Variables and Expressions: Free Response

Terms are the pieces that an expression adds and subtracts together, so a $+$ or $-$ that is not inside a grouping is where one term ends and the next begins.

Breaking the line at those signs leaves three pieces:

$8w, \qquad v, \qquad -13.$

The sign written in front of a piece belongs to that piece. That is why the last term is $-13$ rather than a bare $13$: the expression takes thirteen away, and a term recording only the $13$ would have kept the number and thrown away the instruction attached to it.

A coefficient is the number multiplying the variable inside a term, so read the two variable terms one at a time.

In the first, the number is written in front of the letter, and a number written beside a letter means multiply:

$8w = 8 \times w.$

The second shows no number at all, and that is the case worth care. A single $v$ is one copy of $v$:

$v = 1 \times v.$

So its coefficient is $1$. The remaining term has no letter in it at all, which is what makes it the constant, and it carries the minus sign standing in front of it:

$-13.$

Both questions are about what the notation is already saying, before anybody rewrites anything.

Start with the missing number. Writing a number beside a letter means multiply, so $8w$ is eight copies of $w$. A letter standing alone is the same kind of thing with the count equal to one:

$v = 1 \times v.$

The coefficient is therefore $1$, not nothing. Calling it nothing suggests the letter is not being multiplied by anything, and by the same reading $-v$ would lose its count too, when in fact it is one copy taken away and its coefficient is $-1$.

Now the sign. A term is one of the pieces being combined, so the instruction to subtract has to travel with the piece it applies to. Rewriting the whole line as a sum makes that visible:

$8w + v - 13 = 8w + v + (-13).$

In that form the expression is three terms added together, and the third of them is $-13$. A reader who recorded the constant as $13$ would keep the number and drop the instruction, so anything they built later from their list of terms would add thirteen where the expression takes thirteen away, an error of twenty-six.

Name the unknown once, then take each phrase apart and replace one piece at a time.

The first phrase says to start at the number and take eleven away from it, so the letter is written first even though the eleven is spoken first:

$k - 11.$

The second phrase has two layers. Its inner phrase, the difference of the number and seven, is $k - 7$, and the outer word triples that entire difference rather than any one piece of it. Parentheses are what hold the difference together while the multiplication reaches it:

$3(k - 7).$

The third phrase also has an inner quantity, the sum of nine and the number, and the whole of that sum is divided by four. A fraction bar groups everything written above it, so here the bar does silently what parentheses would otherwise have to do:

$\frac{9 + k}{4}.$

Read each line forward, exactly as it is written, and describe what it does.

The first line starts at twenty and takes the unknown away from that, so in words it is twenty decreased by the number, or the number subtracted from twenty. What it does not say is twenty less than the number, which would start from the letter instead and be written $k - 20$. One value keeps the two apart. Taking $k = 8$:

$20 - 8 = 12, \qquad 8 - 20 = -12.$

The second line multiplies the number by six and then removes five from that product, so in words it is five less than six times the number. The subtraction reaches the product, not the letter on its own, which is why the six belongs to the quantity being reduced:

$6(8) - 5 = 48 - 5 = 43.$

In both lines one question does the work: what quantity does the subtraction take its amount away from? Answer that and the English follows.

A claim like this is settled by testing, not by inspection.

Any value will do; take $k = 10$. With the parentheses, the sum settles first and the whole of it is then multiplied by four:

$4(10 + 3) = 4 \times 13 = 52.$

Without them the tiers take over. Multiplication outranks addition, so the number alone is multiplied by four and the three joins afterwards:

$4 \times 10 + 3 = 40 + 3 = 43.$

The two lines name $52$ and $43$, so the parentheses are not decoration and cannot be dropped.

What the test shows is where grouping earns its place, and it is narrower than it first looks. A group is not needed merely because a sum is present. In $4 + (k + 3)$ the parentheses may be dropped, since $4 + k + 3$ adds the same three quantities and addition may be gathered in any order, and in $4 + (k \times 3)$ they may be dropped too, because the tiers already multiply first. What cannot be dropped is a group whose removal would change which operation acts first, or what it acts on, and that is what happened above: without the parentheses the multiplication reaches the letter alone instead of the whole sum.

Build the total out of the two things actually on the counter.

Each full box holds six muffins, so $b$ boxes hold six copies of that count, written with the number in front of the letter:

$6b = 6 \times b.$

The four loose muffins are there whatever $b$ happens to be, so they join as a fixed amount added on:

$6b + 4.$

The two pieces count different things. The term $6b$ changes from morning to morning, because it depends on how many boxes are out, while the $4$ is a constant and contributes the same amount whatever $b$ is. Writing the rule once in this form covers every morning at once, which is precisely what the letter buys.

Substituting means putting the value where the letter stood, and the parentheses keep the multiplication in plain view.

For nine boxes:

$6(9) + 4 = 54 + 4 = 58.$

For fifteen boxes:

$6(15) + 4 = 90 + 4 = 94.$

So the counter holds $58$ muffins on the first morning and $94$ on the second. Notice what did not change between those two calculations. The rule was identical both times, six muffins for every box and four more on top; the only thing that moved was the number fed into it, which is why one written expression can report every morning's total.

Evaluate the helper's line first, so that the comparison rests on what that line actually reports.

The parentheses group the sum, so the addition settles before the multiplication reaches it:

$6(9 + 4) = 6 \times 13 = 78.$

Against the $58$ muffins on the counter, that is twenty too many.

The arithmetic is faultless. What is wrong is what the line describes. Its parentheses put the four inside the quantity being multiplied by six, so the six is applied to the four as well. In muffin terms the line counts thirteen full boxes at six muffins each, which is what the counter would hold if the four were four more boxes rather than four single muffins.

The gap says the same thing in numbers. The line credits those four muffins with $6 \times 4 = 24$ muffins instead of $4$, and

$78 - 58 = 20 = 24 - 4.$

Where the parentheses close is what decides whether a number joins the quantity being multiplied or stands beside it.

One feature settles every case: an equation carries an equals sign joining two expressions, and an expression carries none at all.

The first two lines hold only numbers, letters and operations, so each names a value once its letter is known:

$5t - 8, \qquad \frac{h}{6} - 2.$

The other two each join two expressions with an equals sign, so each states that the two sides have the same value:

$2(r + 3) = 16, \qquad 8 = g - 5.$

The last line deserves a second look, because its lone number is written first. That changes nothing. A number on its own is a perfectly good expression, so the equals sign is still joining two of them and the line is still an equation. Which side is written first is a matter of style, not of kind.

Substitute, then follow the order of operations.

At $d = 6$ the multiplication comes before the subtraction:

$8(6) - 3 = 48 - 3 = 45.$

At $d = 2$ the same rule runs on a smaller input:

$8(2) - 3 = 16 - 3 = 13.$

The second line divides before it adds, and the fraction bar groups the whole numerator:

$\frac{(30)}{5} + 3 = 6 + 3 = 9.$

The first line came out as two different numbers and nothing has gone wrong. A line carrying a letter is not a number; it is a rule waiting for an input, and it is free to report a different value for each input, which is what this one does. The operations were identical in both of its calculations, which is what makes it one rule, and only the number fed in differed. That freedom is exactly what the word variable is recording.

Each instruction asks its line for something that kind of line has not got, and the reason is the difference between a phrase and a sentence.

Take the first. The line $2(r + 3) = 16$ joins two expressions with an equals sign, so it makes a claim: that the quantity on the left and the number on the right are the same. To evaluate something is to work out the single value it names, and a claim does not name a value. What may be asked of that line is whether it is true at a particular value of $r$. Testing $r = 2$, for instance:

$2((2) + 3) = 2 \times 5 = 10,$

which is not $16$, so the sentence is false at $r = 2$. That is a verdict on a claim, not a value for the line. Finding the value that makes such a sentence true is what solving means, and it is the work of the next few lessons.

Now the second. The line $\dfrac{h}{6} - 2$ carries no equals sign, so it asserts nothing, and there is nothing in it that could be true or false. Solving is finding the value that makes a sentence true, and this line is not a sentence. What it can be asked for is its value once $h$ is given, which is the kind of thing part B asked of its two lines.

The short form is worth keeping: you evaluate a phrase and you solve a sentence, and an equals sign is what tells you which one is in front of you.

Each value replaces the letter, and the parentheses keep the multiplication in view.

$7(4) - 5 = 28 - 5 = 23.$

$7(10) - 5 = 70 - 5 = 65.$

The third input is negative, and the parentheses do the most work here, because they keep the minus sign attached to the three. Seven copies of negative three is negative twenty-one, and taking five from that moves five further down the number line:

$7(-3) - 5 = -21 - 5 = -26.$

Three inputs, three outputs, and one rule used three times.

The mistake is in the notation rather than in the arithmetic: five was correctly subtracted from the wrong number.

A number written beside a letter is the algebraic shorthand for multiplication:

$7y = 7 \times y.$

When the letter is replaced by $4$, that shorthand has to survive the replacement, and writing the digit $4$ hard against the digit $7$ does not let it. Two digits side by side make a two-digit number, so the line says seventy-four. This is the one place the shorthand breaks down, and it breaks down precisely because $7y$ and $74$ look alike while meaning different things.

Parentheses repair it by keeping the input visibly separate from the coefficient:

$7(4) - 5 = 28 - 5 = 23.$

The same habit rescues the negative input in part A, where a bare $-3$ written against the seven would leave $7-3$ on the page, a subtraction where a multiplication was meant.

Both people are looking at the same line, and their disagreement is about what kind of thing it is.

The student is right that a letter holds a place. What they have missed is that holding a place is not the same as saying nothing. The line fixes the operations completely: multiply the input by seven, then take five away. None of that is left open. The only thing left open is which number goes in.

So the line is a rule, and a rule is more than the calculations it produces. A list such as

$7(4) - 5 = 23, \qquad 7(10) - 5 = 65$

covers exactly the inputs written on it, and a reader who wants the value at a fourth input has to go back to the rule to get it. The line covers every input at once, which is why one short expression can stand in for an endless list of arithmetic.

That also answers the objection about the three values. They are three outputs of one fixed rule, not three meanings of one line. All three calculations used identical operations, and the only thing that differed was the number fed into them, so a change in the output is exactly what the word variable allows. A line that reported the same value for every input would be the strange one, since it would not be letting its letter make any difference.