Unit 6 Expressions and Equations — Unit Plan

TitleTakeawaysStudent SummaryAssessment
Lesson 1
Tape Diagrams and Equations
  • 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.

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

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

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:

x+x+x=21\displaystyle x+x+x=21

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

x=21÷3\displaystyle x=21\div3

x=1321\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:

y+3=21\displaystyle y+3=21

y=213\displaystyle y=21-3

3=21y\displaystyle 3=21-y

Complete the Diagrams (1 problem)

Complete the first diagram so it represents 5x=155 \boldcdot x = 15. Complete the second diagram so it represents 5+y =155 + y = 15.

Two blank tape diagrams. Each tape diagram totals 15.

Show Solution

Sample response:

Lesson 2
Truth and Equations
  • 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=427+1=4 \boldcdot 2. An example of a false equation is 7+1=97+1=9

An equation can have a letter in it to represent a value, for example, u+1=8u+1=8. This equation is false if uu is 3, because 3+13+1 does not equal 8. This equation is true if uu is 7, because 7+1=87+1=8.

A letter in an equation that represents an unknown value is called a variable. In u+1=8u+1=8, the variable is uu. 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=8u+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=217x=21 means the same thing as 7x=217 \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=5p+3=5, the coefficient of pp 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 18x=11\frac18 x = 11 by completing the sentences:

The word “solution” means . . .

88 is a solution to 18x=11\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 18x=11\frac18x = 11, because if xx is 88, the equation is 1888=11\frac18 \boldcdot 88=11, which is true.

Lesson 3
Staying in Balance
  • 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.

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

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

Diagram A can be represented by the equation 3x=113x=11.

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

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

  • 3x3x divided by 3 is xx.
  • 11 divided by 3 is 113\frac{11}{3}.
  • If 3x=113x=11 is true, then x=113x=\frac{11}{3} is true.
  • The solution to 3x=113x=11 is 113\frac{11}{3}.

Diagram B can be represented with the equation 11=y+511=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.

  • 11511-5 is 6.
  • y+55y+5-5 is yy.
  • If 11=y+511=y+5 is true, then 6=y6=y is true.
  • The solution to 11=y+511=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.

  1. Write an equation that represents the diagram.
  2. Find the weight of one circle. Explain or show your reasoning.
  3. What is the solution to your equation?
Show Solution
  1. 4w=254w=25
  2. 254\frac{25}{4} or 6146\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 6146\frac14units. This shows that each circle piece weighs 6146\frac14units.
  3. w=254w=\frac{25}{4} or 6146\frac14
Lesson 4
Practice Solving Equations
  • 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+34=78x+\frac{3}{4}=\frac{7}{8}.

The fraction 34\frac{3}{4} is added to the variable xx.

x+34=78\displaystyle x+\frac{3}{4}=\frac{7}{8}

So, we can subtract 34\frac{3}{4} from each side of the equation.

x+3434=7834\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 18\frac{1}{8} is on the other side. 

x=18\displaystyle x=\frac{1}{8}

When we substitute 18\frac{1}{8} for xx in the original equation, the equation is true. So, we know 18\frac{1}{8} is the solution.

18+34=7878=78\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=31.53.5x=31.5.

The variable xx is multiplied by 3.5.

3.5x=31.5\displaystyle 3.5x=31.5

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

3.5x÷3.5=31.5÷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. 

x=9\displaystyle x = 9

When we substitute 9 for xx in the original equation, the equation is true. So, we know 9 is the solution.

3.5(9)=31.531.5=31.5\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.

  1. x+134=10x+1\frac34=10.
  2. 5.7x=17.15.7x = 17.1
  3. 110x=25\frac{1}{10}x=\frac25
Show Solution

Sample responses:

  1. x=814x=8\frac14. Sample reasoning: x+134134=10134x+1\frac34-1\frac34=10-1\frac34
  2. x=3x=3. Sample reasoning: 5.7x÷5.7=17.1÷5.75.7x\div5.7=17.1\div5.7
  3. x=4x=4. Sample reasoning: I divided both sides of the equation by 110\frac{1}{10}.
Lesson 5
Represent Situations with Equations
  • 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:

13.68+x=16.05\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:

x=16.05 13.68\displaystyle x=16.05 - 13.68

x=2.37\displaystyle x=2.37

  • The solution x=2.37x=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 34\frac{3}{4} pound. How many portions can they make?

  • We can represent this situation with the equation:

34x=54\displaystyle \frac34 x = 54

  • We can find the value of xx by dividing each side by 34\frac34. This gives us some new equations that represent the same situation:

x=54÷34\displaystyle x=54\div \frac34

x=72\displaystyle x=72

  • The solution x=72x=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.
  1. Lin needs 10 cups of flour for a bread recipe. She only has 2122\frac12 cups. How much more flour does she need?

    • x+212=10x+2\frac12=10
    • x=10+212x=10+2\frac12
    • 212x=102\frac12x=10
  2. Each notebook costs 5.70. How many notebooks does Diego buy if he spends a total of 17.10?

    • x+5.7=17.1x+5.7=17.1
    • 5.7x=17.15.7x=17.1
    • 17.1x=5.717.1x=5.7
Show Solution
  1. x+212=10x+2\frac12=10, x=712x=7\frac12. Lin needs 7127\frac12 cups of flour.
  2. 5.7x=17.15.7x=17.1, x=3x=3. Diego buys 3 notebooks.
Lesson 6
Percentages and Equations
  • "A%A\% of BB is CC" can be written as the equation A100B=C\frac{A}{100} \cdot B = C
  • To find the whole when you know a percent and its value, solve px=qpx = q by dividing both sides by pp
  • 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 ss represents the total number of students who go to the school, then 70% of ss, or 70100s\frac{70}{100}s, represents the number of students that are in school today, which is 455.

We can write and solve the equation:

70100s=455s=455÷70100s=45510070\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=650s = 650

The equation can also be written using the decimal equivalent of 70100\frac{70}{100}, which is 0.7:

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

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

0.7x=4550.7(650)=455455=455\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?

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

Show Solution
  1. 54=60100x54=\frac{60}{100}x (or equivalent)
  2. $90. Sample reasoning: 
    • Divide both sides of the equation by 60100\frac{60}{100} or 35\frac35 to get x=5453=90x=54\boldcdot \frac53=90.
    • I wrote 0.6x=540.6x=54 and thought about six-tenths of what number would be equal to 54. 69=546 \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.

  1. Is 38\frac38 a solution to y+12=78y+\frac12=\frac78?
  2. Is 15 a solution to 0.4x=200.4x=20?
Show Solution
  1. Yes. Sample reasoning: Substituting 38\frac38 for yy gives a true equation,  38+12=78\frac38+\frac12=\frac78.
  2. No. Sample reasoning:  Substituting 15 for xx gives the equation (0.4)15=20(0.4) \boldcdot 15= 20, which is false.
Problem 2

Solve each equation, and explain or show your reasoning.

  1. a+123=468a + 123 = 468
  2. 2.5b=20.52.5b=20.5
Show Solution
  1. a=345a=345. Sample reasoning: Subtract 123123 from each side.
  2. b=8.2b=8.2. Sample reasoning: Divide each side by 2.52.5.
Lesson 7
Write Expressions with Variables
  • 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 aa represent your age at any time, your neighbor’s age can be expressed a+3a+3.

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

We often use a variable, such as xx or aa, 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+3a+3. We can write the equation a+3=32a+3=32 to represent these relationships. When your neighbor is 32 you will be 29, because a+3=32a+3=32 is true when aa is 29.

Growth (1 problem)
  1. A plant measured xx 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.

    • x8x-8
    • 8x8-x
  2. Each tree needs 1.2 liters of water. Write an expression that represents the amount of water needed for nn trees.
Show Solution
  1. 8x8-x. Sample reasoning: Since the plant grew taller this week, 8 is greater than xx. The difference of 8 and xx is the amount that the plant grew.
  2. 1.2n1.2n (or equivalent)
Lesson 8
Equal and 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+9x+9 and 4x4x are equal when xx is 3, but they are not equal for other values of xx

8 tape diagrams of various sizes and various labels on a grid. Each diagram is matched with an expression.
8 tape diagrams on a grid with matching expressions. First diagram composed of 1 square unit labeled x and 9 square units combined which are blank, matched with x+9 when x=1. Second diagram composed of 4 square units each labeled x matched with 4x when x=1. Third diagram composed 2 combined square units labeled x and 9 combined square units blank, matched with x+9 when x=2. Fourth diagram composed of 2 combined square units labeled x created 4 total times, matched with 4x when x=2. Fifth diagram composed 3 combined square units labeled x and 9 combined square units blank, matched with x+9 when x=3. Sixth diagram composed 3 combined square units labeled x created 4 total times, matched with 4x when x=3. Seventh diagram composed of 4 combined square units labeled x and 9 combined square units blank, matched with x+9 when x=4. Eighth diagram composed of 4 combined square units labeled x created 4 total times, matched with 4x when x=4.

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+3x+3 is equivalent to 3+x3+x because of the commutative property of addition. The order of the values being added doesn’t affect the sum.
  • 4y4\boldcdot {y} is equivalent to y4y\boldcdot 4 because of the commutative property of multiplication. The order of the factors doesn’t affect the product.
  • a+a+a+a+aa+a+a+a+a is equivalent to 5a5\boldcdot {a} because adding 5 copies of something is the same as multiplying it by 5.
  • b÷3b\div3 is equivalent to b13b \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.

  1. x+x+x+xx+x+x+x and 4x4x
  2. 5x5x and x+5x+5
Show Solution
  1. Equivalent. Sample reasoning: The diagrams representing these expressions would have the same length for any value of xx.
  2. Not equivalent. Sample reasoning: if x=1x=1, then 5x=55x=5 and x+5=6x+5=6, so they do not have the same value.
Lesson 9
The Distributive Property, Part 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:

5795(70+9)570+59350+453955 \boldcdot 79\\ 5 \boldcdot (70+9)\\ 5 \boldcdot 70 + 5 \boldcdot 9\\ 350 + 45\\ 395

When we think, “79 is the same as 70+970 + 9. I can just multiply 5705 \boldcdot 70 and 595 \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 5795 \boldcdot 79:

5795(801)5805140053955 \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.

  1. 7(3+5)=3333+33337 \boldcdot (3+5)=\boxed{\phantom{\huge3333}}+\boxed{\phantom{\huge3333}}
  2. 5352=33(32)5\boldcdot3-5\boldcdot2=\boxed{\phantom{\huge33}} \boldcdot (3-2)
Show Solution
  1. 7(3+5)=21+357\boldcdot(3+5)=\boxed{21}+\boxed{35} or 7(3+5)=73+757\boldcdot(3+5)=\boxed{7 \boldcdot 3}+ \boxed{7 \boldcdot 5} (or equivalent)
  2. 5352=5(32)5\boldcdot3-5\boldcdot2=\boxed{5} \boldcdot (3-2)
Section B Check
Section B Checkpoint
Problem 1

Andre says that 2x+52x + 5 and 7x7x are equivalent expressions because they have the same value when xx 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 xx is 0, 2x+52x+5 is 5, but 7x7x is 0. Since they do not have the same value when xx is 0, they are not equivalent expressions.
Problem 2

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

  1. 3(x+4)3(x+4)

  2. a(3.52.7)a(3.5-2.7)

  3. 5y+105y+10

Show Solution

Sample responses:

  1. 3x+123x+12
  2. 0.8a0.8a
  3. 5(y+2)5(y+2)
Lesson 12
Meaning of 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 2n2^n, we call nn the exponent.

If nn is a whole number, it tells how many factors of 2 we should multiply to find the value of the expression. For example, 21=22^1=2, and 25=22 2 222^5=2 \boldcdot 2 \boldcdot 2 \boldcdot 2 \boldcdot 2.

There are different ways to say 252^5. 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)
  1. What is the value of the expression 353^5?
  2. Explain how to use that value to quickly find the value of 363^6.
Show Solution
  1. 243
  2. Sample response: 36=353=7293^6=3^5 \boldcdot 3=729
Lesson 14
Evaluating Expressions with Exponents
  • 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 6426 \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 6426 \boldcdot 4^2 represented the surface area of a cube with edge lengths of 4 units. When computing the surface area, we compute 424^2 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 6426 \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:

64261696\begin{aligned} 6 &\boldcdot 4^2 \\ 6 &\boldcdot 16 \\ &96 \end{aligned}

45+5245+2570\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:

(64)2242576\begin{aligned} (6 &\boldcdot 4)^2 \\ &24^2 \\ &576 \end{aligned}

(45+5)25022,500\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+10320 + 10^3
30330^3
27,00027,000

Noah's method:

20+10320 + 10^3
20+1,00020 + 1,000
1,0201,020

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.

  1. 24=242^4=2 \boldcdot 4

  2. 334=353 \boldcdot 3^4 = 3^5

  3. 232=622 \boldcdot 3^2 = 6^2

Show Solution
  1. False. Sample reasoning:  16=816=8 is not true.
  2. True. Sample reasoning: Each side is 333333 \boldcdot 3 \boldcdot 3 \boldcdot 3 \boldcdot 3.
  3. False. Sample reasoning: 18=3618=36 is not true.
Problem 2
Find the value of 7+x37+x^3 when xx is 2.
Show Solution

15. Sample reasoning:

7+x37+237+8157 + x^3 \\ 7 + 2^3 \\ 7 + 8 \\15

Lesson 16
Two Related Quantities, Part 1
  • 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 rr is always 53\frac53 as large as gg and that gg is always 35\frac35 as large as rr.

green apples (gg) red apples (rr)
3 5
6 10
9 15
12 20

We can write equations to describe the relationship between gg and rr.

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

    r=53g\displaystyle r=\frac53g

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

    We can use this equation with any value of gg to find rr. If 270 green apples are used, then 53(270)\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:

    g=35r\displaystyle g=\frac35r

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

    We can use this equation with any value of rr to find gg. If 275 red apples are used, then 35(275)\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 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.

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.
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 12\frac{1}{2} cup of water. 

cups of baking soda cups of water
1 12\frac12
2 1
3 32\frac32

  1. 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.
    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.
    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.
    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.

  2. Select all equations that can represent the relationship between bb, cups of baking soda, and ww, cups of water, in this situation.

    1. w=12bw = \frac{1}{2}b
    2. b=12wb=\frac{1}{2}w
    3. b=wb = w
    4. b=2wb = 2w
    5. w=2bw = 2b
Show Solution
  1. 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 (12,1)(\frac{1}{2}, 1), (1,2)(1, 2), (32,3)(\frac{3}{2}, 3), and so on. 
    • I matched the coordinates of the points to the values in the table: 12\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.
  2. A, D
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
  1. Complete the table to show the costs for different amounts of seeds.  

  2. Write an equation that shows the relationship between the weight of seeds in ounces, ww, and the cost in dollars, cc.
Show Solution
  1. weight of seeds (ounces) cost (dollars)
    4 10
    10 25
    14 35
    50 125
  2. 2.5w=c2.5w = c or w=0.4cw = 0.4c (or equivalent)
Problem 2

Diego worked out a deal with his parents. For every hour that he reads a book, he earns 14\frac14 hour of screen time. Diego uses the equation s=14rs=\frac14r to represent this relationship.

  1. What does each variable in the equation represent?
  2. Which is the independent variable? Which is the dependent variable? Explain how you know.
Show Solution
  1. The variable ss represents the number of hours of screen time Diego earns and the variable rr represents the number of hours Diego reads.
  2. The independent variable is rr and the dependent variable is ss. Sample reasoning: The number of hours of screen time Diego earns depends on the number of hours he reads.
Unit 6 Assessment
End-of-Unit Assessment