Apply Two-Variable Relationships to Solve Problems
Use equations and representations to solve real-world problems
involving two quantities that change in relationship to one another.
6.EE.C.9 · Two-Variable Relationships
Level
Guided practice with vocabulary support
🟠 Level 0 — Extra Support
Watch first: the worked example and hints are already shown for you.
One at a time: answer one question, check it, then go on.
Read aloud: press 🔊 Read aloud, then click any sentence to hear it.
Sentence starters: “First, I…” · “The answer is… because…” · “I know this because…”
W
Warm-Up
2 questions
Warm-Up 1
A gym charges $25 per month. The equation for the
total cost is c = 25m. How much does it
cost for 6 months?
Vocabulary: To
substitute means to replace a
variable with a number. Replace
m with 6, then multiply: 25 × 6 = ?
✓Correct! Substitute m = 6 into c = 25m: c = 25 × 6 =
$150.
✗Not quite. Substitute m = 6 into c = 25m: c = 25 × 6 = 150.
The answer is B: $150.
Warm-Up 2
True or False: If y = 3x and
x = 10, then
y = 30.
Hint:Substitute 10 for
x in the equation y = 3x. Multiply: 3 × 10 = ? Does it
equal 30?
✓Correct! y = 3(10) = 30. The statement is true.
✗Check again: y = 3 × 10 = 30. Since 30 = 30, the statement is
true.
P
Practice
5 questions
Practice 1
A pool fills at a rate of 50 gallons per hour. The
equation is g = 50h. How many gallons
are in the pool after 8 hours?
Step-by-step:
Write the equation: g = 50h
Substitute h = 8
Multiply: 50 × 8 = ?
g =gallons
Sentence frame: "I substituted h = ___ into g = 50h. So g = 50
× ___ = ___ gallons."
✓Correct! g = 50 × 8 = 400 gallons.
✗Substitute h = 8: g = 50 × 8 =
400 gallons.
Practice 2
A store sells notebooks for $3 each. You have
$20. Which equation AND solution shows the
most notebooks you can buy?
Strategy: The total cost is
c = 3n, where n is the number of
notebooks. You need c ≤ 20. What is the largest whole number
n where 3n does not go over 20?
Explain your reasoning:
Sentence frame: "The equation is c = 3n. The most notebooks I can
buy is ___ because 3 × ___ = ___, which is under $20."
✓Correct! c = 3(6) = $18, with $2 left over. 7 notebooks would cost
$21, which is more than $20. So 6 is the most you can buy.
✗The equation is c = 3n. Try n = 6: 3 × 6 = $18 (under $20).
Try n = 7: 3 × 7 = $21 (over $20). The answer is
B: n = 6.
Practice 3
Match each real-world situation to its correct equation.
Remember: Identify the
variable that changes and the
constant (fixed number). Ask: Is the amount
increasing (add/multiply) or decreasing (subtract)?
p = 12h
m = 50 − 7d
c = 5n + 3
d = 60t
Earn $12 per hour
Start with $50, spend $7 per day
$5 per item plus $3 shipping
Drive 60 miles per hour
✓All correct! Each equation matches its real-world situation.
✗Some matches are incorrect. Think about what operation each
situation describes: earning is multiplying, spending from a
starting amount is subtracting. Try again!
Practice 4
A car travels at 55 miles per hour. The equation is
d = 55t. How long does it take to travel
275 miles?
Step-by-step:
You know d = 275 and the equation is d = 55t
Substitute: 275 = 55t
Solve: divide both sides by 55
Show your work:
Sentence frame: "I substituted d = ___ into d = 55t. Then I divided
___ ÷ 55 = ___. It takes ___ hours."
✓Correct! 275 ÷ 55 = 5. It takes 5 hours to travel 275
miles.
✗Substitute: 275 = 55t. Divide both sides by 55: t = 275 ÷ 55
= 5 hours. The answer is C.
Practice 5
Complete the table for the equation
y = 4x − 2.
How to use the equation:
To find y: substitute the
given x value, then compute 4x − 2
To find x: set y equal to the given value, then
solve by adding 2 and dividing by 4
a. When x = 5, y =
b. When y = 22, x =
Sentence frame: "I substituted ___ into y = 4x − 2 and got
___."
✓All correct! When x = 5: y = 4(5) − 2 = 18. When y = 22: 22 =
4x − 2, so 24 = 4x, and x = 6.
✗Check your work: For x = 5, y = 4(5) − 2 = 20 − 2 =
18. For y = 22: 22 + 2 = 24, then 24 ÷ 4 =
6.
★
Challenge
1 question
Challenge
A phone plan costs $30 per month plus
$0.10 per text message. Answer all three parts below.
Guide:
The equation has a fixed cost ($30)
plus a rate ($0.10) times the number of texts (t)
To find the cost for 200 texts,
substitute t = 200
To find how many texts fit in a $50 budget, set c = 50 and
solve for t
Part A: Write an equation for the monthly cost c based on
the number of texts t:
Part B: What is the cost for 200 texts in one month?
$
Part C: If your budget is $50, how many texts can you send? Show
your work.
texts
Sentence frame: "The equation is c = ___. For 200 texts, c = ___ +
___ = ___. With a $50 budget, I can send ___ texts because ___."
✓Great work! c = 30 + 0.10t. For 200 texts: c = 30 + 0.10(200) = 30
+ 20 = $50. With a $50 budget: 50 = 30 + 0.10t, so 20 = 0.10t, and t
= 200 texts.
✗The equation is c = 30 + 0.10t. For 200 texts: 30 + 0.10(200) = 30
+ 20 = $50. For a $50 budget: 50 − 30 = 20,
then 20 ÷ 0.10 = 200 texts.