Solving Two-Step Equations: Free Response

The left side is the variable multiplied by $6$ with $21$ added on, so the addition is the outer layer and comes off first. Subtract $21$ from both sides:

$6x + 21 - 21 = 57 - 21 \;\Rightarrow\; 6x = 36.$

The variable term now stands alone, so undo the multiplication by dividing both sides by the coefficient $6$:

$\frac{6x}{6} = \frac{36}{6} \;\Rightarrow\; x = 6.$

Check by substituting into the original equation, multiplying before adding as the order of operations requires:

$6(6) + 21 = 36 + 21 = 57.$

Both sides come to $57$, so $x = 6$. Every number met along the way was a whole number.

Dividing both sides by $6$ means dividing every term on each side, not only the term holding the variable. On the left, $6x$ becomes $x$ and $21$ becomes $\frac{21}{6}$; on the right, $57$ becomes $\frac{57}{6}$:

$\frac{6x + 21}{6} = \frac{57}{6} \;\Rightarrow\; x + \frac{21}{6} = \frac{57}{6}.$

Both of those fractions simplify by a factor of $3$:

$x + \frac{7}{2} = \frac{19}{2}.$

One subtraction finishes it. Take $\frac{7}{2}$ from both sides, and since the two fractions already share a denominator the numerators subtract directly:

$x = \frac{19}{2} - \frac{7}{2} = \frac{12}{2} = 6.$

The value is $x = 6$, the same as before. The route is sound; it simply passed through halves to get there.

Read the left side as a set of instructions carried out on $x$. Under the order of operations the multiplication happens first, producing $6x$, and the addition happens after that, producing $6x + 21$. So the multiplication by $6$ is the inner layer, wrapped directly around the variable, and the $21$ is the outer layer, wrapped around everything.

Taking a construction apart means running it backwards, and the last thing wrapped on is the first thing that can be removed. While $21$ is still attached, the variable term is not alone on its side, so nothing done to $6x$ by itself is a legal move on the equation. Subtracting $21$ strips the outer layer and leaves exactly one operation standing between $x$ and its value. That is why the standard order is subtract, then divide.

The second route is legal too, because dividing both sides by $6$ preserves the equality just as subtracting $21$ does. What it costs is arithmetic. The division reaches every term, so the constant and the number on the right are dragged into it as well, and neither is a whole number of sixes:

$\frac{21}{6} = \frac{7}{2}, \qquad \frac{57}{6} = \frac{19}{2}.$

The rest of that work is then done in halves. Nothing goes wrong, but there are two more fractions to carry and two more places to slip.

Both routes are two moves long, here and always, so what is at stake is never the number of steps but whether fractions appear. That gives a condition worth stating exactly. Dividing $6x + b = c$ by $6$ produces $x + \frac{b}{6} = \frac{c}{6}$, and those two quotients are whole precisely when $6$ divides $b$ and $6$ divides $c$. So dividing first brings in no fractions exactly when the constant and the number on the right are both multiples of the coefficient, and it brings in fractions the moment either one is not. Divide $6x + 18 = 48$ by $6$ and you get $x + 3 = 8$ with no fraction anywhere, and $x = 5$ follows at once. It is the divisibility of the particular numbers, not the shape of the equation, that decides which route is tidier.

Name the unknown first. The quantity nobody knows is the trip's own costs, so let $c$ be that amount in dollars.

Those costs are shared equally among the $8$ members, so one member's share of them is the whole amount split eight ways:

$\frac{c}{8}.$

The museum charge is not shared. Every member pays $15$ dollars of it, so it is added to the share rather than divided by anything. One member therefore pays $\frac{c}{8} + 15$ dollars, and the treasurer's figure says that this is $41$ dollars:

$\frac{c}{8} + 15 = 41.$

One equation, one unknown, and every quantity the situation states appears in it exactly once.

The unknown is divided by $8$ and then has $15$ added, so the addition is the outer layer and comes off first. Subtract $15$ from both sides:

$\frac{c}{8} + 15 - 15 = 41 - 15 \;\Rightarrow\; \frac{c}{8} = 26.$

That line already says something readable: one member's share of the trip's own costs is $26$ dollars.

The unknown is now divided by $8$ and nothing else. A division is undone by multiplying, so multiply both sides by $8$:

$8 \cdot \frac{c}{8} = 8 \cdot 26 \;\Rightarrow\; c = 208.$

Check it against the story rather than only against the equation: $208$ dollars split eight ways is $26$ dollars each, and $26 + 15 = 41$ dollars, which is what each member paid. The trip's own costs are $208$ dollars.

Read the product for what it counts. Each member paid $41$ dollars and there are $8$ members, so

$8 \times 41 = 328$

is the total amount of money the members handed over. That arithmetic is right, and it answers a real question. It is simply not the question the note is asking.

Each of those $41$ dollars is made of two amounts: a share of the trip's own costs, and a museum charge that the museum keeps. Multiplying by $8$ multiplies both amounts, so the $328$ dollars contains eight museum charges as well as the whole of the trip's costs:

$8 \times 15 = 120.$

Take those out and what is left is the trip's own costs:

$328 - 120 = 208.$

So the note overstates the trip's own costs by $120$ dollars. The mistake is one of grouping rather than of arithmetic: the entry charge was never part of what the members were sharing, so it has to come out of a member's payment before anything is multiplied by $8$. That is exactly the order the equation in part A records, with the charge added outside the division and therefore removed first.

The constant $14$ is added to the variable term, so it is the outer layer. Subtract it from both sides:

$-5x + 14 - 14 = 59 - 14 \;\Rightarrow\; -5x = 45.$

The coefficient here is $-5$, not $5$. The minus sign belongs to the number multiplying the variable, so it is part of what you divide by. Divide both sides by $-5$, and a positive divided by a negative is negative:

$\frac{-5x}{-5} = \frac{45}{-5} \;\Rightarrow\; x = -9.$

Check in the original equation, multiplying before adding:

$-5(-9) + 14 = 45 + 14 = 59.$

Both sides come to $59$, so $x = -9$.

Take $4x + 9 = 2$ first. Subtract the constant $9$ from both sides. The right side drops below zero, which is nothing unusual:

$4x + 9 - 9 = 2 - 9 \;\Rightarrow\; 4x = -7.$

Divide both sides by the coefficient $4$. The quotient is not a whole number, and $7$ and $4$ share no factor above $1$, so the fraction is already in lowest terms:

$x = \frac{-7}{4} = -\frac{7}{4}.$

Now $0.4x - 1.6 = 2.8$. A constant is subtracted here, so undo it by adding $1.6$ to both sides:

$0.4x - 1.6 + 1.6 = 2.8 + 1.6 \;\Rightarrow\; 0.4x = 4.4.$

Divide both sides by the coefficient $0.4$. Multiplying both numbers by $10$ turns that into a division of whole numbers without changing the quotient:

$x = \frac{4.4}{0.4} = \frac{44}{4} = 11.$

Check the second in its own original equation:

$0.4(11) - 1.6 = 4.4 - 1.6 = 2.8.$

That is what the equation claims, so $x = 11$.

Take the sign first. In $-5x + 14 = 59$ the variable is multiplied by $-5$, which is one number, and the minus sign is part of that number rather than an instruction standing beside it. The second move exists to undo that multiplication, so it has to divide by the whole of it:

$\frac{-5x}{-5} = x.$

Dividing by $5$ instead would undo a multiplication by $5$, and that is not the multiplication that is there. It leaves $-x$ on the left, so the value that comes out carries the wrong sign. The sign travels into the division because it was never separate from the coefficient to begin with.

Now the fractions. A solution is whatever number makes the two sides equal, and nothing in that definition asks it to be whole. When $4x = -7$ is divided by $4$, the quotient $-\frac{7}{4}$ is not a symptom of an error; it is what splitting $-7$ into four equal parts gives, and the $-7$ arrived honestly because $2$ is smaller than $9$. Rounding it to $-1.8$ would replace the solution with a number that does not satisfy the equation at all.

The procedure that settles either worry is the same one, and it requires trusting none of the reasoning above. Substitute the value into the original equation and evaluate both sides under the full order of operations:

$4 \left(-\frac{7}{4}\right) + 9 = -7 + 9 = 2.$

The two sides agree, so that value is the solution, sign and fraction and all. The check is the authority here, not the tidiness of the number.

Name the unknown: let $t$ be the price of one toolbox in dollars.

One kit is a toolbox together with a pair of gloves, so a single kit costs

$t + 12$

dollars. That sum is one quantity, the price of a kit, and the workshop bought seven of them. Seven copies of a quantity is that quantity multiplied by $7$, and the parentheses are what say that both items are bought seven times rather than the toolbox alone:

$7(t + 12) = 175.$

Every number the situation states appears once: the number of kits, the glove price, and the invoice total.

The left side is a number multiplied by a sum, so clear the parentheses first. The distributive property multiplies the $7$ across both terms inside:

$7(t + 12) = 7t + 84 \;\Rightarrow\; 7t + 84 = 175.$

That is the standard two-step form. The constant $84$ is the outer layer, so subtract it from both sides:

$7t + 84 - 84 = 175 - 84 \;\Rightarrow\; 7t = 91.$

Now divide both sides by the coefficient $7$:

$\frac{7t}{7} = \frac{91}{7} \;\Rightarrow\; t = 13.$

Check in the original equation, settling the parentheses first as the order of operations demands:

$7(13 + 12) = 7(25) = 175.$

One toolbox costs $13$ dollars.

Run the colleague's route. The whole left side is $7$ times the quantity $t + 12$, so dividing both sides by $7$ removes that factor at a stroke:

$\frac{7(t + 12)}{7} = \frac{175}{7} \;\Rightarrow\; t + 12 = 25.$

One subtraction finishes it:

$t = 25 - 12 = 13.$

Same price, fewer lines.

Both routes are legitimate, and for the same reason: it is not that one is a rule and the other a trick. Every balance move in each route is applied to both sides in full, so at no point do the two sides stop holding equal amounts, and each line therefore has the same solution as the line above it. Distributing is not even a balance move. It rewrites one side into an equal form, changing how that side is written rather than what it is worth.

Count the moves before looking at the numbers. Dividing first clears the factor from the whole side at once and leaves $t + 12 = 25$, which one subtraction finishes: two moves. Distributing turns the left side into $7t + 84$, and that equation still needs its constant cleared and then its coefficient divided out: three moves. That gap does not depend on the invoice total at all. Had the invoice read $170$ dollars, dividing first would still be two moves and distributing still three.

What the particular numbers decide is not the length of the route but the tidiness of its arithmetic. Dividing by $7$ reaches every term on both sides, so it stays in whole numbers only when everything it touches is a whole number of sevens. On the left it meets a single factor of $7$, which cancels exactly; on the right it meets the invoice total, and

$\frac{175}{7} = 25$

comes out whole. With $170$ that same first move would have produced $\frac{170}{7}$ and the rest of the work would have been done in sevenths. Distributing divides nothing until the very last move, so it postpones the only division to a point where a fraction, if one is coming, is unavoidable anyway.

The general reading: when a side is a factor times a sum, dividing by that factor first is the shorter route, and the divisibility of the other side decides only whether that route stays in whole numbers along the way.

Test the lines one at a time, asking of each whether it must be true given the line immediately above it.

Line 1 collects the like terms on the left. The terms $12x$ and $-5x$ count the same thing, and $12 - 5 = 7$, so

$12x - 5x + 23 = 7x + 23,$

which makes line 1 a correct rewriting of the original equation. Nothing was done to both sides there; one side was rewritten into an equal form.

Line 2 is where it breaks. Going from $7x + 23 = 93$ to $x + 23 = \frac{93}{7}$, the term $7x$ has been divided by $7$ and the right side has been divided by $7$, but the $23$ has been copied down as it stood. Dividing an equation by $7$ divides every term on both sides, so the line that division actually produces is

$x + \frac{23}{7} = \frac{93}{7}.$

One side was scaled in full and the other only in part, so from line 2 onward the two sides no longer hold equal amounts and the student is working on a different equation from the one they were given.

Line 3 does follow from line 2: subtracting $23$ from both sides of line 2 gives exactly what the student wrote. That is why the reported value is wrong even though every subtraction and division in the work is correct. Faithfully solving the wrong equation is still solving the wrong equation.

Simplify the left side before any inverse operation. The like terms $12x$ and $-5x$ combine:

$12x - 5x + 23 = 93 \;\Rightarrow\; 7x + 23 = 93.$

Now it is a standard two-step equation. The constant $23$ is the outer layer, so subtract it from both sides:

$7x + 23 - 23 = 93 - 23 \;\Rightarrow\; 7x = 70.$

Divide both sides by the coefficient $7$:

$\frac{7x}{7} = \frac{70}{7} \;\Rightarrow\; x = 10.$

Check in the original equation, before any simplification, so that the check tests the whole solution including the combining step:

$12(10) - 5(10) + 23 = 120 - 50 + 23 = 93.$

Both sides come to $93$, so $x = 10$.

Take the two routes in turn on $ax + b = c$, with $a$ not zero.

Clearing the constant first: subtract $b$ from both sides, which leaves the variable term alone because $b - b = 0$, and then divide both sides by $a$:

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

Dividing first: divide both sides by $a$. That division reaches every term, so the left side becomes $\frac{ax}{a} + \frac{b}{a}$, which is $x + \frac{b}{a}$, and the right side becomes $\frac{c}{a}$:

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

Now subtract $\frac{b}{a}$ from both sides. The two fractions already share the denominator $a$, so their numerators subtract directly:

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

The routes end on the same expression, and no step assumed anything about $a$, $b$ and $c$ beyond $a$ not being zero, which both routes need in order to divide at all. So the choice between them is a choice about arithmetic and never about the answer.

Apply that to the student's work. Their route was the second one, and the line it produces is

$x + \frac{23}{7} = \frac{93}{7},$

with the constant divided along with everything else. Had they written that line, the general result guarantees the same value as the standard route, and finishing it confirms as much:

$x = \frac{93}{7} - \frac{23}{7} = \frac{70}{7} = 10.$

What defeated them was not the route they chose but a division that reached only part of a side.