Evaluating and Simplifying Expressions: Free Response

Each term carries the sign printed in front of it, so the four terms are $+10m$, $+8$, $-3m$ and $-13$. Two of them are multiples of $m$ and two are plain numbers, and only terms of the same kind may be gathered.

Collect the multiples of $m$ and the plain numbers separately, adding and subtracting the coefficients:

$10m - 3m = 7m, \qquad 8 - 13 = -5.$

Writing the two results side by side gives

$10m + 8 - 3m - 13 = 7m - 5.$

Nothing further can be done. One term is a multiple of $m$ and the other is a plain number, so the two are unlike and the expression stands at two terms.

The distributive property says that $a(b + c) = ab + ac$. That is an equation, so it may be read from either end, and read from right to left it says something equally true: a factor common to two terms can be pulled out in front of them.

Both terms here are built on the same factor $m$, since $10m = m \cdot 10$ and $3m = m \cdot 3$. Subtracting is adding the opposite, so the pair is $m \cdot 10$ added to $m \cdot (-3)$, and pulling the shared $m$ out in front gives

$10m - 3m = m(10 - 3).$

Inside the parentheses sits ordinary arithmetic with no letter in it at all, and $10 - 3 = 7$, so

$m(10 - 3) = m(7) = 7m.$

No new rule was invented anywhere in that line. Combining like terms is the distributive property collecting a shared factor, with the coefficients doing the only arithmetic in sight.

Two separate facts are at work, and they are worth separating.

First, the reordering. A subtraction is an addition of the opposite, so the expression is really a pure sum:

$10m + 8 + (-3m) + (-13).$

A sum may be added in any order and in any grouping without changing its total, which is the property proved back in chapter 1. So the four terms may be shuffled freely, provided each keeps the sign it arrived with. Placing the two multiples of $m$ side by side is exactly such a shuffle, and that is why the simplified form does not depend on the order in which the terms happened to be written.

Second, the refusal. Collecting works by pulling out a factor the two terms share, and $8$ and $-13$ contain no $m$ whatever, so a group such as $10m + 8$ has no factor of $m$ to pull out. A number can still come out, since $10m + 8 = 2(5m + 4)$, but that leaves a sum inside the parentheses rather than a single term, so the two have been rewritten and not merged. Treating the group as $18m$ would count the plain $8$ as though it were eight copies of $m$, and one value shows that it is not:

$10(4) + 8 = 48, \qquad 18(4) = 72.$

So the two multiples of $m$ join, the two plain numbers join, and the two groups stay where they are.

Substitute $-2$ for every $w$, each one inside its own parentheses so that no minus sign drifts away from the number it belongs to:

$4w^2 - 7w + 3 = 4(-2)^2 - 7(-2) + 3.$

Exponents run before multiplications. Squaring $-2$ multiplies a negative by a negative, which is positive, so $(-2)^2 = 4$ and the first term is $4 \cdot 4 = 16$. Notice that the exponent reached the $-2$ only, never the coefficient $4$.

The middle term is a multiplication of two negatives as well, because the sign in front of the $7$ belongs to that term:

$-7(-2) = +14.$

Only additions are left, and they run left to right:

$16 + 14 + 3 = 33.$

The value is $33$. Both places where a negative met a negative produced a positive, which is why the result is larger than the expression looks at first glance.

Take the expressions one at a time and let the order of operations decide what each exponent is attached to.

In $-t^2$ the exponent sits on the letter alone, so the squaring happens first and the leading minus is applied last. At $t = 7$,

$-t^2 = -(7)^2 = -(49) = -49,$

and at $t = -7$ the square is positive again, so the leading minus still hands back a negative:

$-t^2 = -(-7)^2 = -(49) = -49.$

In $(-t)^2$ the parentheses gather the minus sign in with the letter, so the whole of $-t$ is squared. At $t = 7$ that is $(-7)^2 = 49$, and at $t = -7$ the opposite of $-7$ is $7$, so it is $(7)^2 = 49$:

$(-7)^2 = 49, \qquad (7)^2 = 49.$

The four values are $-49$, $49$, $-49$ and $49$. The first expression came out negative at both inputs and the second came out positive at both.

A pair of tested values can refute a claim about every $t$, but it can never establish one, so argue about the expressions themselves.

Squaring destroys a sign. A positive times itself is positive, a negative times itself is positive, and zero times itself is zero, so no square is ever negative. Since $-t$ and $t$ differ only in sign, their squares agree, while in the other expression the leading minus is applied after the squaring:

$(-t)^2 = t^2 \qquad \text{and} \qquad -t^2 = -(t^2).$

Now read off what each one is capable of. The first is never negative. The second is the opposite of something never negative, so it is never positive. Two numbers, one of them never negative and the other never positive, can be equal only if both are zero.

That leaves a single candidate. A square is zero only when the number squared is zero, since any other number multiplied by itself gives a positive amount. So the two expressions agree at $t = 0$ and nowhere else, and there both come to $0$. Everywhere else they are opposites of each other, and the gap grows quickly: at $t = 7$ the two values already stand $98$ apart.

What the club is left paying is the quoted charge less the rebate, so the whole rebate arrives with a minus sign in front of its parentheses:

$9(t + 2) - 4(t - 5).$

Distribute each factor across its own group. The $9$ reaches both terms of the first group:

$9(t + 2) = 9t + 18.$

The factor in front of the second group is $-4$, not $4$, and it too reaches both terms inside. A negative times a negative is positive, so the $-5$ comes back as a gain:

$-4(t - 5) = -4t + 20.$

With both sets of parentheses cleared, gather the like terms:

$9t + 18 - 4t + 20 = 5t + 38.$

The club's net cost is $5t + 38$ dollars.

The simplified form makes short work of it. At $t = 14$,

$5(14) + 38 = 70 + 38 = 108.$

The long route is what the shop and the supplier actually do. The quoted charge treats the order as though it were two shirts larger, and the parentheses are finished first:

$9(14 + 2) = 9(16) = 144.$

The rebate pays four dollars for every shirt beyond the first five:

$4(14 - 5) = 4(9) = 36.$

Taking the rebate off the charge gives

$144 - 36 = 108.$

Both routes report $108$ dollars, as they must: simplifying rewrote the expression without changing what it computes at any input, so the shorter route is a saving of effort and nothing else.

Read each number back into the shop.

The number multiplying $t$ answers what one more shirt does to the bill. Each extra shirt adds nine dollars to the quoted charge and brings four dollars back as rebate, so the club ends up five dollars worse off for every shirt it adds:

$9 - 4 = 5.$

The number standing alone is what the club would owe if the count of shirts contributed nothing at all, and two pieces make it up. The quoted charge carries a setup amount, written as two extra shirts at nine dollars each, which is $9 \cdot 2 = 18$ dollars. The rebate is four dollars a shirt but only beyond the first five, so the supplier holds back $4 \cdot 5 = 20$ dollars of it, and rebate not paid is money the club keeps paying:

$18 + 20 = 38.$

Neither piece mentions $t$. The setup charge is agreed before a single shirt is printed, and the five shirts excluded from the rebate are the same five whatever the size of the order, so both amounts sit still while the order size moves. That is exactly what the simplified form reports: a fixed part, and a part that grows five dollars at a time.

Read the chain one expression at a time and stop at the first that does not equal the one before it.

The opening expression is the problem as it was set, so nothing can be wrong there yet. The second is where the parentheses were removed, and that is the step to inspect. A minus sign in front of a group is a factor of $-1$ multiplying everything the group holds, so it reaches both terms:

$-(5h - 8) = -1(5h) + (-1)(-8) = -5h + 8.$

The student produced $-5h - 8$ instead. The $5h$ was handled correctly and the $8$ was copied down with the sign it already had, as though the minus had run out of reach halfway through the group. So the second expression is the first wrong one, and the third only carries that error forward.

The size of the slip can be read off as well. Writing $-8$ where $+8$ was owed loses $16$, and the student's constant sits exactly $16$ below the correct one.

Clear the parentheses first, giving both terms inside the sign that the leading minus hands them:

$12h - (5h - 8) + 2 = 12h - 5h + 8 + 2.$

Nothing has been gathered yet. The expression simply has no parentheses left in it, and the sign work is finished and out of the way.

Now collect the like terms. The multiples of $h$ combine through their coefficients, and the two plain numbers combine on their own:

$12h - 5h = 7h, \qquad 8 + 2 = 10.$

Putting the two groups together,

$12h - (5h - 8) + 2 = 7h + 10.$

Any value will do; take $h = 3$. The original expression finishes its parentheses first:

$12(3) - (5 \cdot 3 - 8) + 2 = 36 - 7 + 2 = 31.$

The student's result at the same input is

$7(3) - 6 = 21 - 6 = 15.$

The two numbers differ, and one disagreement is enough. Equivalent expressions agree at every value without exception, so a single input at which they disagree rules the equivalence out completely. The corrected simplification passes the same test, since $7(3) + 10 = 31$ matches the original.

What a check like this cannot do is confirm. Suppose the two had agreed at $h = 3$: that would have left the student's work unproven rather than proven, because expressions that are not equivalent can still cross at an input. The pair $2h + 4$ and $6h$ shows it, both coming to $6$ at $h = 1$ while at $h = 2$ they give $8$ and $12$.

So the asymmetry is the point. One value can refute a simplification outright, while agreement at a chosen input cannot certify one; for that you need the algebra.

Four terms are present, each carrying the sign in front of it: $+9a^2$, $+4a$, $-2a^2$ and $-7a$. Sort them by variable part. Two are multiples of $a^2$ and two are multiples of $a$, and those are different kinds, because the powers differ.

Combine each kind through its coefficients, leaving the variable part exactly as it is:

$9a^2 - 2a^2 = 7a^2, \qquad 4a - 7a = -3a.$

The second group came out negative, which is ordinary enough: $4 - 7 = -3$, and the term is written $-3a$. Putting the groups together,

$9a^2 + 4a - 2a^2 - 7a = 7a^2 - 3a.$

Two terms remain, one with variable part $a^2$ and one with variable part $a$. They are unlike, so nothing further combines.

The classmate's rule asks only that a letter be shared, so any pair with matching letters and different powers will test it. Take $4a$ and $4a^2$.

An evaluation shows what would go wrong. At $a = 3$ the two terms are $4(3) = 12$ and $4(3)^2 = 36$, so the pair is worth

$4a + 4a^2 = 12 + 36 = 48.$

Merging them into $8a$ would report $8(3) = 24$, and merging them into $8a^3$ would report $8(27) = 216$. Neither lands on $48$, so neither of the mergers the classmate's rule invites is equal to the pair.

The repair is to compare the whole variable part instead of merely spotting a letter. Two terms are of the same kind when their variable parts are identical, the same letters each raised to the same power, and only their coefficients differ. By that test $4a$ and $4a^2$ are unlike, while $4a$ and $9a$ are like and merge into $13a$.

Combining like terms is the distributive property read from right to left, so the question is what that reading leaves behind.

Take two multiples of $a^2$, say $8a^2$ and $3a^2$, and pull out the shared $a^2$:

$8a^2 - 3a^2 = a^2(8 - 3) = a^2(5) = 5a^2.$

What sits inside the parentheses is $8 - 3$, plain arithmetic with no letter in it, so it collapses to the single number $5$ and the pair really does become one term.

Now try the same move on a mixed pair, $5a^2$ and $2a$. They do share a factor: both contain at least one $a$, since $5a^2 = a \cdot 5a$ and $2a = a \cdot 2$. Pulling that shared $a$ out in front is perfectly legal:

$5a^2 - 2a = a(5a - 2).$

But look at what is left inside. It is $5a - 2$, which still contains the letter, so it is not a number and cannot be folded into a coefficient. The expression has been rewritten, not shortened into one term, and the two kinds are still visible inside the parentheses.

That is the whole distinction. Terms are of the same kind exactly when the variable factor they share is the entire variable part of each, so that pulling it out leaves nothing but numbers behind, and it is those numbers, the coefficients, that the arithmetic then finishes off. Whenever the letter survives inside the parentheses, no amount of rearranging will produce a single term.