Unit 4 Linear Equations And Linear Systems — Unit Plan

Title	Takeaways	Assessment
Lesson 1 Writing Equivalent Equations	—	Explain the Reasoning (1 problem) Label all 4 arrows to describe what happens in each move. Are the equations equivalent? Explain your reasoning. Show Solution Subtract 3 (or add -3) and divide by 2 (or multiply by $\frac{1}{2}$ ) Yes. Sample reasoning: As long as the same operations are done correctly to each side, the equations remain equivalent.
Lesson 2 Keeping the Equation Balanced	—	Changing Blocks (1 problem) Here is a hanger that is in balance. We don’t know how much any of its shapes weigh. How could you remove shapes from the hanger and keep it in balance? Describe in words or draw a new diagram. How could you add shapes to the hanger and keep it in balance? Describe in words or draw a new diagram. Show Solution Sample response: I can remove 2 circles from each side. I can add 1 triangle to each side.
Lesson 3 Balanced Moves	—	More Matching Moves (1 problem) Match these pairs of equations with the description of what is done in each step. Step 1: $\begin{align} 12x-6&=10\\ 6x-3&=5 \end{align}$ A: Add 3 to each side Step 2: $\begin{align} 6x-3&=5\\ 6x&=8 \end{align}$ B: Multiply each side by $\frac16$ Step 3: $\begin{align} 6x&=8\\ x&=\frac43 \end{align}$ C: Divide each side by 2 You are given the equation $3(x-2) = 8$ . Is your first step to distribute or divide? Explain your reasoning. Show Solution Step 1: C, Step 2: A, Step 3: B Sample responses: I would distribute the 3. That way I do not need to deal with fractions like $\frac{8}{3}$ until the end. I would divide each side by 3. Then there are fewer terms to manage while solving.
Lesson 4 More Balanced Moves	—	Mis-Steps (1 problem) Examine Lin’s solution to $8(x-3) + 7 = 2x(4-17)$ . Lin’s solution: For each step, determine if the 2 equations are equivalent. If they are not, describe the error. What is the correct solution to the original equation? Show Solution Sample response: $x = \frac{1}{2}$ or equivalent
Lesson 5 Solving Any Linear Equation	—	Check It (1 problem) Noah tries to solve the equation $\frac{1}{2}(7x-6)=6x-10$ . Check Noah’s work. If it is not correct, describe what is wrong and show the correct work. $\begin{align} \frac{1}{2}(7x - 6) &=6x - 10 \\[2ex] 7x - 6 &=12x - 10 \\[2ex] 7x &= 12x - 4 \\[2ex] \text{-}5x &= \text{-}4 \\[2ex] x &= \frac{4}{5} \end{align}$ Show Solution Sample response: Going from line 1 to line 2, Noah tried to multiply each side of the equation by 2, but did not multiply the 10. When you double each side of an equation, each term needs to be multiplied by 2. $\begin{align} \frac{1}{2}(7x - 6) &=6x - 10 \\ 7x - 6 &=12x - 20 \\ 7x &= 12x - 14 \\ \text{-}5x &= \text{-}14 \\ x &= \frac{14}{5} \end{align}$
Lesson 6 Strategic Solving	—	Think Before You Step (1 problem) Without solving, identify whether this equation has a solution that is positive, negative, or zero. Explain your reasoning. $3x-5=\text-3$ Solve the equation. $x-5(x-1)=x-(2x-3)$ Show Solution Positive. Sample reasoning: If $3x-5=\text-3$ , then the $x$ must be positive. If $x$ is negative, then subtracting 5 from $3x$ would result in a number less than $\text-3$ . For similar reasons, $x$ cannot be zero. $x=\frac23$ (or equivalent)
Section A Check Section A Checkpoint		Problem 1 Label the arrows to describe the moves that create equivalent equations. Are these 2 equations equivalent? Explain your reasoning. $\begin{align} 4x + 2 &= 20x\\ x + 2 &= 5x \end{align}$ Show Solution Add 3, distributive property No. Sample reasonings: Each term on the left should be divided by 4, but the 2 was not divided. The solutions are not the same. For the second equation, the solution is $x = \frac{1}{2}$ , but that does not solve the first equation because $4 \boldcdot \frac{1}{2} + 2 \neq 20\boldcdot \frac{1}{2}$ . Problem 2 Solve the equation. Show your reasoning by describing any moves that you make to write equivalent equations. $4(3 - x) = 3x-2$ Show Solution Sample response: $12 - 4x = 3x - 2$ , I applied the distributive property $12 = 7x - 2$ , I added $4x$ to each side $14 = 7x$ , I added 2 to each side $2 = x$ , I divided each side by 7

Lesson 7 All, Some, or No Solutions	—	Choose Your Own Solution (1 problem) $\displaystyle 3x + 8 = 3x + \underline{\hspace{.5in}}$ What value could you write in after " $3x$ + " that would make the equation true for: no values of $x$ ? all values of $x$ ? just one value of $x$ ? Show Solution Any value other than 8. 8 Any variable term. like $x$ or $2x$ , in order to create an equation with one solution.
Lesson 8 How Many Solutions?	—	How Does She Know? (1 problem) Elena begins to solve this equation: $\begin{align} \dfrac{12x+6(4x+3)}3 &\,=\,2(6x+4)-2 \\[2ex]12x+6(4x+3) &\,=\,3(2(6x+4)-2)\\[2ex] 12x+6(4x+3) &\,=\,6(6x+4)-6 \\[2ex] 12x+24x+18 &\,=\,36x+24-6 \end{align}$ When she gets to the last line she stops and says the equation is true for all values of $x$ . How can Elena tell? Show Solution Sample response: Elena can see that there are the same number of $x$ 's and the same constant terms on each side of the equation.
Lesson 9 When Are They the Same?	—	Printers and Ink (1 problem) To own and operate a home printer, it costs $100 for the printer and an additional $0.05 per page for ink. To print out pages at an office store, it costs $0.25 per page. Let $p$ represent number of pages. What does the equation $100+0.05p=0.25p$ represent? The solution to that equation is $p=500$ . What does the solution mean? Show Solution The equation represents when the cost for owning and operating a home printer is equal to the cost for printing at an office store. The solution of $p=500$ means that the costs are equal for printing 500 pages.
Section B Check Section B Checkpoint		Problem 1 $3x + 7 = 5x + 7$ How many solutions does the equation have? Explain how you know without solving. Change 1 number in the equation $2x + 4 = 2x + 6$ so that it has infinitely many solutions. Show Solution 1 solution. Sample reasoning: The coefficients of $x$ on each side of the equation are not equal. Sample responses: $2x + 4 = 2x+4$ $2x + 6 = 2x+6$ Problem 2 Two friends go out for a run. Friend A runs at a steady pace of 160 meters per minute so that their distance from the starting line is represented by $160t$ . Friend B gets started later and begins running a little further along the route so that their distance from the starting line is represented by $180(t-3)+100$ . Solve the equation $160t = 180(t-3)+100$ . Show your reasoning. What does the solution mean in this situation? Show Solution $t = 22$ . Sample reasoning: $160t=180t - 540 + 100$ by distributive property. $160t = 180t - 440$ by combining like terms. $\text{-}20t = \text{-}440$ by subtracting $180t$ from each side. $t = 22$ by dividing each side by -20. Sample response: 22 minutes after Friend A started running the friends are the same distance from the starting line.

Lesson 10 On or Off the Line?	—	Another Pocket Full of Change (1 problem) On the coordinate plane shown, one line shows combinations of dimes and quarters that are worth $3. The other line shows combinations of dimes and quarters that total to 12 coins. Graph of two intersecting lines in the x y plane, origin 0, with grid. Horizontal axis, number of quarters, scale 0 to 20, by 1s. Vertical axis, number of dimes, scale 0 to 20 by 1s. A line, labeled 12 coins all together, crosses the y axis at 12 and slants downward and to the right. It passes through the points 1 comma 11, 2 comma 10, 3 comma 9, 4 comma 8, 5 comma 7, 6 comma 6, 7 comma 5, 8 comma 4, 9 comma 3, 10 comma 2, 11 comma 1, and 12 comma 0. Another line, labeled dimes and quarters that total to 3 dollars slants downward and to the right. It passes through the points 6 comma 15, 8 comma 10 and 10 comma 5. Name one combination of 12 coins shown on the graph. How does the graph show that the combination is true? Name one combination of coins shown on the graph that total to $3. How many quarters and dimes would you need to have both 12 coins and $3 at the same time? How does the graph show that this is true? Show Solution Sample responses: 6 quarters and 6 dimes. 11 quarters and 1 dime. The point $(6,6)$ (or $(11,1)$ ) is on the graph of the line representing 12 coins all together.0 Sample responses: 6 quarters and 15 dimes. 10 quarters and 5 dimes. 12 quarters and 0 dimes because the point $(12,0)$ is on both lines.
Lesson 11 On Both of the Lines	—	Saving Cash (1 problem) Andre and Noah start tracking their savings at the same time. Andre starts with $15 and deposits $5 per week. Noah starts with $2.50 and deposits $7.50 per week. The graph of Noah's savings is given, and his equation is $y=7.5x+2.5$ , where $x$ represents the number of weeks and $y$ represents his savings. Write the equation for Andre's savings, and graph it alongside Noah's. What does the intersection point mean in this situation? <p>Graph of a line in the x y plane, origin O, with grid. Horizontal axis, weeks, scale 0 to 12, by 1’s. Vertical axis, savings in dollars, scale 0 to 45, by 5’s. The line passes through the points 1 comma 10 and 3 comma 25. </p> Show Solution Sample response: The intersection at $(5,40)$ means that after 5 weeks, Noah and Andre each have $40.
Lesson 12 Systems of Equations	—	Finishing Their Water Again (1 problem) Lin’s glass has 12 ounces of water and she drinks it at a rate of $\frac{1}{3}$ ounce per second. Diego’s glass has 20 ounces and he drinks it at a rate of $\frac{2}{3}$ ounce per second. Graph this situation on the axes provided. What does the graph tell you about the situation and how many solutions there are? Show Solution Sample response: There is one solution at $(24,4)$ meaning that after 24 seconds both of them have 4 ounces of water left.
Lesson 13 Solving Systems of Equations	—	Two Lines (1 problem) Given the lines shown here, what are two possible equations for this system of equations? How many solutions does this system of equations have? Explain your reasoning. <p>Two lines in an x y plane. One line slants upward and right. It crosses the x axis to the left of the origin. It crosses the y axis above the origin. A second line slants upward and right. It crosses the y axis below the origin. The lines do not intersect. </p> Show Solution Any two equations with the same positive slope for each linear equation yet one with a negative $y$ -intercept and the other a positive $y$ -intercept 0. Sample reasoning: Since the lines are parallel and do not intersect, there are no solutions to the system of equations.
Lesson 14 Solving More Systems	—	Solve It (1 problem) Solve this system of equations: $\begin{cases} y=2x \\[2ex] x = \text-y+6 \end{cases}$ Show Solution $(2,4)$ . Sample Reasoning: Use the substitution method to rewrite the system as the one variable equation $x = \text{-} (2x)+6$ , then solve.
Lesson 15 Writing Systems of Equations	—	Solve This (1 problem) Solve. $\begin{cases} y= \dfrac34x \\[2ex] \dfrac52x+2y = 5 \end{cases}$ Show Solution $x = \frac{5}{4}, y = \frac{15}{16}$
Section C Check Section C Checkpoint		Problem 1 $\begin{cases} y = 3x + 5 \\ y = 3(x + 1) \end{cases}$ How many solutions does this system have? Explain your reasoning without solving the system. Based on the number of solutions, describe the graph of this system. Show Solution No solutions. Sample reasoning: The second equation is equivalent to $y = 3x + 3$ . This shows that the 2 equations have the same slope and different $y$ -intercepts, so there is no solution. The graphs of the lines are parallel. Problem 2 In a card game, each round you earn either 3 points or 5 points depending on the cards you play. After 5 rounds you have 19 points. Use $x$ for the number of 3 point rounds and $y$ for the number of 5 point rounds. Write a system of 2 equations that describes this situation. Another system is solved by the point $(7,10)$ . Explain how you can check that this solution is correct. Show Solution $\begin{cases} 3x + 5y &= 19\\ x + y &= 5 \end{cases}$ (or equivalent) Sample response: The values make both equations true. Substitute 7 for $x$ and 10 for $y$ in the original equations and check that each side of the equations are equal to the other side.

Lesson 16 Solving Problems with Systems of Equations	—	No cool-down
Unit 4 Assessment End-of-Unit Assessment