Unit 6 Plan - Grade 6

Addition tape diagrams have different-sized parts; multiplication tape diagrams have equal-sized parts
A variable like x stands for a number we don't know yet
You can build a tape diagram from an equation, or write an equation from a tape diagram

Tape diagrams can help us understand relationships between quantities and how operations describe those relationships.

Tape diagram A, 3 equal parts labeled x, x, x. Total, 21. — A

Tape diagram B, 2 parts, labeled y, 3. Total, 21. — B

Diagram A has 3 parts that add to 21. Each part is labeled with the same letter, so we know the 3 parts are equal. Here are some equations that all represent Diagram A:

$\displaystyle x+x+x=21$

$\displaystyle 3\boldcdot {x}=21$

$\displaystyle x=21\div3$

$\displaystyle x=\frac13\boldcdot {21}$

Notice that the number 3 is in the equations, but it's not written in the diagram. The 3 comes from counting 3 boxes representing 3 equal parts in 21.

Diagram B has 2 parts that add to 21. Here are some equations that all represent Diagram B:

$\displaystyle y+3=21$

$\displaystyle y=21-3$

$\displaystyle 3=21-y$

Complete the Diagrams (1 problem)

Complete the first diagram so it represents $5 \boldcdot x = 15$ . Complete the second diagram so it represents $5 + y = 15$ .

Two blank tape diagrams. Each tape diagram totals 15.

Show Solution

Sample response:

—

Lesson 2

Truth and Equations

6.EE.5

Understand solving an equation or inequality as a process of answering a question: which values from a specified set, if any, make the equation or inequality true? Use substitution to determine whether a given number in a specified set makes an equation or inequality true.

6.EE.6

Use variables to represent numbers and write expressions when solving a real-world or mathematical problem. Understand that a variable can represent an unknown number, or, depending on the purpose at hand, any number in a specified set.

An equation can be true or false depending on the value of the variable
A coefficient is the number written next to a variable (e.g., 30 in 30x)
A solution is a value that makes the equation true — you check by substituting it in

An equation can be true or false. An example of a true equation is $7+1=4 \boldcdot 2$ . An example of a false equation is $7+1=9$ .

An equation can have a letter in it to represent a value, for example, $u+1=8$ . This equation is false if $u$ is 3, because $3+1$ does not equal 8. This equation is true if $u$ is 7, because $7+1=8$ .

A letter in an equation that represents an unknown value is called a variable. In $u+1=8$ , the variable is $u$ . A number that can be used in place of the variable that makes the equation true is called a solution to the equation. In $u+1=8$ , the solution is 7.

When a number is written next to a variable, it means the number and the variable are being multiplied. For example, $7x=21$ means the same thing as $7 \boldcdot x = 21$ . A number written next to a variable is called a coefficient. If no coefficient is written, the coefficient is 1. For example, in the equation $p+3=5$ , the coefficient of $p$ is 1.

How Do You Know a Solution Is a Solution? (1 problem)

Explain how you know that 88 is a solution to the equation $\frac18 x = 11$ by completing the sentences:

The word “solution” means . . .

88 is a solution to $\frac18x = 11$ because . . .

Show Solution

Sample responses:

The word “solution” means a value for the variable that makes the equation true.

88 is a solution to $\frac18x = 11$ , because if $x$ is 88, the equation is $\frac18 \boldcdot 88=11$ , which is true.

—

Lesson 3

Staying in Balance

6.EE.5

Understand solving an equation or inequality as a process of answering a question: which values from a specified set, if any, make the equation or inequality true? Use substitution to determine whether a given number in a specified set makes an equation or inequality true.

6.EE.6

Use variables to represent numbers and write expressions when solving a real-world or mathematical problem. Understand that a variable can represent an unknown number, or, depending on the purpose at hand, any number in a specified set.

6.EE.7

Solve real-world and mathematical problems by writing and solving equations of the form x + p = q; x - p = q; px = q; and x/p = q for cases in which p, q, and x are all nonnegative rational numbers.

A balanced hanger means both sides weigh the same, just like a true equation
You can write an equation to represent a balanced hanger diagram
To find the unknown, reason about what keeps both sides equal

A hanger stays balanced when the weights on both sides are equal. We can change the weights and the hanger will stay balanced as long as both sides are changed in the same way. For example, adding 2 pounds to each side of a balanced hanger will keep it balanced. Removing half of the weight from each side will also keep it balanced.

An equation can be compared to a balanced hanger. We can change the equation, but for a true equation to remain true, the same thing must be done to both sides of the equal sign. If we add or subtract the same number on each side, or multiply or divide each side by the same number, the new equation will still be true.

This way of thinking can help us find solutions to equations. Instead of checking different values for the variable, we can think about subtracting the same amount from each side or dividing each side by the same number.

Balanced Hanger, left side, 3 identical squares, x, right side, 1 rectangle, 11. — A

Balanced Hanger, , left side 1 rectangle, 11, right side, 1 triangle, y, and 1 circle, 5. — B

Diagram A can be represented by the equation $3x=11$ .

If we break the 11 into 3 equal parts, each part will have the same weight as 1 block with an $x$ .

Splitting each side of the diagram into 3 equal parts is the same as dividing each side of the equation by 3.

$3x$ divided by 3 is $x$ .
11 divided by 3 is $\frac{11}{3}$ .
If $3x=11$ is true, then $x=\frac{11}{3}$ is true.
The solution to $3x=11$ is $\frac{11}{3}$ .

Diagram B can be represented with the equation $11=y+5$ .

If we remove a weight of 5 from each side of the diagram, it will stay in balance.

Removing 5 from each side of the diagram is the same as subtracting 5 from each side of the equation.

$11-5$ is 6.
$y+5-5$ is $y$ .
If $11=y+5$ is true, then $6=y$ is true.
The solution to $11=y+5$ is 6.

Weight of the Circle (1 problem)

Here is a balanced hanger diagram.

Balanced hanger. Left side, 4 identical circles, w. Right side, 1 rectangle, 25.

Write an equation that represents the diagram.
Find the weight of one circle. Explain or show your reasoning.
What is the solution to your equation?

Show Solution

$4w=25$
$\frac{25}{4}$ or $6\frac14$ units. Sample reasoning: The left side of the diagram has 4 circles, so I divided the right side into 4 equal pieces. Each of those pieces weighs $6\frac14$ units. This shows that each circle piece weighs $6\frac14$ units.
$w=\frac{25}{4}$ or $6\frac14$

—

Lesson 4

Practice Solving Equations

6.EE.5

Understand solving an equation or inequality as a process of answering a question: which values from a specified set, if any, make the equation or inequality true? Use substitution to determine whether a given number in a specified set makes an equation or inequality true.

6.EE.7

Solve real-world and mathematical problems by writing and solving equations of the form x + p = q; x - p = q; px = q; and x/p = q for cases in which p, q, and x are all nonnegative rational numbers.

6.EE.B

No additional information available.

6.NS.3

Fluently add, subtract, multiply, and divide multi-digit decimals using a standard algorithm for each operation.

Addition and subtraction are inverse operations; multiplication and division are inverse operations
To solve x + p = q, subtract p from both sides; to solve px = q, divide both sides by p
This strategy works with whole numbers, fractions, decimals, and mixed numbers

When we solve an equation with a variable, we find the value for the variable that makes the equation true. One way to solve the equation is to do the same thing to each side until the variable is alone on one side of the equal sign, and see what is on the other side.

Solve the equation $x+\frac{3}{4}=\frac{7}{8}$ .

The fraction

\frac{3}{4}

is added to the variable

x

.

$\displaystyle x+\frac{3}{4}=\frac{7}{8}$

So, we can subtract

\frac{3}{4}

from each side of the equation.

$\displaystyle x+\frac{3}{4}-\frac{3}{4}=\frac{7}{8}-\frac{3}{4}$

The variable is alone on one side of the equal sign, and

\frac{1}{8}

is on the other side.

$\displaystyle x=\frac{1}{8}$

When we substitute

\frac{1}{8}

for

x

in the original equation, the equation is true. So, we know

\frac{1}{8}

is the solution.

$\begin{aligned}\displaystyle \frac{1}{8} +\frac{3}{4} &= \frac{7}{8} \\ \displaystyle \frac{7}{8} &= \frac{7}{8}\end{aligned}$

Solve the equation

3.5x=31.5

.

The variable

x

is multiplied by 3.5.

$\displaystyle 3.5x=31.5$

So, we can divide each side of the equation by 3.5.

$\displaystyle 3.5x \div 3.5=31.5 \div 3.5$

The variable is alone on one side of the equal sign, and 9 is on the other side.

$\displaystyle x = 9$

When we substitute 9 for

x

in the original equation, the equation is true. So, we know 9 is the solution.

$\begin{aligned}\displaystyle 3.5(9)&=31.5 \\ \displaystyle 31.5 &= 31.5\end{aligned}$

Solve It! (1 problem)

Solve each equation. Explain or show your reasoning.

$x+1\frac34=10$ .
$5.7x = 17.1$
$\frac{1}{10}x=\frac25$

Show Solution

Sample responses:

$x=8\frac14$ . Sample reasoning: $x+1\frac34-1\frac34=10-1\frac34$
$x=3$ . Sample reasoning: $5.7x\div5.7=17.1\div5.7$
$x=4$ . Sample reasoning: I divided both sides of the equation by $\frac{1}{10}$ .

—

Lesson 5

Represent Situations with Equations

6.EE.5

Understand solving an equation or inequality as a process of answering a question: which values from a specified set, if any, make the equation or inequality true? Use substitution to determine whether a given number in a specified set makes an equation or inequality true.

6.EE.6

Use variables to represent numbers and write expressions when solving a real-world or mathematical problem. Understand that a variable can represent an unknown number, or, depending on the purpose at hand, any number in a specified set.

6.EE.7

Solve real-world and mathematical problems by writing and solving equations of the form x + p = q; x - p = q; px = q; and x/p = q for cases in which p, q, and x are all nonnegative rational numbers.

Identify the unknown, the total, and the operation to write an equation for a word problem
Solve the equation, then explain what the answer means in the real-world situation

Writing and solving equations can help us answer questions about situations.

A scientist has 13.68 liters of oil and needs 16.05 liters for an experiment. How many more liters of oil does she need for the experiment?

We can represent this situation with the equation:

$\displaystyle 13.68 + x=16.05$

We can solve the equation by subtracting 13.68 from each side. This gives us some new equations that also represent the situation:

$\displaystyle x=16.05 - 13.68$

$\displaystyle x=2.37$

The solution $x=2.37$ means the scientist needs 2.37 more liters of oil.

Volunteers at a food pantry divide a 54-pound bag into portions that each weigh $\frac{3}{4}$ pound. How many portions can they make?

We can represent this situation with the equation:

$\displaystyle \frac34 x = 54$

We can find the value of $x$ by dividing each side by $\frac34$ . This gives us some new equations that represent the same situation:

$\displaystyle x=54\div \frac34$

$\displaystyle x=72$

The solution $x=72$ means the volunteers can make 72 portions.

More Storytime (1 problem)

For each situation:

Choose an equation that represents it.
Solve the equation.
Explain what the solution means in the situation.

Lin needs 10 cups of flour for a bread recipe. She only has $2\frac12$ cups. How much more flour does she need?
- $x+2\frac12=10$
- $x=10+2\frac12$
- $2\frac12x=10$
Each notebook costs 5.70. How many notebooks does Diego buy if he spends a total of 17.10?
- $x+5.7=17.1$
- $5.7x=17.1$
- $17.1x=5.7$

Show Solution

$x+2\frac12=10$ , $x=7\frac12$ . Lin needs $7\frac12$ cups of flour.
$5.7x=17.1$ , $x=3$ . Diego buys 3 notebooks.

—

Lesson 6

Percentages and Equations

6.EE.6

Use variables to represent numbers and write expressions when solving a real-world or mathematical problem. Understand that a variable can represent an unknown number, or, depending on the purpose at hand, any number in a specified set.

6.EE.7

Solve real-world and mathematical problems by writing and solving equations of the form x + p = q; x - p = q; px = q; and x/p = q for cases in which p, q, and x are all nonnegative rational numbers.

6.RP.3.c

Use ratio and rate reasoning to solve real-world and mathematical problems.

" $A\%$ of $B$ is $C$ " can be written as the equation $\frac{A}{100} \cdot B = C$
To find the whole when you know a percent and its value, solve $px = q$ by dividing both sides by $p$
You can check your answer by substituting it back into the equation

We can write equations to help us solve percentage problems.

Example: There are 455 students in school today, which is 70% school attendance. How many students go to the school?

The number of students in school today is known in two different ways: as 70% of the students in the school, and also as 455. If $s$ represents the total number of students who go to the school, then 70% of $s$ , or $\frac{70}{100}s$ , represents the number of students that are in school today, which is 455.

We can write and solve the equation:

$\begin{aligned} \frac{70}{100} s &= 455 \\ s &= 455 \div \frac{70}{100} \\ s &= 455 \boldcdot \frac{100}{70}\\ \end{aligned}$

There are 650 students in the school.

$s = 650$

The equation can also be written using the decimal equivalent of $\frac{70}{100}$ , which is 0.7:

$\begin{aligned} 0.7s &= 455 \\ s &= 455 \div 0.7\\ s & = 650 \end{aligned}$

We can check this answer by substituting 650 for $x$ in the equation and seeing if the equation is true.

$\displaystyle \begin{aligned} 0.7x&=455\\0.7(650)&=455\\ 455&=455\end{aligned}$

Fundraising for the Animal Shelter (1 problem)

Noah raised $54 to support the animal shelter, which is 60% of his fundraising goal. What is Noah’s fundraising goal?

Write an equation with a variable to represent the situation.
Answer the question. Show or explain your reasoning.

Show Solution

$54=\frac{60}{100}x$ (or equivalent)
$90. Sample reasoning:
- Divide both sides of the equation by $\frac{60}{100}$ or $\frac35$ to get $x=54\boldcdot \frac53=90$ .
- I wrote $0.6x=54$ and thought about six-tenths of what number would be equal to 54. $6 \boldcdot 9 = 54$ , so 0.6 needs to be multiplied by 90 to get 54.

—

Section A Check

Section A Checkpoint

Problem 1

Answer each question, and explain or show your reasoning.

Is $\frac38$ a solution to $y+\frac12=\frac78$ ?
Is 15 a solution to $0.4x=20$ ?

Show Solution

Yes. Sample reasoning: Substituting $\frac38$ for $y$ gives a true equation, $\frac38+\frac12=\frac78$ .
No. Sample reasoning: Substituting 15 for $x$ gives the equation $(0.4) \boldcdot 15= 20$ , which is false.

Problem 2

Solve each equation, and explain or show your reasoning.

$a + 123 = 468$
$2.5b=20.5$

Show Solution

$a=345$ . Sample reasoning: Subtract $123$ from each side.
$b=8.2$ . Sample reasoning: Divide each side by $2.5$ .

Lesson 7

Write Expressions with Variables

6.EE.2

Write, read, and evaluate expressions in which letters stand for numbers.

6.EE.2.a

Write, read, and evaluate expressions in which letters stand for numbers.

6.EE.6

Use variables to represent numbers and write expressions when solving a real-world or mathematical problem. Understand that a variable can represent an unknown number, or, depending on the purpose at hand, any number in a specified set.

A variable lets you write one expression that works for many different numbers
To write an expression, figure out what calculation you repeat and replace the changing number with a variable
You can use the expression to find unknown values by writing and solving an equation

Suppose you were born on the same day as your neighbor, but she is 3 years older than you. When you were 1, she was 4. When you were 9, she was 12. When you are 42, she will be 45.

If we let $a$ represent your age at any time, your neighbor’s age can be expressed $a+3$ .

your age	1	9	42	$a$
neighbor's age	4	12	45	$a+3$

We often use a variable, such as $x$ or $a$ , as a placeholder for a number in expressions. Variables make it possible to write expressions that represent a calculation even when we don't know all the numbers in the calculation.

How old will you be when your neighbor is 32? We know your neighbor is 32. We also know your neighbor’s age is your age plus 3, or $a+3$ . We can write the equation $a+3=32$ to represent these relationships. When your neighbor is 32 you will be 29, because $a+3=32$ is true when $a$ is 29.

Growth (1 problem)

A plant measured $x$ inches tall last week and 8 inches tall this week.

Circle the expression that represents the number of inches the plant grew this week. Explain how you know.
- $x-8$
- $8-x$
Each tree needs 1.2 liters of water. Write an expression that represents the amount of water needed for $n$ trees.

Show Solution

$8-x$ . Sample reasoning: Since the plant grew taller this week, 8 is greater than $x$ . The difference of 8 and $x$ is the amount that the plant grew.
$1.2n$ (or equivalent)

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Lesson 8

Equal and Equivalent

6.EE.4

Identify when two expressions are equivalent.

Two expressions can be equal for one value of a variable but not others
Equivalent expressions are always equal, no matter what value the variable takes
Properties like commutative (order doesn't matter) and the meaning of operations let us know expressions are equivalent

We can use tape diagrams to see when expressions are equal. For example, the expressions $x+9$ and $4x$ are equal when $x$ is 3, but they are not equal for other values of $x$ .

8 tape diagrams of various sizes and various labels on a grid. Each diagram is matched with an expression.

Sometimes two expressions are equal for only one particular value of their variable. Other times, they seem to be equal no matter what the value of the variable.

Expressions that are always equal for the same value of their variable are called equivalent expressions. However, it would be impossible to test every possible value of the variable. How can we know for sure that expressions are equivalent?

We can use the meaning of operations and properties of operations to know that expressions are equivalent. Here are some examples:

$x+3$ is equivalent to $3+x$ because of the commutative property of addition. The order of the values being added doesn’t affect the sum.
$4\boldcdot {y}$ is equivalent to $y\boldcdot 4$ because of the commutative property of multiplication. The order of the factors doesn’t affect the product.
$a+a+a+a+a$ is equivalent to $5\boldcdot {a}$ because adding 5 copies of something is the same as multiplying it by 5.
$b\div3$ is equivalent to $b \boldcdot {\frac13}$ because dividing by a number is the same as multiplying by its reciprocal.

In the coming lessons, we will see how another property, the distributive property, can show that expressions are equivalent.

Decisions about Equivalence (1 problem)

Decide if the expressions in each pair are equivalent. Explain or show how you know.

$x+x+x+x$ and $4x$
$5x$ and $x+5$

Show Solution

Equivalent. Sample reasoning: The diagrams representing these expressions would have the same length for any value of $x$ .
Not equivalent. Sample reasoning: if $x=1$ , then $5x=5$ and $x+5=6$ , so they do not have the same value.

—

Lesson 9

The Distributive Property, Part 1

6.NS.4

Find the greatest common factor of two whole numbers less than or equal to 100. Find the least common multiple of two whole numbers less than or equal to 12. Use the distributive property to express a sum of two whole numbers 1–100 with a common factor as a multiple of a sum of two whole numbers with no common factor other than 1.

The distributive property lets you break a factor into parts, multiply each part, then add the products
For example, 5 times 79 equals 5 times 70 plus 5 times 9
This works with subtraction too: 5 times 79 equals 5 times 80 minus 5 times 1

When we need to do mental calculations, we often come up with ways to make the calculation easier to do mentally.

Suppose we are grocery shopping and need to know how much it will cost to buy 5 cans of beans at 79 cents a can. We may calculate mentally in this way:

$5 \boldcdot 79\\ 5 \boldcdot (70+9)\\ 5 \boldcdot 70 + 5 \boldcdot 9\\ 350 + 45\\ 395$

When we think, “79 is the same as $70 + 9$ . I can just multiply $5 \boldcdot 70$ and $5 \boldcdot 9$ and add the products together” we are using the distributive property.

In general, when we multiply two factors, we can break up one of the factors into parts, multiply each part by the other factor, and then add the products. The result will be the same as the product of the two original factors. When we break up one of the factors and multiply the parts we are using the distributive property of multiplication.

The distributive property also works with subtraction. Here is another way to find $5 \boldcdot 79$ :

$5 \boldcdot 79\\ 5 \boldcdot (80-1)\\ 5 \boldcdot 80 - 5 \boldcdot 1\\ 400 - 5\\ 395$

Complete the Equation (1 problem)

Write a number or expression in each empty box to create true equations.

$7 \boldcdot (3+5)=\boxed{\phantom{\huge3333}}+\boxed{\phantom{\huge3333}}$
$5\boldcdot3-5\boldcdot2=\boxed{\phantom{\huge33}} \boldcdot (3-2)$

Show Solution

$7\boldcdot(3+5)=\boxed{21}+\boxed{35}$ or $7\boldcdot(3+5)=\boxed{7 \boldcdot 3}+ \boxed{7 \boldcdot 5}$ (or equivalent)
$5\boldcdot3-5\boldcdot2=\boxed{5} \boldcdot (3-2)$

—

Section B Check

Section B Checkpoint

Problem 1

Andre says that $2x + 5$ and $7x$ are equivalent expressions because they have the same value when $x$ is 1. Do you agree with Andre’s reasoning? Explain your reasoning. Use a diagram if it helps.

Show Solution

No, I do not agree. Sample reasoning: To be equivalent expressions, they need to have the same value for every value of the variable. When

x

is 0,

2x+5

is 5, but

7x

is 0. Since they do not have the same value when

x

is 0, they are not equivalent expressions.

Problem 2

Use the distributive property to write an expression that is equivalent to each expression.

$3(x+4)$
$a(3.5-2.7)$
$5y+10$

Show Solution

Sample responses:

$3x+12$
$0.8a$
$5(y+2)$

Lesson 12

Meaning of Exponents

6.EE.1

Write and evaluate numerical expressions involving whole-number exponents.

An exponent tells you how many times to multiply the base by itself
For example, 2 to the fifth power means 2 times 2 times 2 times 2 times 2
Exponent notation is a shorthand for repeated multiplication

When we write an expression like $2^n$ , we call $n$ the exponent.

If $n$ is a whole number, it tells how many factors of 2 we should multiply to find the value of the expression. For example, $2^1=2$ , and $2^5=2 \boldcdot 2 \boldcdot 2 \boldcdot 2 \boldcdot 2$ .

There are different ways to say $2^5$ . We can say “two raised to the power of five” or “two to the fifth power” or just “two to the fifth.”

More 3's (1 problem)

What is the value of the expression $3^5$ ?
Explain how to use that value to quickly find the value of $3^6$ .

Show Solution

243
Sample response: $3^6=3^5 \boldcdot 3=729$

—

Lesson 14

Evaluating Expressions with Exponents

6.EE.1

Write and evaluate numerical expressions involving whole-number exponents.

6.EE.2.c

Write, read, and evaluate expressions in which letters stand for numbers.

When an expression has exponents and other operations, evaluate the exponent first
Parentheses override this: evaluate what's inside the parentheses before applying the exponent
The full order of operations is parentheses, then exponents, then multiply/divide, then add/subtract

Exponents give us a new way to describe operations with numbers, so we need to understand how exponents work with other operations.

When we write an expression such as $6 \boldcdot 4^2$ , we want to make sure everyone agrees about how to find its value. Otherwise, some people might multiply first and others compute the exponent first, and different people would get different values for the same expression!

Earlier we saw situations in which $6 \boldcdot 4^2$ represented the surface area of a cube with edge lengths of 4 units. When computing the surface area, we compute $4^2$ first (or find the area of one face of the cube first) and then multiply the result by 6 (because the cube has 6 faces).

In many other expressions that use exponents, the part with an exponent is intended to be computed first.

To make everyone agree about the value of expressions like $6 \boldcdot 4^2$ , we follow the convention to find the value of the part of the expression with the exponent first. Here are a couple of examples:

$\begin{aligned} 6 &\boldcdot 4^2 \\ 6 &\boldcdot 16 \\ &96 \end{aligned}$

$\begin{aligned} 45 &+ 5^2 \\ 45 &+ 25 \\ &70 \end{aligned}$

If we want to communicate that 6 and 4 should be multiplied first and then squared, then we can use parentheses to group parts of the expression together:

$\begin{aligned} (6 &\boldcdot 4)^2 \\ &24^2 \\ &576 \end{aligned}$

$\begin{aligned} (45 &+ 5)^2 \\ &50^2 \\ 2,&500 \end{aligned}$

In general, to find the value of expressions, we use this order of operations:

Do any operations in parentheses.
Apply any exponents.
Multiply or divide from left to right in the expression.
Add or subtract from left to right in the expression.

Calculating Volumes (1 problem)

Jada and Noah want to find the combined volume of two gift boxes. One is shaped like a cube and the other is shaped like a rectangular prism that is not a cube. The prism has a volume of 20 cubic inches. The cube has edge lengths of 10 inches.

Jada says the total volume is 27,000 cubic inches. Noah says it is 1,020 cubic inches. Here is how each of them reasoned:

Jada's method:

$20 + 10^3$
$30^3$
$27,000$

Noah's method:

$20 + 10^3$
$20 + 1,000$
$1,020$

Do you agree with either of them? Explain your reasoning.

Show Solution

I agree with Noah. Sample reasoning: The cube has a volume of 1,000 cubic inches, and the additional 20 cubic inches from the prism makes the total volume 1,020 cubic inches. The exponent calculation comes before addition.

—

Section C Check

Section C Checkpoint

Problem 1

Decide whether each equation is true or false. Explain how you know.

$2^4=2 \boldcdot 4$
$3 \boldcdot 3^4 = 3^5$
$2 \boldcdot 3^2 = 6^2$

Show Solution

False. Sample reasoning: $16=8$ is not true.
True. Sample reasoning: Each side is $3 \boldcdot 3 \boldcdot 3 \boldcdot 3 \boldcdot 3$ .
False. Sample reasoning: $18=36$ is not true.

Problem 2

Find the value of

7+x^3

when

x

is 2.

Show Solution

15. Sample reasoning:

$7 + x^3 \\ 7 + 2^3 \\ 7 + 8 \\15$

Lesson 16

Two Related Quantities, Part 1

6.EE.9

Use variables to represent two quantities in a real-world problem that change in relationship to one another; Given a verbal context and an equation, identify the dependent variable, in terms of the other quantity, thought of as the independent variable. Analyze the relationship between the dependent and independent variables using graphs and tables, and relate these to the equation.

6.RP.3.a

Use ratio and rate reasoning to solve real-world and mathematical problems.

Two related quantities can be described with tables, equations, and graphs
The independent variable is the one you choose; the dependent variable is the one that changes based on it
You can write two equations for the same relationship, depending on which variable you solve for

Equations are very useful for representing the relationship in a set of equivalent ratios. Here is an example.

A cider recipe calls for 3 green apples for every 5 red apples. We can create a table to show some equivalent ratios.

We can see from the table that $r$ is always $\frac53$ as large as $g$ and that $g$ is always $\frac35$ as large as $r$ .

green apples ( $g$ )	red apples ( $r$ )
3	5
6	10
9	15
12	20

We can write equations to describe the relationship between $g$ and $r$ .

When we know the number of green apples and want to find the number of red apples, we can write:

$\displaystyle r=\frac53g$

In this equation, if $g$ changes, $r$ is affected by the change, so we refer to $g$ as the independent variable and $r$ as the dependent variable.

We can use this equation with any value of $g$ to find $r$ . If 270 green apples are used, then $\frac53 \boldcdot (270)$ or 450 red apples are used.

When we know the number of red apples and want to find the number of green apples, we can write:

$\displaystyle g=\frac35r$

In this equation, if $r$ changes, $g$ is affected by the change, so we refer to $r$ as the independent variable and $g$ as the dependent variable.

We can use this equation with any value of $r$ to find $g$ . If 275 red apples are used, then $\frac35 \boldcdot (275)$ or 165 green apples are used.

To help us see the relationship between the two quantities, we can also create two graphs, one graph that corresponds to each equation.

A graph that represents a ratio of two quantities. The graph has a horizontal axis labeled number of green apples and the numbers 1 through 15 are indicated, The vertical axis is labeled number of red apples and the numbers 1 through 20 are indicated. The following four points are indicated on the graph: 3 comma 5, 6 comma 10, 9 comma 15, and 12 comma 20.<br>

A graph that represents a ratio of two quantities. The graph has a horizontal axis labeled number of red apples and the numbers 1 through 15 are indicated. The vertical axis is labeled number of green apples and the numbers 1 through 20 are indicated. The following three points are indicated on the graph: 5 comma 3, 10 comma 6, and 15 comma 9.

Kitchen Cleaner (1 problem)

To remove grease from kitchen surfaces, a recipe says to use 1 cup of baking soda for every $\frac{1}{2}$ cup of water.

cups of baking soda	cups of water
1	$\frac12$
2	1
3	$\frac32$

Which graph represents the relationship between cups of baking soda and cups of water? Explain how you know.

A
Five points graphed on a coordinate plane with the origin labeled O. The horizontal axis is labeled “cups of water and the numbers 0 through 5 are indicated. The vertical axis is labeled “cups of baking soda” and the numbers 0 through 5 are indicated. The data are as follows: 1 comma one half, 2 comma 1, 3 comma 1 and one half, 4 comma 2, and 5 comma 2 and one half.

B
Five points graphed on a coordinate plane with the origin labeled O. The horizontal axis is labeled “cups of water” and the numbers 0 through 5 are indicated. The vertical axis is labeled “cups of baking soda” and the numbers 0 through 5 are indicated. The data are as follows: one half comma 1, 1 comma 2, 1 and one half comma 3, and 2 comma 4.

C
Five points graphed on a coordinate plane with the origin labeled O. The horizontal axis is labeled “cups of water” and the numbers 0 through 5 are indicated. The vertical axis is labeled “cups of baking soda” and the numbers 0 through 5 are indicated. The data are as follows: 1 comma 1, 2 comma 2, 3 comma 3, and 4 comma 4.
Select all equations that can represent the relationship between $b$ , cups of baking soda, and $w$ , cups of water, in this situation.
1. $w = \frac{1}{2}b$
2. $b=\frac{1}{2}w$
3. $b = w$
4. $b = 2w$
5. $w = 2b$

Show Solution

Graph B. Sample reasoning:
- In all graphs, the first value of the coordinates represents the amount of water. The amount of baking soda is twice the amount of water, so the coordinates of the points should be $(\frac{1}{2}, 1)$ , $(1, 2)$ , $(\frac{3}{2}, 3)$ , and so on.
- I matched the coordinates of the points to the values in the table: $\frac{1}{2}$ cup of water goes with 1 cup of baking soda, 1 cup of water goes with 2 cups of baking soda, and so on.
- The ratio of cups of water to cups of baking soda is 2 to 1, so I looked at the coordinate points that show the same ratio.
A, D

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Section D Check

Section D Checkpoint

Problem 1

Tyler bought 4 ounces of vegetable seeds for $10 from an online store that sells seeds in bulk.

weight of seeds (ounces)	cost (dollars)
4	10
10
	35
50

Complete the table to show the costs for different amounts of seeds.
Write an equation that shows the relationship between the weight of seeds in ounces, $w$ , and the cost in dollars, $c$ .

Show Solution

weight of seeds (ounces) cost (dollars)

4 10

10 25

14 35

50 125
$2.5w = c$ or $w = 0.4c$ (or equivalent)

Problem 2

Diego worked out a deal with his parents. For every hour that he reads a book, he earns $\frac14$ hour of screen time. Diego uses the equation $s=\frac14r$ to represent this relationship.

What does each variable in the equation represent?
Which is the independent variable? Which is the dependent variable? Explain how you know.

Show Solution

The variable $s$ represents the number of hours of screen time Diego earns and the variable $r$ represents the number of hours Diego reads.
The independent variable is $r$ and the dependent variable is $s$ . Sample reasoning: The number of hours of screen time Diego earns depends on the number of hours he reads.

Unit 6 Expressions and Equations — Unit Plan