Math Class Didn’t Show Its Work, companion reader · Chapter 13

A Variable Is a Name

A letter saves a place for any value the context allows, and the grouping tells you what to do with it.

These four demonstrations follow the chapter's own examples. Pick an input, watch the same rule run, and check each output with a pencil before trusting the picture.

Every example in these readers is a constructed teaching example taken from or modeled on the book's own worked examples. The numbers are declared inputs chosen to make the mathematics visible. They are not measurements of any real person, product or study.

Demonstration 1 of 4

Add two, then triple

Why do 3(n + 2) and 3n + 2 give different outputs from the same input?

Each row follows one instruction from left to right. In 3(n + 2) the addition happens inside the group before the triple. In 3n + 2 only n is tripled, so the 2 is added once instead of three times.

\[3(n + 2)\]

\[3n + 2\]

n names the starting value, here a nonnegative whole number. The 3 and the 2 are constants. Writing 3 next to the parentheses means 3 times the whole group.

Predict first. At n = 4, the chapter gets 18 from one instruction and 14 from the other. Will the gap still be 4 at n = 0?

Choose an example

Figure: Add two, then triple. Top row: three outlined groups, each with 4 circles and 2 squares, 18 marks in all. Bottom row: three groups of 4 circles and one separate pair of squares, 14 marks in all. The top row is highlighted.
Input n: 4, Which instruction?: 3(n + 2): add 2, then triple
Constructed example: the chapter's add-two-then-triple rule and its inputs 0, 1 and 4.

Calculated values

Input n
4
3(n + 2)
18
3n + 2
14
Difference
18 - 14 = 4

With n = 4, add 2 first: 4 + 2 = 6, then triple: 3 x 6 = 18. The other instruction triples first: 3 x 4 = 12, then 12 + 2 = 14. The outputs differ by 18 - 14 = 4: the grouped rule copies the pair of added units three times (6 squares), the other adds one pair (2 squares).

Use the idea

When you turn words into symbols, read your symbols back aloud. If "then triple" applies to the whole result, it needs parentheses.

Where the conclusion applies

The allowed inputs are nonnegative whole numbers, as in the chapter's rule. The gap of 4 comes from the two missing copies of 2, so it does not depend on n.

Check your understanding: Using n = 1, what does each instruction give, and what is the gap?
3(1 + 2) = 3 x 3 = 9, and 3(1) + 2 = 3 + 2 = 5. The gap is 9 - 5 = 4.

Chapter 13 source: section "An expression is an instruction". Demonstration C13-D01.

Demonstration 2 of 4

Which terms can combine

When does a shorter expression really name the same quantity?

Each bar is the value of one form at the chosen x. A correct simplification matches the original at every x. A wrong one can still match at a lucky input, so one agreement is not enough.

\[4x+3-2x+5=2x+8\]

\[4x-2x=(4-2)x\]

A term is a signed part separated by addition or subtraction. In 4x the coefficient is 4. Like terms have the same variable part, so 4x and -2x combine, while 2x and 3x² do not.

Predict first. Keep x = 3 and switch to the unlike pair 2x + 3x². Will either shortcut, 5x or 5x², match the original?

Choose an example

Figure: Which terms can combine. Tiles: 4 x strips, 3 units, 2 negative strips and 5 units; two strip pairs are crossed out, leaving 2 strips and 8 units. Bars of the values at x = 3: 4x + 3 - 2x + 5 is 14, 2x + 8 is 14.
Which expression?: Like terms: 4x + 3 - 2x + 5, Input x: 3
Constructed example: the chapter's expressions 4x + 3 - 2x + 5 and 2x + 3x², at chosen inputs.

Calculated values

x
3
4x + 3 - 2x + 5
14
2x + 8
14
Like terms
4x and -2x (same variable part)

At x = 3: 4 x 3 + 3 - 2 x 3 + 5 = 12 + 3 - 6 + 5 = 14, and 2 x 3 + 8 = 6 + 8 = 14. The tiles show why they agree at every input: two x strips cancel two -x strips, so 4x - 2x = (4 - 2)x = 2x, and 3 + 5 = 8 units remain.

Use the idea

After simplifying, test the old and new forms at one or two inputs. A single mismatch proves an error; a match is only a check, not a proof.

Where the conclusion applies

x may be any number. The proof that like terms combine is distribution, 4x - 2x = (4 - 2)x, not the bars. At x = 0 every term with x vanishes, which is why the wrong forms appear to work there.

Check your understanding: At x = 3, what do 2x + 3x² and 5x give?
2 x 3 + 3 x 9 = 6 + 27 = 33, while 5 x 3 = 15. They differ, so 2x + 3x² is not 5x.

Chapter 13 source: section "Terms and coefficients identify what can combine". Demonstration C13-D02.

Demonstration 3 of 4

Four regions in one rectangle

Why does (x + 2)(x + 5) have a middle term 7x, and what does x² + 10 leave out?

Reading across the whole rectangle gives (x + 2)(x + 5). Adding its four regions gives x² + 5x + 2x + 10. The two side strips are the cross-products that combine into 7x.

\[(x+2)(x+5)\]

\[x^2+7x+10\]

\[x^2+10\]

x is a positive length for the picture. The width is x + 5 and the height is x + 2. Each region's area is one width part times one height part.

Predict first. At x = 1, how much area does x² + 10 lose compared with the whole rectangle?

Choose an example

Figure: Four regions in one rectangle. A rectangle 6 wide and 3 tall split into four regions with areas 1, 5, 2 and 10.
Length x: 1, Which expanded form?: x² + 7x + 10
Constructed example: the chapter's product (x + 2)(x + 5) drawn at whole-number x.

Calculated values

x
1
(x + 2)(x + 5)
3 x 6 = 18
x² + 7x + 10
1 + 7 + 10 = 18
x² + 10
1 + 10 = 11
Missing from x² + 10
5x + 2x = 5 + 2 = 7

At x = 1, the whole rectangle is (x + 2)(x + 5) = 3 x 6 = 18. Adding the four regions: 1 + 5 + 2 + 10 = 18. Both ways measure the same rectangle, so (x + 2)(x + 5) = x² + 7x + 10.

Use the idea

When multiplying two sums, list every pairing: each term of one group meets each term of the other. Four products before anything combines.

Where the conclusion applies

Positive x is needed only for the picture. Distribution proves the identity for every real x. The shortcut x² + 10 fails at every positive x because it drops 7x.

Check your understanding: At x = 2, what is (x + 2)(x + 5), and how much does x² + 10 miss?
4 x 7 = 28. The proposal gives 4 + 10 = 14, missing 5 x 2 + 2 x 2 = 10 + 4 = 14.

Chapter 13 source: section "Two grouped factors require every cross-product". Demonstration C13-D03.

Demonstration 4 of 4

A fixed fee and a rate

In the example cost rule, which part stays put and which part moves with the count?

The navy part of the bar is the fixed fee, the same at every count. The teal part is 1.50 dollars times the count. Dots mark the named cost at each allowed count.

\[C(n) = \$4 + (\$1.50/\text{item})n\]

\[C(6) = \$4 + \$9 = \$13\]

n is the number of items, a nonnegative whole number. $4 is the fixed fee. $1.50/item is the rate, dollars per item. C(n) names the cost at input n; C does not multiply anything.

Predict first. Going from 2 items to 6 items, by how many dollars should the cost rise?

Choose an example

Figure: A fixed fee and a rate. Dots show the cost at 0 through 8 items. At 6 items a bar splits into the fixed $4 and the per-item $9, totaling $13.
Number of items n: 6
Constructed example: the chapter's example service with a $4 fee and $1.50 per item.

Calculated values

Items n
6
Fixed term
$4
Per-item term
$1.50 x 6 = $9
Named cost
C(6) = $13

By hand: 4 + 1.50 x 6 = 4 + 9 = 13. With units, C(6) = $4 + ($1.50/item)(6 items) = $4 + $9 = $13. The item unit cancels, so dollars are added to dollars. The fixed $4 stays the same at every count; only the per-item term moves.

Use the idea

Any price of the form "a fixed amount plus so much per item" splits this way. Carry the units: dollars per item times items leaves dollars.

Where the conclusion applies

An example service, not a real price. Counts are whole numbers from 0. At n = -3 the arithmetic still runs, but the result is not a cost in this model.

Check your understanding: What is C(2), and how much more is C(6)?
C(2) = $4 + $1.50 x 2 = $4 + $3 = $7. C(6) - C(2) = $13 - $7 = $6, which is $1.50 x 4 items.

Chapter 13 source: section "Worked examples: expansion and a cost rule". Demonstration C13-D04.