Unit 2 Linear Equations and Systems — Unit Plan

Title	Takeaways	Assessment
Lesson 1 Planning a Party	—	School Supply Reward (1 problem) As a reward for achieving their goals, all students in the ninth grade are awarded a pack of school supplies. Write an expression that could represent an estimated cost for the reward. Use at least one variable. State what each part of the expression represents. Choose a variable in your expression. Describe the values that would be reasonable for the quantity that the letter represents. Show Solution Sample responses: $1.50n + 0.08(1.50n)$ . There are $n$ students in the ninth grade. A notebook with the school logo is about \$1.50, and tax is about 8%. $\frac {125}{6} \boldcdot 5 + 6(3) + p$ . There are 125 students in the ninth grade. Suppose a pack of notebooks costs \$5 and has 6 notebooks. A pack of 24 erasers costs about \$3 and 6 packs are needed. A pack of 150 bookmarks costs $p$ dollars. Sample responses: A whole number between 90 and 150 is a reasonable value for the number of students in ninth grade. A number less than $4 but more than $1 is reasonable for the cost of a pack of erasers.
Lesson 2 [PRIORITY] Writing Equations to Model Relationships (Part 1)	—	Shirt Colors (1 problem) A school choir needs to make T-shirts for its 75 members and has set aside some money in their budget to pay for them. The members of the choir decided to order from a printing company that charges $3 per shirt, plus a $50 set-up fee for each color to be printed on the shirts. Write an equation that represents the relationship between the number of T-shirts ordered, the number of colors on the shirts, and the total cost of the order. If you use a variable, specify what it represents. In this situation, which quantities do you think can vary? Which might be fixed? Show Solution $75(3) + 50x = D$ (or equivalent). $x$ represents the number of colors on the shirts. $D$ represents the total cost in dollars. Sample response: The cost per shirt and the set-up fee-per-color are fixed (they are set by the printing company). The number of colors on the shirts can vary, so can the total cost, depending on the number of colors being printed.
Lesson 3 [PRIORITY] Writing Equations to Model Relationships (Part 2)	—	Labeling Books (1 problem) Clare volunteers at a local library during the summer. Her work includes putting labels on 750 books. How many minutes will she need to finish labeling all books if she takes no breaks and labels: 10 books a minute 15 books a minute Suppose Clare labels the books at a constant speed of $s$ books per minute. Write an equation that represents the relationship between her labeling speed and the number of minutes it would take her to finish labeling. Show Solution 75 minutes 50 minutes $s \boldcdot m = 750$ , or $m = \frac{750}{s}$ , where $m$ is the number of minutes and $s$ is the speed in books per minute.
Lesson 4 Equations and Their Solutions	—	Box of T-shirts (1 problem) An empty shipping box weighs 250 grams. The box is then filled with T-shirts. Each T-shirt weighs 132.5 grams. The equation $W = 250 + 132.5T$ represents the relationship between the quantities in this situation, where $W$ is the weight, in grams, of the filled box, and $T$ is the number of shirts in the box. Name two possible solutions to the equation $W = 250 + 132.5T$ . What do the solutions mean in this situation? Consider the equation $2,900 = 250 + 132.50T$ . In this situation, what does the solution to this equation tell us? Show Solution Sample response: $T=2$ and $W = 515$ $T= 10$ and $W=1,575$ Each solution tells us the number of T-shirts in the box and the corresponding total weight in grams. It tells us the number of T-shirts in the box that results in a total weight of 2,900 grams.
Lesson 5 Equations and Their Graphs	—	A Spoonful of Sugar (1 problem) A ceramic sugar bowl weighs 340 grams when empty. It is then filled with sugar. One tablespoon of sugar weighs 12.5 grams. Write an equation to represent the relationship between the total weight of the bowl in grams, $W$ , and the tablespoons of sugar, $T$ . When the sugar bowl is full, it weighs 740 grams. How many tablespoons of sugar can the bowl hold? Plot a point on the graph that represents this situation. Show Solution $W=340 + 12.5T$ , or $340 + 12.5T = 740$ 32 tablespoons. A point plotted near $(32,740)$

Lesson 6 Equivalent Equations	—	Box of Beans and Rice (1 problem) A cardboard box, which weighs 0.6 pound when empty, is filled with 15 bags of beans and a 4-pound bag of rice. The total weight of the box and the contents inside it is 25.6 pounds. One way to represent this situation is with the equation $0.6 + 15b + 4 = 25.6$ . In this situation, what does the solution to the equation represent? Select all equations that are also equivalent to $0.6 + 15b + 4=25.6$ . Equation A: $15b + 4 = 25.6$ Equation B: $15b + 4 = 25$ Equation C: $3(0.6 + 15b + 4) = 76.8$ Equation D: $15b = 25.6$ Equation E: $15b = 21$ Show Solution The weight, in pounds, of one bag of beans B, C, and E
Lesson 8 [PRIORITY] Which Variable to Solve for? (Part 1)	—	A Rectangular Relationship (1 problem) The perimeter of a rectangle is 48 centimeters. The relationship between the length, the width, and the perimeter of the rectangle can be described with the equation $2 \boldcdot \text{length} +2 \boldcdot \text{width}=48$ . Find the length, in centimeters, if the width is: 10 centimeters 3.6 centimeters $w$ centimeters Show Solution 14 20.4 $\dfrac {48-2w}{2}$ , or $24-w$
Lesson 10 [PRIORITY] Connecting Equations to Graphs (Part 1)	—	Kiran at the Carnival (1 problem) Kiran is spending $12 on games and rides at another carnival, where a game costs $0.25 and a ride costs $1. Write an equation to represent the relationship between the dollar amount Kiran is spending and the number of games, $x$ , and the number of rides, $y$ , that he could purchase tickets for. Which graph represents the relationship between the quantities in this situation? Explain how you know. A Graph of a line, origin O. Horizontal axis, number of games, scale is 0 to 18, by 2’s. Vertical axis, number of rides, scale is 0 to 18, by 2’s. Line starts at 0 comma 12, and passes through 2 comma 4. B Graph of a line, origin O. Horizontal axis, number of games, scale is 0 to 18, by 2’s. Vertical axis, number of rides, scale is 0 to 18, by 2’s. Line starts at 0 comma 12, and passes through 2 comma 4. C Graph of a line, origin O. Horizontal axis, number of games, scale is 0 to 18, by 2’s. Vertical axis, number of rides, scale is 0 to 18, by 2’s. Line starts at 0 comma 12, passes through 8 comma 10, and 16 comma 8. Show Solution $0.25x + y = 12$ Graph C (for $y=12-\frac14 x$ ). Sample explanations: If Kiran plays 0 games, he could get on 12 rides. If he plays 8 games, he could get on 10 rides. Both the points $(0,12)$ and $(8,10)$ are on the line in Graph C. Rearranging the equation into slope-intercept form gives $y=12-\frac14 x$ , so the graph has a slope of $\text-\frac 14$ and intersects the $y$ -axis at 12, which matches Graph C.
Lesson 11 Connecting Equations to Graphs (Part 2)	—	Features of a Graph (1 problem) Consider the equation $1.5x + 4.5y = 18$ . For each question, explain or show your reasoning. If we graph the equation, what is the slope of the graph? Where does the graph intersect the $y$ -axis? Where does it intersect the $x$ -axis? Show Solution Sample responses: The slope is $\text-\frac13$ . Rearranging the equation to isolate $y$ gives $y=\text- \frac13 x+4$ . The graph intersects the $y$ -axis at $(0,4)$ . The 4 in $y=\text- \frac13 x+4$ tells us where the graph crosses the $y$ -axis. The graph intersects the $x$ -axis at $(12,0)$ . Substituting 0 for $y$ in the original equation and solving for $x$ gives $x=12$ .

Lesson 12 Writing and Graphing Systems of Linear Equations	—	Fabric Sale (1 problem) At a fabric store, fabrics are sold by the yard. A dressmaker spent $36.35 on 4.25 yards of silk and cotton fabrics for a dress. Silk is $16.90 per yard and cotton is $4 per yard. Here is a system of equations that represent the constraints in the situation. $\displaystyle \begin{cases} \begin {align} x + \hspace{2.2mm} y &= \hspace{2mm} 4.25\\ 16.90x + 4y &= 36.35 \end{align} \end{cases}$ What does the solution to the system represent? Find the solution to the system of equations. Explain or show your reasoning. Show Solution Sample reasoning: The solution represents the combination of lengths of silk and cotton (in yards) that meet both constraints—they add up to 4.25 yards and the cost of buying both is $36.35. 1.5 yards of silk and 2.75 yards of cotton. Sample reasoning: The graph of the equations intersect at $(1.5, 2.75)$ . Solving by substitution gives $x=1.5$ and $y=2.75$ .
Lesson 13 Solving Systems by Substitution	—	A System to Solve (1 problem) Solve this system of equations without graphing and show your reasoning: $\begin{cases} \begin {align} 5x + y&=7 \\20x+2&= y \end {align}\end{cases}$ Show Solution $x = \frac15, y = 6$ . Sample strategies: Substituting $20x + 2$ for $y$ in the first equation, $5x+y=7$ , gives $5x + (20x+2) = 7$ , or $25x + 2 = 7$ , which then gives $x=\frac15$ . Substituting $\frac15$ for $x$ in either equation and solving for $y$ gives $y=6$ . The first equation, $5x+y=7$ , can be rearranged to $y = 7 - 5x$ . Substituting $7-5x$ for $y$ in the second equation gives $20x + 2 = 7-5x$ . Solving this equation leads to $x=\frac15$ , and substituting $\frac15$ for $x$ in either equation gives $y=6$ .
Lesson 14 [PRIORITY] Solving Systems by Elimination (Part 1)	—	What to Do with This System? (1 problem) Here is a system of linear equations: $\begin{cases} \begin {align} 2x + \frac12y &=7\\6x - \frac12y&=5 \end{align} \end{cases}$ Which would be more helpful for solving the system: adding the two equations or subtracting one from the other? Explain your reasoning. Solve the system without graphing. Show your reasoning. Show Solution Adding the equations. Sample explanation: $\frac12y + \text-\frac12y =0$ , so adding the equations would eliminate the $y$ -variable and enable solving for $x$ . $x = \frac32$ (or equivalent) and $y = 8$ . Sample reasoning: $\begin {align}2x + \frac12y &= 7\\ 6x - \frac12y &= 5 \quad+\\ \overline {8x +\hspace{2.5mm}0\hspace{1mm}} &\overline{ \hspace{1mm}= 12 \qquad}\\8x &= 12\\x &=\frac32 \end{align}$ $\begin {align}2x +\frac12 y&= 7\\ 2\left(\frac32\right) +\frac12y &= 7\\ 3+\frac12y&= 7\\ \frac12y &= 4\\ y&=8 \end{align}$
Lesson 16 [PRIORITY] Solving Systems by Elimination (Part 3)	—	Make Your Move (1 problem) Lin and Priya are working to solve this system of equations. $\begin{cases} \begin {align} \frac13x+2y&=4 \\ x+ \hspace{2mm}y&=\text-3 \end {align} \end{cases}$ Lin's first move is to multiply the first equation by 3. Priya's first move is to multiply the second equation by 2. Explain why either move creates a new equation with the same solutions as the original equation. Whose first move would you choose to do to solve the system? Explain your reasoning. Show Solution Sample responses: Multiplying the two sides of an equation by the same factor creates an equivalent equation, which has the same solution as the original equation. Multiplying the two sides of an equation by the same number keeps the two sides equal, so the solution of the first equation still works for the second one. Sample responses: I would choose Priya's move. Multiplying the second equation by 2 gives $2x + 2y= \text-6$ , it can be subtracted from the first to eliminate the $y$ -variable. I would choose Lin's move. Multiplying the first equation by 3 gives $x+6y=12$ . Subtracting the second equation from this equation eliminates the $x$ -variable. Either person's move would work. Priya's move eliminates the $y$ -variable and Lin's move eliminates the $x$ -variable.
Lesson 17 Systems of Linear Equations and Their Solutions	—	No Graphs, No Problem (1 problem) Mai is given these two systems of linear equations to solve: System 1: $\begin {cases} \begin{align}5x +\hspace{2.2mm}y &=13\\20x + 4y &=64\end{align} \end{cases}$ System 2: $\begin {cases} 5x +y =13\\20x =52 - 4y \end{cases}$ She analyzed them for a moment, and then—without graphing the equations—said, "I got it! One of the systems has no solution, and the other has infinitely many solutions!" Mai is right! Which system has no solution, and which one has many solutions? Explain or show how you know (without graphing the equations). Show Solution The first system has no solutions, and the second system has infinitely many solutions. Sample explanations: Multiplying the first equation by 4 gives $20x + 4y=52$ . Subtracting the second equation from the first equation gives $0 = \text-12$ , which is a false equation and tells us that the system has no solutions. $20x + 4y$ (in the second equation) is 4 times $5x + y$ (in the first equation), but 64 is not 4 times 13. This means there is no pair of $x$ - and $y$ -values that could make both equations simultaneously true. If we isolate $y$ in the first equation, we have $y = 13-5x$ . If we do the same with the second equation, we have $4y = 64-20x$ or $y = 16 - 5x$ . The graphs of the two equations have the same slope (-5), and they intersect the $y$ -axis at different points (13 and 16), so the lines are parallel and have no points (solutions) in common. The second system has infinitely many solutions. Sample explanations: Multiplying the first equation by 4 gives $20x + 4y=52$ . Rearranging the second equation so that the variables are on the left side also gives $20x + 4y=52$ . The two equations are identical, so they have the same solutions. The second equation can be rearranged to $20x + 4y=52$ . We can see that it is 4 times the first equation, which means the two equations, $5x + y = 13$ and $20x + 4y =52$ , are equivalent and have all the same solutions. If we isolate $y$ in each equation, we'd have the exact same equation, $y=13 - 5x$ , so they have all the same solutions.
Unit 2 Assessment End-of-Unit Assessment		Problem 1 Select all equations that are equivalent to $\frac{(5x+6)}2=3-(4x+12)$ . Show Solution B, C, E Problem 2 A phone company charges a base fee of $12 per month plus an additional charge per minute. The monthly phone cost $C$ can be represented by this equation: $C=12+a \boldcdot m$ , where $a$ is the additional charge per minute, and $m$ is the number of minutes used. Which equation can be used to find the number of minutes a customer used if we know $a$ and $C$ ? A. $m=\frac{(C-12)}{a}$ ✓ B. $m = (C-12) - a$ C. $m = C - 12a$ D. $m = \frac {C}{a} - 12$ Show Solution A Problem 3 Tickets to the zoo cost $12 for adults and $8 for children. The school has a budget of $240 for the field trip. An equation representing the budget for the trip is $240=12x+8y$ . Here is a graph of this equation. Graph of a line. Horizontal axis from 0 to 32, by 2's, number of adults. Vertical axis from 0 to 32, by 2's, number of children. Line begins on the y axis at 30, goes through 4 comma 24, 12 comma 12, and ends on the x axis at 20. Select all the true statements. Show Solution A, C, E Problem 4 A pizza shop sells pizzas that are 10 inches (in diameter) or larger. A 10-inch cheese pizza costs $8. Each additional inch costs $1.50, and each additional topping costs $0.75. Write an equation that represents the cost of a pizza. Be sure to specify what the variables represent. Show Solution Sample response: $C$ =cost of pizza, $s$ =additional inches above 10 for the pizza diameter, $t$ =number of toppings. $C=8+1.50s+0.75t$ . Problem 5 Consider this system of equations: $\displaystyle \begin {cases} y = \text{-}\frac{1}{2}x + 5 \\ 6x - 7y = 22 \end{cases}$ Solve the system by graphing. Label each graph and the solution. Show Solution $(6, 2)$ or $x=6, \, y=2$ Problem 6 Solve the system of equations without graphing. Show your reasoning. $\displaystyle \begin{cases} 2y=x-4 \\ 4x+3y=5 \end{cases}$ Show Solution $(2, -1)$ or $x=2, \, y=-1$ Problem 7 The system of equations $\begin{cases} 4x+6y=24\\ 2x+y=8 \end{cases}$ has exactly one $(x,y)$ pair for its solution. If we double each side of the second equation, $2x+y=8$ , we have $4x+2y=16$ . Explain why the same $(x,y)$ pair that is the solution to the system is also a solution to this new equation. If we add the two equations in the original system, we have $6x +7y=32$ . Explain why the same $(x,y)$ pair is also a solution to this equation. Show Solution Multiplying both sides of an equation by 2 means that the amount on each side is twice as large, but the two sides are still equal. Since the original equation had the solution pair, the new one will too. Since $4x+6y=24$ , adding this equation to $2x+y=8$ is like adding 24 to each side. Adding the same thing to both sides doesn't change the solutions.