Math 6 · 6.EE.A.2 · 6.EE.A.2C

Evaluating expressions and formulas

An expression such as 6d + 15 has a letter in it. Once you are told which number the letter stands for, how do you find the value of the expression?

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Evaluating expressions and formulas

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When you reach for it

An expression with a letter in it, such as 6d + 15, is a calculation waiting for one number. The letter marks the place where that number goes (what a letter stands for). Until the number is known, the expression has no single value; once it is known, it has exactly one. Putting the number in the letter’s place is called substituting, and working out the result is called evaluating the expression.

You reach for it whenever an expression or a formula has to give an actual number:

a rule for pay, cost or time, written in letters from words, and one particular case of it: how much for 8 deliveries?
a formula for a shape, such as A = lw for the area of a rectangle, and the measurements of one real rectangle;
a table of values, which is one expression evaluated again and again, once for each number in the top row.

The lesson runs over two days. Part 1 puts one number into one letter, including a letter under a power. Part 2 uses formulas, expressions with two letters, and tables.

The procedure

Part 1: one letter, one number

Write the expression again with the number in place of the letter. Take 6d + 15 when d = 7. The letter d stood for a number that was not yet known; now it is known, so the expression with 7 in it describes the same quantity:

6d + 15 \quad\text{becomes}\quad 6(7) + 15

Nothing else changes. Every operation stays where it was, and only the letter is swapped out.

Put a bracket round the number when it meets a number or a power. In algebra, 6d means 6 \times d: a number written straight next to a letter multiplies it. Two digits written next to each other mean something else entirely: 67 is sixty-seven, not 6 \times 7. So 6d at d = 7 is written 6(7) or 6 \times 7, never 67. A number written against a bracket multiplies what is inside it, so 6(7) = 42.

The bracket matters just as much under a power. In n^2 the exponent says how many factors of the base to multiply (Base, exponent and the value of a power). When \displaystyle n = \frac{2}{5}, the whole of \displaystyle \frac{2}{5} is the base, so write \displaystyle \left(\frac{2}{5}\right)^2; without the bracket, \displaystyle \frac{2}{5^2} squares only the 5. A decimal is treated the same way: n^2 at n = 0.3 is (0.3)^2.

A bracket round a single number never changes that number. It only keeps it in one piece, so that the operations around it act on all of it.

Use the same number every time the letter appears. In w(10 - w) the letter turns up twice, and within one expression a letter stands for one number. If w = 2.5, both places get 2.5.

Then work it out in the usual order. Once the letter has gone, the expression is plain arithmetic, done in the agreed order: grouping first, then powers, then multiplying and dividing from left to right, then adding and subtracting from left to right (Order of operations with powers). In 6(7) + 15 the multiplication comes before the addition:

6(7) + 15 = 42 + 15 = 57

The same order explains why 2x^2 and (2x)^2 are different expressions. In 2x^2 the power acts before the multiplication, so only x is squared and the 2 multiplies afterwards. In (2x)^2 the bracket makes 2x the base, so the 2 is squared as well. Example 3 puts numbers to this.

A different number for the letter usually gives a different value, and that is the point of the letter: one expression covers every case, and substituting picks out one of them. Example 1 works three cases of the same rule.

Part 2: formulas, several letters and tables

A formula is an expression whose result has a name. P = 2l + 2w says that the perimeter P of a rectangle comes from its length l and width w (Perimeter); A = lw gives its area (Area of rectangles and squares); V = lwh gives the volume of a box with length l, width w and height h (Volume of a box). A cube with edge s has volume V = s^3 and surface area A = 6s^2. The lesson on speed, distance and time said in words that distance is speed times time; as a formula, an object moving at a steady speed r for a time t covers the distance d = rt. Evaluating a formula means evaluating the expression on its right side; the letter on the left names what comes out.

Figure 1The letters in P = 2l + 2w and A = lw name the sides of the rectangle; the letters in V = lwh name the three edges of the box.

Each letter gets its own number, in every place it appears. Match each number to its letter by what it measures. In A = lw the order makes no difference, because multiplication is commutative, but in a pay rule such as 4n + 0.75d, where n counts deliveries and d counts kilometers, putting the numbers in the wrong places gives a different answer.

Units come along with the numbers. The measurements must be in the same unit before they go in, and the answer’s unit comes from the formula: centimeters times centimeters is square centimeters, and three lengths multiplied give cubic centimeters. Kilometers per hour times hours gives kilometers.

The result of one formula can be the number for another. If a pay rule needs a distance and only the speed and time are known, d = rt supplies the distance first. Example 6 does this.

A table of values evaluates one expression at many numbers. The top row lists the numbers for the letter, the row below lists the values. Reading along the row shows how the value changes as the number changes, which a single substitution cannot show.

Worked examples

Examples 1 to 4 go with Part 1, and Examples 5 to 7 with Part 2.

Example 1

Pay for a day of deliveries

Rina works as a bike courier. Each day she is paid 15 dollars for being on call, plus 6 dollars for each delivery, so her pay in dollars for a day with d deliveries is

6d + 15

Find her pay on a day with 8 deliveries, on a day with 14, and on a day with none.

Solution

Put each number where the d is, with a bracket so that the 6 multiplies it, then multiply before adding.

\begin{aligned} d = 8:&\quad 6(8) + 15 = 48 + 15 = 63 \\ d = 14:&\quad 6(14) + 15 = 84 + 15 = 99 \\ d = 0:&\quad 6(0) + 15 = 0 + 15 = 15 \end{aligned}

She earns 63 dollars, 99 dollars and 15 dollars. The day with no deliveries shows what the 15 in the rule means: it is the pay before any delivery is made. Written without the bracket, the first case would have read 68 + 15, with the 6 and the 8 run together into sixty-eight, and that is not what the rule says.

Example 2

Fractions and decimals

Evaluate (a) 5x - 2 when x = 1.6 \qquad (b) w(10 - w) when w = 2.5 \qquad (c) \dfrac{k}{4} + k when k = \dfrac{2}{3}

Solution

(a) The multiplication comes before the subtraction. Five lots of 1.6 is 8:

5(1.6) - 2 = 8 - 2 = 6

(b) The letter appears twice, and both places get 2.5. The bracket is worked first:

2.5(10 - 2.5) = 2.5(7.5) = 18.75

To multiply, 2 \times 7.5 = 15 and half of 7.5 is 3.75, so 2.5 \times 7.5 = 18.75.

(c) The fraction bar in \displaystyle \frac{k}{4} means divide by 4, and dividing by 4 is the same as multiplying by \displaystyle \frac{1}{4}, so \displaystyle \frac{2}{3} \div 4 = \frac{2}{12} = \frac{1}{6}. The division comes before the addition:

\frac{2}{3} \div 4 + \frac{2}{3} = \frac{1}{6} + \frac{4}{6} = \frac{5}{6}

Writing \displaystyle \frac{2}{3} as \displaystyle \frac{4}{6} gives the two fractions the same denominator, so they can be added.

Example 3

Powers

(a) Evaluate s^3 when s = 6, and when s = 0.3.

(b) Evaluate 2x^2 and (2x)^2 when x = 4.

Solution

(a) The exponent 3 asks for three factors of the base:

\begin{aligned} 6^3 &= 6 \times 6 \times 6 = 216 \\ (0.3)^3 &= 0.3 \times 0.3 \times 0.3 = 0.027 \end{aligned}

For the decimal, 0.3 \times 0.3 = 0.09 and 0.09 \times 0.3 = 0.027.

(b) In 2x^2 the power comes first and belongs to the 4 alone; the bracket also stops the 2 and the 4 from running together into 24:

2(4)^2 = 2 \times 16 = 32

In (2x)^2 the bracket comes first, so the whole product 2 \times 4 is the base:

(2 \times 4)^2 = 8^2 = 64

The values differ because the expressions hold different factors. 2x^2 is 2 \times x \times x, with one factor of 2. (2x)^2 is 2x \times 2x, with two factors of 2, which is why it came out twice as large here.

\left(\frac{2}{5}\right)^2 = \frac{2}{5} \times \frac{2}{5} = \frac{4}{25}

Without the bracket, \displaystyle \frac{2}{5^2} would mean \displaystyle \frac{2}{25}, with only the 5 squared.

Example 4

Baking time in a pizza oven

The guide for Selma’s pizza oven gives the baking time, in minutes, for a pizza d centimeters across as

\frac{d^2}{100} + 3

(a) How long does a pizza 30 centimeters across take? (b) How long does a 25 centimeter pizza take? (c) How much longer does the 30 centimeter pizza take than a 20 centimeter one?

Solution

The fraction bar groups everything above it, so the top is worked out first: square d, then divide by 100, then add 3.

(a) \dfrac{30^2}{100} + 3 = \dfrac{900}{100} + 3 = 9 + 3 = 12 minutes.

(b) \dfrac{25^2}{100} + 3 = \dfrac{625}{100} + 3 = 6.25 + 3 = 9.25 minutes. A quarter of a minute is 15 seconds, so this is 9 minutes and 15 seconds.

The 30 centimeter pizza is 1.5 times as wide as the 20 centimeter one, but its time is not 1.5 \times 7 = 10.5 minutes. The width is squared in the rule, and the 3 minutes are added whatever the size, so the times do not grow in step with the widths.

Example 5

A tray and a box

Selma’s baking tray is a rectangle 42 cm long and 30.5 cm wide. Find its perimeter and its area. Her pizza boxes are 35 cm long, 35 cm wide and 4 cm high. Find the volume of one box.

Solution

For the perimeter, put l = 42 and w = 30.5 into P = 2l + 2w. Each multiplication comes before the addition:

P = 2(42) + 2(30.5) = 84 + 61 = 145

The perimeter is 145 cm. For the area, A = lw = 42(30.5). Split the decimal: 42 \times 30 = 1260 and 42 \times 0.5 = 21, so A = 1281. The unit is centimeters times centimeters, so the area is 1281 square centimeters, 1281 \text{ cm}^2.

For the box, put l = 35, w = 35 and h = 4 into V = lwh:

V = 35(35)(4)

The factors may be multiplied in any order (the commutative and associative properties), and 35 \times 4 = 140 is the easier start: 140 \times 35 = 4900. The volume is 4900 \text{ cm}^3.

Example 6

Distance, then pay

Linus rides for a courier firm that pays 4 dollars for each delivery and 0.75 dollars for each kilometer ridden, so his pay in dollars for n deliveries over d kilometers is 4n + 0.75d. One afternoon he rides at a steady 16 kilometers per hour for 2.5 hours and makes 13 deliveries. What is he paid?

Solution

The pay rule needs d, which is not given directly. The distance formula d = rt supplies it, with r = 16 and t = 2.5:

d = 16(2.5) = 40

Kilometers per hour times hours gives kilometers, so he rides 40 km. Now both letters of the pay rule have numbers, n = 13 and d = 40:

4(13) + 0.75(40) = 52 + 30 = 82

He is paid 82 dollars. The 13 went to n because n counts deliveries, and the 40 to d because d counts kilometers. Placed the other way round, the same arithmetic would give 4(40) + 0.75(13) = 169.75, which is the pay for 40 deliveries over 13 kilometers, a different afternoon.

Example 7

Two price plans in a table

Zola’s dance studio offers two plans. Plan A has a 25 dollar sign-up fee and then costs 9 dollars a class, so c classes cost 9c + 25 dollars. Plan B has no fee and costs 14 dollars a class, so c classes cost 14c dollars. Make a table of both costs for c = 1, 3, 5, 7, 9, and use it to say when each plan is cheaper.

Solution

Each cell is one substitution. For c = 3, Plan A costs 9(3) + 25 = 27 + 25 = 52 and Plan B costs 14(3) = 42. The others go the same way:

Classes c	1	3	5	7	9
Plan A, 9c + 25	34	52	70	88	106
Plan B, 14c	14	42	70	98	126

Plan B is cheaper for 1 and 3 classes, the two cost the same 70 dollars at 5, and Plan A is cheaper for 7 and 9.

The table also shows why. Each extra class adds 9 dollars to Plan A and 14 dollars to Plan B, so Plan B closes 5 dollars of the gap with every class. It starts 25 dollars cheaper, since with no classes Plan A already costs the 25 dollar fee, and 25 \div 5 = 5 classes use that lead up. So Plan B is cheaper for fewer than 5 classes and Plan A for more than 5. A number not in the table agrees: for 4 classes Plan A costs 9(4) + 25 = 61 and Plan B costs 14(4) = 56.

Boundaries

The number has to make sense for the letter. Rina’s d counts deliveries, so it is a whole number; 6d + 15 at d = 2.5 is good arithmetic but describes no day of hers. Selma’s oven rule is meant for pizzas that fit in the oven; at d = 0 it gives 3 minutes for a pizza that does not exist. Substituting always produces a number, and the setting decides whether that number means anything.

Measurements go in with the same unit. For a rectangle 2 m long and 50 cm wide, change 2 m to 200 cm first: P = 2(200) + 2(50) = 500 cm. Putting in 2 and 50 as they stand gives 104, which is neither meters nor centimeters.

Some numbers give no value. Division by zero has no answer, so \dfrac{12}{y} has no value when y = 0. Every other number gives it a value.

A formula belongs to its shape. P = 2l + 2w and A = lw are for rectangles, and V = lwh is for boxes whose faces are all rectangles. Putting a triangle’s sides into them gives a number, but not the triangle’s perimeter or area.

Numbers that are not negative. This lesson substitutes whole numbers, fractions and decimals that are not negative. Negative numbers are substituted in exactly the same way, with a bracket round them, but the arithmetic then needs signed numbers: see Order of operations with negative numbers.

One substitution tells you about one number. At x = 2, both x + x and x^2 equal 4, but at x = 3 they are 6 and 9. Two expressions that agree at one number can disagree at the next, so a single substitution never shows that two expressions are the same. Showing that they agree for every number is the work of equivalent expressions.

Where this goes next

Evaluating is the first thing done with almost any expression, so it turns up nearly everywhere after this. An expression has a value, while an equation claims that two values are equal (Expression, equation and inequality); putting a number into both sides of an equation and seeing whether they come out equal is how a solution is checked.

A table of values becomes a picture once its pairs are plotted as points, which is graphing from a table. Writing the rule as f(x) and the value at 3 as f(3) is evaluating a function, the same move under a new name. And many formulas met later are evaluated exactly as here, among them simple interest, I = Prt, and the rule for converting temperatures.

Exercises

Part 1: one letter

Evaluate each expression for the value given.

7m + 3 when m = 6
40 - 3k when k = 11
9 + 2t when t = 0
\dfrac{a}{5} + 6 when a = 35

Part 1: fractions and decimals

4p + 1.5 when p = 2.25
6x - 1 when x = \dfrac{2}{3}
y(y + 2) when y = 0.5
\dfrac{10}{q} + q when q = \dfrac{1}{2}

Part 1: powers

b^2 + 5 when b = 7
s^3 when s = 4, and when s = 0.1
3x^2 and (3x)^2 when x = 2
(n - 1)^3 + n when n = 3
Zola orders costumes for a show. For k dancers the order costs 18k + 30 dollars. What does it cost for 7 dancers?

Part 2: formulas

A square tile has side s = 7.5 cm. Use P = 4s to find its perimeter.
The floor of Zola’s main studio is a rectangle 14 m long and 9.5 m wide. Find its perimeter, using P = 2l + 2w, and its area, using A = lw.
Selma lets dough rise in a cube-shaped box with edge s = 0.5 m. Find its volume, V = s^3, and its surface area, A = 6s^2.
Dora’s courier bag is a box 40 cm long, 35 cm wide and 30 cm high. Use V = lwh to find its volume.
Dora rides at 15 kilometers per hour. Use d = rt to find how far she goes in t = \dfrac{1}{3} hour, and in t = 2.4 hours.

Part 2: several letters and tables

3a + 2b when a = 5 and b = 1.5
x^2 + xy when x = 4 and y = 2.5
(m + n)^2 and m^2 + n^2 when m = 6 and n = 1
Selma’s oven takes 18 minutes to heat up, and then each batch of pizzas takes 7.5 minutes, so b batches take 18 + 7.5b minutes from a cold start. Make a table of the times for b = 0, 2, 4, 6. By how much does the time grow from each column to the next?
A courier firm pays Rina 3.5 dollars for each delivery and 0.4 dollars for each kilometer, so her pay is 3.5n + 0.4d dollars for n deliveries over d kilometers. What is she paid for 18 deliveries over 45 kilometers?

Reasoning

Explain why 5n at n = 3 is written 5(3), and why writing 53 instead gives the wrong value.
Find the values of 2x^2 and (2x)^2 when x = 5. Explain, using what a power means, why they differ, and find the one number x for which they are equal.

Answers

45. 7(6) + 3 = 42 + 3.
7. 40 - 3(11) = 40 - 33.
9. 9 + 2(0) = 9 + 0.
13. 35 \div 5 = 7, and 7 + 6 = 13.
10.5. 4(2.25) + 1.5 = 9 + 1.5.
3. \displaystyle 6\left(\frac{2}{3}\right) - 1 = 4 - 1.
1.25. 0.5(0.5 + 2) = 0.5(2.5).
\displaystyle \frac{41}{2}, that is \displaystyle 20\frac{1}{2}. There are 20 halves in 10, so \displaystyle 10 \div \frac{1}{2} = 20, and then \displaystyle 20 + \frac{1}{2}.
54. 7^2 + 5 = 49 + 5.
64 and 0.001. 4 \times 4 \times 4 = 64, and (0.1)^3 = 0.1 \times 0.1 \times 0.1 = 0.001.
12 and 36. 3(2)^2 = 3 \times 4 = 12, while (3 \times 2)^2 = 6^2 = 36.
11. (3 - 1)^3 + 3 = 2^3 + 3 = 8 + 3.
156 dollars. 18(7) + 30 = 126 + 30.
30 cm. 4(7.5) = 30.
Perimeter 47 m, area 133 \text{ m}^2. 2(14) + 2(9.5) = 28 + 19, and 14(9.5) = 126 + 7.
Volume 0.125 \text{ m}^3, surface area 1.5 \text{ m}^2. (0.5)^3 = 0.125, and 6(0.5)^2 = 6(0.25) = 1.5.
42000 \text{ cm}^3. 40(35)(30) = 1400 \times 30.
5 km and 36 km. \displaystyle 15\left(\frac{1}{3}\right) = 5, and 15(2.4) = 36.
18. 3(5) + 2(1.5) = 15 + 3.
26. 4^2 + 4(2.5) = 16 + 10.
49 and 37. (6 + 1)^2 = 7^2 = 49, while 6^2 + 1^2 = 36 + 1 = 37.
The times are 18, 33, 48 and 63 minutes, and each column is 15 minutes more than the one before. Each column adds 2 batches, and 2(7.5) = 15.
81 dollars. 3.5(18) + 0.4(45) = 63 + 18.
A number written next to a letter multiplies it, so 5n at n = 3 is 5 \times 3 = 15, and the bracket in 5(3) shows that product. Digits written side by side make one numeral instead, so 53 is fifty-three, a different number from 15.
50 and 100; they are equal only when x = 0. 2(5)^2 = 2 \times 25 = 50, and (2 \times 5)^2 = 10^2 = 100. The power in 2x^2 belongs to x alone, so it is 2 \times x \times x, with one factor of 2. In (2x)^2 the base is 2x, so it is 2x \times 2x = 2 \times 2 \times x \times x, with two factors of 2: always twice 2x^2. A number equals twice itself only when it is 0, which happens when x = 0.