Lesson 4
Equations and Their Solutions
4.1: What is a Solution? (5 minutes)
Warm-up
This warm-up prompts students to recall what they know about the solution to an equation in one variable. Students interpret a given equation in the context of a situation, explain why certain values are not solutions to the equation, and then find the value that is the solution.
Student Facing
A granola bite contains 27 calories. Most of the calories come from \(c\) grams of carbohydrates. The rest come from other ingredients. One gram of carbohydrate contains 4 calories.
The equation \(4c + 5= 27\) represents the relationship between these quantities.
- What could the 5 represent in this situation?
- Priya said that neither 8 nor 3 could be the solution to the equation. Explain why she is correct.
- Find the solution to the equation.
Student Response
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Activity Synthesis
Focus the discussion on how students knew that 8 and 3 are not solutions to the equation and how they found the solution. Highlight strategies that are based on reasoning about what values make the equation true.
Ask students: "In general, what does a solution to an equation mean?" Make sure students recall that the solution to an equation in one variable is a value for the variable that makes the equation a true statement.
4.2: Weekend Earnings (15 minutes)
Activity
In this activity, students write equations in one variable to represent the constraints in a situation. They then reason about the solutions and interpret the solutions in context.
To solve the equation, some students may try different values of \(h\) until they find one that gives a true equation. Others may perform the same operations to each side of the equation to isolate \(h\). Identify students who use different strategies and ask them to share later.
Launch
Arrange students in groups of 2 and provide access to calculators. Give students a few minutes of quiet work time, and then time to discuss their responses. Ask them to share with their partner their explanations for why 4 and 7 are or are not solutions.
If students are unsure how to interpret “take-home earnings,” clarify that it means the amount Jada takes home after paying job-related expenses (in this case, the bus fare).
Design Principle(s): Maximize meta-awareness; Support sense-making
Supports accessibility for: Organization; Attention
Student Facing
Jada has time on the weekends to earn some money. A local bookstore is looking for someone to help sort books and will pay $12.20 an hour. To get to and from the bookstore on a work day, however, Jada would have to spend $7.15 on bus fare.
- Write an equation that represents Jada’s take-home earnings in dollars, \(E\), if she works at the bookstore for \(h\) hours in one day.
- One day, Jada takes home $90.45 after working \(h\) hours and after paying the bus fare. Write an equation to represent this situation.
- Is 4 a solution to the last equation you wrote? What about 7?
- If so, be prepared to explain how you know one or both of them are solutions.
- If not, be prepared to explain why they are not solutions. Then, find the solution.
- In this situation, what does the solution to the equation tell us?
Student Response
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Student Facing
Are you ready for more?
Jada has a second option to earn money—she could help some neighbors with errands and computer work for $11 an hour. After reconsidering her schedule, Jada realizes that she has about 9 hours available to work one day of the weekend.
Which option should she choose—sorting books at the bookstore or helping her neighbors? Explain your reasoning.
Student Response
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Anticipated Misconceptions
If students struggle to write equations in the first question, ask them how they might find out Jada's earnings if she works 1 hour, 2 hours, 5 hours, and so on. Then, ask them to generalize the computation process for \(h\) hours.
Activity Synthesis
Ask a student to share the equation that represents Jada earning $90.45. Make sure students understand why \(90.45 = 12.20h - 7.15\) describes that constraint.
Next, invite students to share how they knew if 4 and 7 are or are not solutions to the equation. Highlight that substituting those values into the equation and evaluating them lead to false equations.
Then, select students using different strategies to share how they found the solution. Some students might notice that the solution must be greater than 7 (because when \(h=7\), the expression \(12.20h-7.15\) has a value less than 90.45) and start by checking if \(h=8\) is a solution. If no students mention this, ask them about it.
Make sure students understand what the solution means in context. Emphasize that 8 is the number of hours that meet all the constraints in the situation. Jada gets paid $12.20 an hour, pays $7.15 in bus fare, and takes home $90.45. For all of these to be true, she must have worked 8 hours.
4.3: Calories from Protein and Fat (15 minutes)
Activity
In the previous activity, students recalled what it means for a number to be a solution to an equation in one variable. In this activity, they review the meaning of a solution to an equation in two variables.
Launch
Give students continued access to calculators.
Student Facing
One gram of protein contains 4 calories. One gram of fat contains 9 calories. A snack has 60 calories from \(p\) grams of protein and \(f\) grams of fat.
The equation \(4p+9f = 60\) represents the relationship between these quantities.
-
Determine if each pair of values could be the number of grams of protein and fat in the snack. Be prepared to explain your reasoning.
- 5 grams of protein and 2 grams of fat
- 10.5 grams of protein and 2 grams of fat
- 8 grams of protein and 4 grams of fat
- If there are 6 grams of fat in the snack, how many grams of protein are there? Show your reasoning.
- In this situation, what does a solution to the equation \(4p+9f = 60\) tell us? Give an example of a solution.
Student Response
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Activity Synthesis
The goal of the discussion is to make sure students understand that a solution to an equation in two variables is any pair of values that, when substituted into the equation and evaluated, make the equation true. Discuss questions such as:
- “In this situation, what does it mean when we say that \(x=12\) and \(y=1.5\) are not solutions to the equation?” (They are not a combination of protein and fat that would produce 60 calories. Substituting them for the variables in the equation leads to a false equation of \(61.5=60\).)
- “How did you find out the grams of protein in the snack given that there are 6 grams of fat?” (Substitute 6 for \(y\) and solve the equation.)
- “Can you find another combination that is a solution?”
- “How many possible combinations of grams of protein and fat (or \(x\) and \(y\)) would add up to 60 calories?” (Many solutions)
As a segue to the next lesson, solicit some ideas on how we know that there are many solutions to the equation. If no one mentions using a graph, bring it up and tell students that they will explore the graphs of two-variable equations next.
Lesson Synthesis
Lesson Synthesis
To summarize the lesson, refer back to the activity about protein and fat. Remind students that a gram of protein has 4 calories and a gram of fat has 9 calories. Discuss questions such as:
- "What does the equation \(4x + 9y=110\) tell us about the calories in a snack?" (It has 110 calories from some grams of protein and some grams of fat.)
- "In this situation, what does it mean to solve the equation?" (To find the combination of grams of protein and fat that produce 110 calories.)
- "Is the combination of 11 grams of protein and 5 grams of fat a solution to the equation? Why or why not?" (No, they don't add up to 110 calories. Substituting 11 for \(x\) and 5 for \(y\) into the equation doesn't lead to a true equation.)
- "Consider the equation \(4(5) + 9y=110\). What does it tell us about the snack?" (The snack has 5 grams of protein and a total of 110 calories.)
- "What does it mean to solve this equation?" (To find the grams of fat that, when combined with 5 grams of calories, give a total of 110 calories. To find the value of \(y\) that would make the equation true.)
4.4: Cool-down - Box of T-shirts (5 minutes)
Cool-Down
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Student Lesson Summary
Student Facing
An equation that contains only one unknown quantity or one quantity that can vary is called an equation in one variable.
For example, the equation \(2\ell + 2w = 72\) represents the relationship between the length, \(\ell\), and the width, \(w\), of a rectangle that has a perimeter of 72 units. If we know that the length is 15 units, we can rewrite the equation as:
\(2(15) + 2w = 72\).
This is an equation in one variable, because \(w\) is the only quantity that we don't know. To solve this equation means to find a value of \(w\) that makes the equation true.
In this case, 21 is the solution because substituting 21 for \(w\) in the equation results in a true statement.
\(\begin {align}2(15) + 2w &=72\\ 2(15)+2(21) &= 72\\ 30 + 42 &=72\\ 72&=72 \end{align}\)
An equation that contains two unknown quantities or two quantities that vary is called an equation in two variables. A solution to such an equation is a pair of numbers that makes the equation true.
Suppose Tyler spends \$45 on T-shirts and socks. A T-shirt costs \$10 and a pair of socks costs \$2.50. If \(t\) represents the number of T-shirts and \(p\) represents the number of pairs of socks that Tyler buys, we can can represent this situation with the equation:
\(10t + 2.50p = 45\)
This is an equation in two variables. More than one pair of values for \(t\) and \(p\) make the equation true.
\(t=3\) and \(p=6\)
\(\begin {align} 10(3) + 2.50(6) &= 45\\ 30 + 15 &=45\\ 45&=45 \end{align}\)
\(t=4\) and \(p=2\)
\(\begin {align} 10(4) + 2.50(2) &= 45\\ 40 + 5 &=45\\ 45&=45 \end{align}\)
\(t=2\) and \(p=10\)
\(\begin {align} 10(2) + 2.50(10) &= 45\\ 20 + 25 &=45\\ 45&=45 \end{align}\)
In this situation, one constraint is that the combined cost of shirts and socks must equal \$45. Solutions to the equation are pairs of \(t\) and \(p\) values that satisfy this constraint.
Combinations such as \(t=1\) and \(p = 10\) or \(t=2\) and \(p=7\) are not solutions because they don’t meet the constraint. When these pairs of values are substituted into the equation, they result in statements that are false.