrac{5}{0}$, which names no number at all, so $x = 5$ is not a solution of it and could not be.

The two equations therefore do not have the same solutions, and the claim as stated is false. Where it broke is visible in the multiplier: $x - 5$ is zero at $x = 5$, and multiplying both sides of an equation by zero makes any two things agree. At every other value the multiplier is a number that is not zero, and there the step can be undone, which is why $5$ is the only value at which these two equations part company.

The two directions are not symmetric, and separating them is what turns a warning into a rule.

Nothing is lost. Suppose some value genuinely solves the original equation. Every denominator in it is then nonzero at that value, or the equation would say nothing there, so the multiplier $x - 5$ is a number that is not zero at that value. Multiplying both sides of a true equality by a nonzero number keeps it true, so the value satisfies the cleared equation as well. Every genuine solution survives the clearing.

Something can be gained. Run the same reasoning backwards on a candidate produced by the clearing. If the multiplier is not zero at that candidate, the multiplication can be undone by dividing both sides by it, and the candidate solves the original too. That leaves exactly one way for a candidate to be spurious: the multiplier has to be zero there, which for this multiplier means

$x - 5 = 0, \qquad x = 5,$

and that is the excluded value itself.

So the condition is plain. The claim as written is true whenever the multiplier cannot be zero, and a number that is not zero cannot: $12$ is not zero once and for all, whatever $x$ turns out to be. That is why clearing the fractions in the earlier questions of this set needed no rejection step, while these two equations do.

And it says what the comparison in parts A and B is for. It is not a ritual to be performed on every answer. It is the one place where a step you have already taken can hand you a value the original never had, and setting the candidate against the excluded values is what catches it.