Solving Linear Equations: Free Response

Two operations are wrapped around the variable: it is multiplied by $4$, and then $7$ is subtracted. Undo them in the reverse of that order.

First the addition property of equality, used with the number $7$:

$4x - 7 + 7 = 13 + 7, \qquad 4x = 20.$

Then the division property of equality, used with the number $4$, which is legal because $4$ is not zero:

$\frac{4x}{4} = \frac{20}{4}, \qquad x = 5.$

Check in the equation as it was written: $4(5) - 7 = 20 - 7 = 13$, which is what the right side says.

Start from the statement whose solution is to be preserved, and apply each move to both sides.

Multiply both sides by $3$, which is the multiplication property of equality used with the number $3$:

$3 \cdot x = 3 \cdot (-2), \qquad 3x = -6.$

Add $5$ to both sides, which is the addition property of equality used with the number $5$:

$3x + 5 = -6 + 5, \qquad 3x + 5 = -1.$

The finished equation is $3x + 5 = -1$, which is of the form $ax + b = c$ with $a = 3$, $b = 5$ and $c = -1$. Substituting confirms the build: $3(-2) + 5 = -6 + 5 = -1$.

Nothing here depended on the particular numbers $3$ and $5$. Any nonzero multiplier and any constant produce an equation of the same form built around the same value, so this part has many correct answers rather than one.

Take the two moves in turn and ask whether each one can be run backwards.

Multiplying both sides by $3$ is undone by dividing both sides by $3$, and that division is legal because $3$ is not zero. Adding $5$ to both sides is undone by subtracting $5$ from both sides. So the chain can be read in either direction:

$x = -2 \quad \longleftrightarrow \quad 3x = -6 \quad \longleftrightarrow \quad 3x + 5 = -1.$

Read downwards, every value satisfying $x = -2$ satisfies the built equation, so nothing was lost. Read upwards, every value satisfying the built equation satisfies $x = -2$, so nothing was gained. Two equations with exactly the same solutions are equivalent, and the statement $x = -2$ is satisfied by one number and by no other. The built equation is therefore satisfied by that one number and no other either. Notice that no value was ever tested: the argument settles every number at once, which testing could not do.

Now the condition. Adding a number $c$ to both sides is undone by subtracting $c$, for every $c$ including zero, so no addition or subtraction can ever damage an equation. Multiplication is different. Multiplying both sides of any equation at all by $0$ produces

$0 = 0,$

which every value of the variable satisfies. There is no move that recovers what stood there before, because that same line comes from every equation, so the step gains solutions instead of preserving them. That is why the property is written for a nonzero number, and it is the one move the balance forbids.

Subtract $3x$ from both sides. The variable terms cancel on one side, and on the other $8x - 3x = 5x$:

$3x + 14 - 3x = 8x - 6 - 3x, \qquad 14 = 5x - 6.$

Add $6$ to both sides:

$20 = 5x.$

Divide both sides by $5$:

$x = 4.$

Check in the equation as given, evaluating each side on its own: the left is $3(4) + 14 = 26$ and the right is $8(4) - 6 = 26$. The two agree, so $x = 4$.

Subtract $8x$ from both sides. Now the variable terms cancel on the other side, and $3x - 8x = -5x$:

$3x + 14 - 8x = 8x - 6 - 8x, \qquad -5x + 14 = -6.$

Subtract $14$ from both sides:

$-5x = -20.$

Divide both sides by $-5$, and a negative divided by a negative is positive:

$x = 4.$

This is the same value the other route produced, which is what two chains of licensed moves on one equation have to deliver.

Why they agree. Each route is a chain of moves, and every move is one of the properties of equality applied to both sides. Such a move never loses a solution: whatever made the two sides equal still makes them equal once the same thing has been done to each. It never gains one either, because the opposite move takes you straight back. So every equation in a chain has exactly the same solutions as the one above it, and therefore as the equation both chains started from. Two routes that both finish with the variable standing alone are both reporting that one solution set, so they cannot report different numbers.

Notice that this argument never mentions which term was subtracted, so it settles the matter for every equation with the variable on both sides, not only for this one.

Which to recommend. The two routes are correct but not equally comfortable. Subtracting the smaller variable term leaves

$14 = 5x - 6,$

with a positive coefficient, and no negative number appears anywhere after it. Subtracting the larger one leaves

$-5x + 14 = -6,$

and finishes with

$\frac{-5x}{-5} = \frac{-20}{-5}.$

That route asks you to get two more signs right than the other one does. This is the whole case for the habit: not that the route through the negative coefficient is wrong, but that it offers more places to go wrong.

Simplify each side on its own before moving anything across the equals sign. Distribute on the left, then combine the two variable terms:

$4(x - 3) + 2x = 4x - 12 + 2x = 6x - 12.$

Distribute on the right:

$3(x + 1) = 3x + 3.$

The equation now reads $6x - 12 = 3x + 3$, with the variable on both sides. Subtract $3x$ from both sides, then add $12$, then divide by $3$:

$3x - 12 = 3, \qquad 3x = 15, \qquad x = 5.$

Check in the equation as originally written: the left is $4(5 - 3) + 2(5) = 8 + 10 = 18$, and the right is $3(5 + 1) = 18$.

Substitute $x = 9$ before solving anything. That leaves a statement whose only unknown is $k$.

The left side then contains no unknown at all:

$4(9 - 3) + 2(9) = 24 + 18 = 42.$

The right side becomes $3(9 + k)$, so what has to hold is

$3(9 + k) = 42.$

Clear the parentheses and isolate $k$ exactly as you would isolate $x$, undoing the constant and then the coefficient:

$27 + 3k = 42, \qquad 3k = 15, \qquad k = 5.$

Check by putting that number back into the right side at $x = 9$: $3(9 + 5) = 3(14) = 42$, which is what the left side came to.

What the two tasks share. Nothing in the properties of equality mentions which letter is unknown. They say that if two expressions are equal, then adding the same number to both, subtracting the same number from both, multiplying both by the same nonzero number, or dividing both by the same nonzero number leaves them equal. So the moves that isolate $k$ are the moves that isolate $x$, used for the same reason and in the same reverse order: undo the constant, then undo the coefficient.

$3(9 + k) = 42 \quad \longrightarrow \quad 27 + 3k = 42 \quad \longrightarrow \quad 3k = 15.$

Line for line, that is the routine from part A with a different letter standing in the hot seat.

Why one unknown is left. An equation is a statement about the letters written in it. Before any substitution, $4(x - 3) + 2x = 3(x + k)$ mentions two letters, so it is not a question with a single answer; it is a condition relating them, and many pairs satisfy it. Supplying $x = 9$ replaces every occurrence of one of those letters by a number. Evaluating what that produces leaves a statement in $k$ alone, which is a linear equation in one unknown, and the lesson's routine takes such an equation down to a single value.

So find the missing number, given the solution is never a new technique. It is the technique from part A with the substitution done first instead of last.

Work each side separately, and resolve what is inside the parentheses before multiplying by the factor outside.

The left side:

$5(4 - 2) = 5(2) = 10.$

The right side:

$3(4) + 6 = 12 + 6 = 18.$

The two sides come to $10$ and $18$. A solution is a value that makes the two sides equal, and these are not equal, so $x = 4$ is not a solution of the equation as it was given.

Clear the parentheses first, and the factor outside has to reach both terms inside:

$5(x - 2) = 5x - 10.$

The equation then reads $5x - 10 = 3x + 6$, with the variable on both sides. Subtract $3x$ from both sides, add $10$, and divide by $2$:

$2x - 10 = 6, \qquad 2x = 16, \qquad x = 8.$

Check in the equation as written, each side on its own: the left is $5(8 - 2) = 5(6) = 30$ and the right is $3(8) + 6 = 30$. The two agree, so $x = 8$.

Her check was honest arithmetic and it established something true. The trouble is what it established something about.

Substituting $x = 4$ into $5x - 2 = 3x + 6$ really does give $18$ on both sides, so $x = 4$ is a solution of that line. A check can only ever report on the equation it was carried out in. It says nothing whatever about a different equation, unless the two are already known to have the same solutions.

Are they? The rewriting replaced $5(x - 2)$ by $5x - 2$, and those two expressions do not take the same value. Distributing correctly gives

$5(x - 2) = 5x - 10,$

which sits $8$ below $5x - 2$ at every number: at $x = 0$ one is $-10$ and the other $-2$, at $x = 1$ one is $-5$ and the other $3$, and the gap never closes. So the line Maya checked is not the equation she was asked about, and a value satisfying one of them is under no obligation to satisfy the other. Part A is that fact in numbers.

The condition, stated generally: a rewritten line certifies a solution of the original only when the two are equivalent, meaning they have exactly the same solutions. Two things guarantee that. Either the line came from applying a property of equality to both sides, each of which is undone by the opposite move, or it came from replacing one side by an expression that equals it at every value, which is what distributing correctly does and what distributing carelessly does not.

That is why the routine ends by substituting into the original equation. The original is the one line nobody could have damaged on the way down.

The licence. Read the property carefully. It is a statement about numbers, and it gets applied to one value of the variable at a time.

An equation asks which values of $x$ make its two sides equal, so fix any candidate value $s$ and look at what stands there. The left side is the number $7s + 1$, the right side is the number $2s + 16$, and the quantity $2x$ is the number $2s$. If that candidate is a solution, those two side values are equal, and the subtraction property (used with the number $2s$) says that taking $2s$ from each leaves them equal:

$7s + 1 - 2s = 2s + 16 - 2s, \qquad 5s + 1 = 16.$

That last line says precisely that $s$ solves $5x + 1 = 16$. Nothing was subtracted that failed to be a number at the moment it was subtracted. The letter is shorthand for whatever value is being tested, and at every value being tested, $2x$ is a number.

Same solutions. The argument above sends every solution of $7x + 1 = 2x + 16$ to a solution of $5x + 1 = 16$, so nothing was lost. Nothing was gained either, because the move is undone: adding $2x$ back to both sides of $5x + 1 = 16$ returns the equation it came from, and the same reasoning run in that direction sends solutions back the other way. Solving confirms what is left:

$5x + 1 = 16, \qquad 5x = 15, \qquad x = 3,$

and in the equation as given the left is $7(3) + 1 = 22$ and the right is $2(3) + 16 = 22$.

The claim is about every equation of that shape, so the argument runs in letters and never picks particular numbers.

Subtract $px$ from both sides. On one side the two variable terms combine, because the distributive law lets the common factor $x$ be taken out of $ax - px$:

$ax + b - px = px + q - px, \qquad (a - p)x + b = q.$

Subtract $b$ from both sides:

$(a - p)x = q - b.$

Divide both sides by $a - p$. This is the step the assumption was for: $a$ and $p$ are different numbers, so $a - p$ is not zero, and division by a nonzero number is a property of equality:

$x = \frac{q - b}{a - p}.$

That gives at MOST one solution: each line followed from the line above it, so any solution of the equation we started from has to be that one number. It also gives at LEAST one, because every move can be undone, by adding $b$ to both sides, multiplying both sides by $a - p$, and adding $px$ to both sides. Read upwards, the chain turns the statement about that number back into the equation, so the number really does satisfy it. Exactly one solution, and it is that quotient.

Notice what the argument never needed. It never asked whether $a$ was bigger than $p$, whether any of the four numbers was negative, or whether the equation had carried parentheses before it was tidied. The one assumption used anywhere is that $a - p$ is not zero.

Match the equation to the shape, which fixes $a = 9$, $b = 4$, $p = 2$ and $q = 25$. The coefficients $9$ and $2$ are different, so the result applies. Substituting the four numbers into it:

$x = \frac{q - b}{a - p} = \frac{25 - 4}{9 - 2} = \frac{21}{7} = 3.$

Verify in the equation as written, each side on its own:

$9(3) + 4 = 31, \qquad 2(3) + 25 = 31.$

The two agree, so $x = 3$.

The point of the exercise is the accounting. The three moves were carried out once, in letters, and every equation of this shape now costs one substitution instead of three lines of algebra.

The two pieces of work are exposed to different kinds of failure, and a substitution check tests for only one of them.

What a check detects. Substituting a value into the original equation and finding the two sides equal tells you that this value satisfies that equation. It is a test of the ANSWER, and it is blind to how the answer arrived. It catches a mis-multiplication, a dropped minus sign, a $15$ written where $51$ was meant, wherever above it that happened, because a value produced by such a slip almost never satisfies the equation you began with.

Why part B has nothing for it to catch. No number is computed anywhere in that argument. Each line is the line above it with a property of equality applied, and the property is what guarantees the new line, so there is no separate step at which a digit could come out wrong. What can go wrong in an argument of that kind is a different thing entirely: a move that was not actually licensed, or an assumption used without being stated. A substitution would not detect either, because a formula reached by an unlicensed move can still satisfy the equation in the cases anyone happens to try.

Why the check goes back to the top. A numerical solve produces a ladder of lines, and a slip anywhere above puts a wrong equation on every line below it. Checking against one of those lines tests the value against an equation that may already carry the mistake, so the substitution goes back to the equation as given:

$9(3) + 4 = 31 = 2(3) + 25.$

That line is the only one you know for certain was written correctly.