6 times what number equals 24?
Turn & Talk (with your partner)
Since area is calculated by multiplying the width by the length, how does setting up an equation help us isolate the unknown length \(x\)?
Sentence Frames to use (Answers):
"To find the missing length \(x\), we can set up the equation \(6x = 24\). To isolate \(x\), we must perform the inverse operation of division (dividing by 6)."
Area Model
Visual showing space as product of sides.
Distributive
Multiply term outside by terms inside.
Length/Width
Perpendicular dimensions.
Inverse Op
Operation that reverses another.
TEACHER FACILITATION COMPASS PAGE 2 FOCUS
What to Do Directions: Project Page 2. Walk students through the connection between the geometric rectangle model on the left and the algebraic equation on the right. Facilitate the 90-second Turn & Talk and check that students are physically speaking the frames.
Script: "Scholars, look at this. We have a rectangle with a width of 6 inches and a missing length \(x\). If the total area inside is 24, we set up our equation as \(6 \cdot x = 24\). Turn to your partner and explain: what inverse operation do we do to get \(x\) completely by itself?"
Check for Understanding (CFU):
"Why is division the correct step here? (Because the width 6 is multiplied by \(x\), and division undos multiplication)."
Anticipated Error Highlight:
Students may answer "\(x = 18\)" thinking they need to subtract 6 from 24. Guide them: "Is the 6 added to \(x\)? No, it's length times width! We must divide."
Solving Area Problems with Equations (TEACHER EDITION) Page 2
Teacher Facilitation Guide & Key — Replica Format
Time: 0:15 - 0:25 (10 min)
We are solving: \(5(2x + 3) = 45\)
STEP 1: Set Up the Parentheses Equation Area = Width × Length
Identify dimensions:
Width = 5
Length = 2x + 3
Area = 45
Fill in the equation template (Key):
\(\underline{\mathbf{5(2x + 3) = 45}}\)
STEP 2: Distribute to Remove Parentheses Multiply width by both terms
Calculate each section's area (Key):
10x
15
2x 3 5
Write distributed equation (Key):
\(\underline{\mathbf{10x + 15 = 45}}\)
STEP 3: Isolate and Solve for \(x\) Subtract, then divide
Inverse operations (Key):
Subtract: - \(\underline{\mathbf{15}}\)
Divide: ÷ \(\underline{\mathbf{10}}\)
Solving steps & Answer (Key):
Simple: \(\underline{\mathbf{10x = 30}}\)
Answer: \(x = \) \(\underline{\mathbf{3}}\)
TEACHER FACILITATION COMPASS PAGE 3 FOCUS
What to Do Directions: Lead direct instruction. Model solving this slightly harder problem. Emphasize that there is a coefficient of 2 attached to \(x\). This means when we distribute, we multiply 5 times \(2x\) to get \(10x\). Walk students step-by-step through filling out each empty box.
Script: "Scholars, we are leveling up! Our length is now \(2x + 3\). In Step 2, look at our green box in the area model. We multiply width 5 times \(2x\). What is 5 times 2 boxes? Yes, 10x! What is 5 times the green 3? Yes, 15! So our distributed equation is \(10x + 15 = 45\). Let's write that down."
Break It Down Question:
"If the total area is 45, and we subtract the 15 from the numbers box, how much area is left for the variable box? (30). Since \(10x = 30\), 10 times what number equals 30? (3!). So \(x = 3\)!"
Check for Understanding (CFU):
"Why didn't we just write \(5x + 3 = 45\)? Because both terms inside the parentheses must be multiplied by the width of 5! Distributing is giving to everyone."
Solving Area Problems with Equations (TEACHER EDITION) Page 3
Teacher Facilitation Guide & Key — Replica Format
Time: 0:25 - 0:35 (10 min)
PROBLEM 1: Area = 48. Width = 6. Length = \(2x + 4\). Solve for \(x\). 6(2x + 4) = 48
Step Checklist (Completed): [✓] 1. Distribute width (\(6 \cdot 2x\) and \(6 \cdot 4\)) [✓] 2. Subtract constant from both sides [✓] 3. Divide by coefficients to solve for \(x\)
Your Workspace (Key):
\(6(2x + 4) = 48\)
\(12x + 24 = 48\)
\(12x + 24 - 24 = 48 - 24\)
\(12x = 24 \rightarrow x = 24 \div 12\)
\(x = 2\)
PROBLEM 2: Area = 20. Width = 4. Length = \(3x - 1\). Solve for \(x\). 4(3x - 1) = 20
Step Checklist (Completed): [✓] 1. Distribute width (\(4 \cdot 3x\) and \(4 \cdot -1\)) [✓] 2. Add constant to undo subtraction [✓] 3. Divide by coefficients to solve for \(x\)
Your Workspace (Key):
\(4(3x - 1) = 20\)
\(12x - 4 = 20\)
\(12x - 4 + 4 = 20 + 4\)
\(12x = 24 \rightarrow x = 24 \div 12\)
\(x = 2\)
Let's Discuss Blanks (Answers):
"On Problem 2, distributing a positive number into parentheses containing subtraction leaves us with a subtraction sign, so we must add to both sides because the inverse of subtraction is addition."
TEACHER MONITORING & GUIDANCE PAGE 4 FOCUS
Monitoring Laps (What to Do):
Break It Down Scaffold:
"If students are stuck on Problem 1, draw arrows from the 6 to the \(2x\) and the 6 to the 4. Ask: 'What is 6 times 2 boxes? (12 boxes). What is 6 times 4 singles? (24). That gives us \(12x + 24 = 48\).'"
Script: "Scholars, as we solve Problem 2, remember that a negative or subtraction inside means we distribute to get a subtraction! What is the inverse of subtraction? Yes, addition! That is why we add 4 to both sides."
Solving Area Problems with Equations (TEACHER EDITION) Page 4
Teacher Facilitation Guide & Key — Replica Format
Time: 0:35 - 0:47 (12 min)
Problem 1 (Bronze): Area = 18. Width = 3. Length = \(x + 2\). Find \(x\). \(3(x + 2) = 18\)
\(3x + 6 = 18\)
\(3x = 12 \rightarrow x = 12 \div 3 \rightarrow \underline{\mathbf{x = 4}}\)
Problem 2 (Silver): Area = 30. Width = 5. Length = \(2x - 2\). Find \(x\). \(5(2x - 2) = 30\)
Distribute to both: \(10x - 10 = 30\)
\(10x - 10 + 10 = 30 + 10 \rightarrow 10x = 40\)
\(\frac{10x}{10} = \frac{40}{10} \rightarrow \underline{\mathbf{x = 4}}\)
Problem 3 (Gold): Area = 40. Width = 4. Length = \(4x + 2\). Find \(x\). \(4(4x + 2) = 40\)
Distribute first: \(16x + 8 = 40\)
\(16x = 32 \rightarrow x = 32 \div 16 \rightarrow \underline{\mathbf{x = 2}}\)
My Area Self-Check (All Checked):
[✓] Distributed Width
[✓] Isolated Variable
[✓] Divided Coefficient
[✓] Checked Area
TEACHER MONITORING TRACKS PAGE 5 FOCUS
Anticipated Errors (Highlights):
Prob 2 Error: Students might write \(10x - 2 = 30\) (forgetting to distribute 5 to the 2). Remind them: "Draw both distribution arrows!"
Prob 3 Error: Students might divide \(16 \div 32\) incorrectly and write \(x = 0.5\). Emphasize: "We are dividing the right side *by* the coefficient. \(32 \div 16 = 2\)."
CFU Checkups:
"For Problem 1, once you find \(x = 4\), what is the actual length of the rectangle? (4 + 2 = 6). Does \(3 \times 6 = 18\)? Yes, our work is verified!"
"Why do we need algebraic solving when numbers get larger or include negatives? (It prevents mistakes!)."
Script: "Geometers, show your independent solving skills! Go through each level. Make sure your arrows are drawn clearly on your paper. Check your final answers by plugging them back into length!"
Solving Area Problems with Equations (TEACHER EDITION) Page 5
Teacher Facilitation Guide & Key — Replica Format
Time: 0:47 - 0:50 (3 min)
Your Goal: Solve for \(x\) Area = 12. Width = 2. Length = \(3x - 3\)
Equation Setup: \(2(3x - 3) = 12\)
Distribute Width: \(6x - 6 = 12\)
Isolate Variable: \(6x - 6 + 6 = 12 + 6 \rightarrow 6x = 18\)
Solve (Divide by 6): \(\frac{6x}{6} = \frac{18}{6} \rightarrow \underline{\mathbf{x = 3}}\)
Scholars rate their self-confidence level. Goal is 😊! [✓] Smiling Emoji Selected
Self-Reflection Blanks Answers:
1. One specific algebraic step I feel really confident about is:
distributing the width to both terms inside the parentheses.
2. One area concept I want to keep practicing is:
double checking that my final side length multiplication equals the area.
EXIT TICKET STRATEGY PAGE 6 FOCUS
What to Do Directions: Prompt students to complete the Exit Ticket in absolute silence. Do not answer questions. Move around the room and do a "cold scan" (spot check answers without grading). Collect pages at the 3-minute mark.
Script: "Scholars, this is your independent runway. Pencils up. Show me how you distribute the width of 2, solve for \(x\), and reflect on your growth. Silent work starts now."
Anticipated Error:
Students may write \(6x - 3 = 12\) (forgetting to distribute 2 to the -3). Write on board: "Remember, distributing is like rain—it must fall on *every* flower inside the garden!"
Immediate CFRP Check:
If a student gets \(x = 3\), verify: \(2(3(3)-3) = 2(6) = 12\). This is correct! Prompt them to select the happy face.
Solving Area Problems with Equations (TEACHER EDITION) Page 6
Teacher Facilitation Guide & Key — Replica Format
Time: 0:50 - 0:55 (5 min)
Problem 1 Answers
Length: x + 1 Width: 3 Area = 15
\(3(x + 1) = 15 \rightarrow 3x + 3 = 15\)
\(3x = 12 \rightarrow x = 12 \div 3 \rightarrow \underline{\mathbf{x = 4}}\)
Problem 2 Answers
Length: 2x - 2 Width: 4 Area = 24
\(4(2x - 2) = 24 \rightarrow 8x - 8 = 24\)
\(8x = 32 \rightarrow x = 32 \div 8 \rightarrow \underline{\mathbf{x = 4}}\)
Problem 3 Answers
Length: 4x + 1 Width: 2 Area = 18
\(2(4x + 1) = 18 \rightarrow 8x + 2 = 18\)
\(8x = 16 \rightarrow x = 16 \div 8 \rightarrow \underline{\mathbf{x = 2}}\)
HOMEWORK ROUTINE & WRAP-UP PAGE 7 FOCUS
What to Do Directions: With 5 minutes remaining, direct students to write down today's Homework. Explain that each homework question now features its own visual rectangle area model to guide their setup! Do a final wrap-up cheer.
Script: "Scholars, look at your homework page! Each problem features a customized, colorful area model helper box to help you visualize length, width, and area. You are all completely equipped to crush this tonight! Keep your skills sharp, and I will see you tomorrow morning."
Key Homework Message:
"Homework is your shield. Bring these 3 problems completed tomorrow. They will be scanned at the door for 100% homework completion marks."
Daily Goal Met Check:
"We solved area problems using linear equations! Scholars proved that geometric formulas and algebra equations are two sides of the exact same coin!"
Solving Area Problems with Equations (TEACHER EDITION) Page 7
\(4 - 1 = \)
Vertical (\(y_2 - y_1\)):
\(6 - 2 = \)
STEP 4 & 5: Square Your Differences & Add Them Multiply by itself, then sum
Square them:
Horizontal: \(\text{result}^2 = \)
Vertical: \(\text{result}^2 = \)
Add them together:
STEP 6: Take the Square Root Solve for final segment length
\(\text{Distance} = \sqrt{\vphantom{A}\smash{?}}\) \(\rightarrow\) \(d = \sqrt{\rule{24px}{0.5px}}\) \(\rightarrow\) \(d = \)
Segment Length & Distance Formula Page 3
Time: 10 Mins
Let's solve these together. Fill out the coordinate templates on the left before attempting your final calculation!
PROBLEM 1: Segment connecting \(E(3, 2)\) and \(F(9, 10)\) Find Length
Fill-in Setup:
\(x_1 = 3\), \(y_1 = 2\)
\(x_2 = 9\), \(y_2 = 10\)
Horizontal: \(9 - 3 = \)
Vertical: \(10 - 2 = \)
Square, Sum, & Square Root:
PROBLEM 2: Segment connecting \(G(1, 1)\) and \(H(6, 13)\) Find Length
Fill-in Setup:
\(x_1 = 1\), \(y_1 = 1\)
\(x_2 = 6\), \(y_2 = 13\)
Horizontal: \(6 - 1 = \)
Vertical: \(13 - 1 = \)
Square, Sum, & Square Root:
Segment Length & Distance Formula Page 4
Time: 12 Mins
Calculate the segment lengths on your own. Remember: subtract to find flat roads, square them, sum them, and square root!
Problem 1 (Bronze): Points \(A(1, 1)\) and \(B(4, 5)\) X-Change = 3, Y-Change = 4
Problem 2 (Silver): Points \(P(2, 3)\) and \(Q(8, 11)\) Hint: Find horizontal/vertical differences first!
Problem 3 (Gold): Points \(R(1, 2)\) and \(S(13, 7)\) Hint: Squares are large numbers! Think \(12^2\) and \(5^2\). Find segment length \(RS\)
Did I correctly label first and second coordinates? Did I subtract correctly to get horizontal & vertical changes? Did I square both numbers (multiply by itself!) before adding? Did I take the final square root of the sum?
Segment Length & Distance Formula Page 5
Time: 3 Mins
Complete this final coordinate segment length problem independently. Show every step in your map clearly!
Your Goal: Find length of segment \(XY\) \(X(3, 4)\) and \(Y(9, 12)\)
How do you feel about applying the Distance Formula today?
😊 😐 ☹️
Distance Formula Reflection:
1. To keep my steps straight, I find it most helpful to:
2. One coordinate trick I have to watch out for is:
Segment Length & Distance Formula Page 6
Week 2 Day 5
Complete these three segment distance questions tonight. Use the helper diagrams on the left to set up coordinates correctly!
Problem 1
Point B: (4, 6) Point A: (1, 2) Length AB = ?
Your Step-by-Step Solving:
Problem 2
Point Q: (8, 9) Point P: (2, 1) Length PQ = ?
Your Step-by-Step Solving:
Problem 3
Point S: (5, 12) Point R: (0, 0) Length RS = ?
Your Step-by-Step Solving:
Segment Length & Distance Formula Page 7
Can just count units!
VS
Diagonal (Uncountable)
Requires formula!
Distance Formula
\(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)
Segment
Straight path connecting two endpoints.
Horiz Change
Subtract X coordinates: \((x_2-x_1)\).
Inverse Op
Operation that reverses another.
TEACHER FACILITATION COMPASS PAGE 2 FOCUS
What to Do Directions: Project Page 2. Walk students through the connection between the geometric rectangle model on the left and the algebraic equation on the right. Facilitate the 90-second Turn & Talk and check that students are physically speaking the frames.
Script: "Scholars, look at this. We have a rectangle with a width of 6 inches and a missing length \(x\). If the total area inside is 24, we set up our equation as \(6 \cdot x = 24\). Turn to your partner and explain: what inverse operation do we do to get \(x\) completely by itself?"
Check for Understanding (CFU):
"Why is division the correct step here? (Because the width 6 is multiplied by \(x\), and division undos multiplication)."
Anticipated Error Highlight:
Students may answer "\(x = 18\)" thinking they need to subtract 6 from 24. Guide them: "Is the 6 added to \(x\)? No, it's length times width! We must divide."
Segment Length & Distance Formula (TEACHER EDITION) Page 2
Teacher Facilitation Guide & Key — Replica Format
Time: 0:15 - 0:25 (10 min)
We are solving: Segment between \(A(1, 2)\) and \(B(4, 6)\)
STEP 1: Label Your Coordinates Endpoint coordinates
\(x_1\) 1
\(y_1\) 2
\(x_2\) 4
\(y_2\) 6
STEP 2 & 3: Find Horizontal & Vertical Changes Subtract coordinates
Horizontal (\(x_2 - x_1\)):
\(4 - 1 = \) \(\underline{\mathbf{3}}\)
Vertical (\(y_2 - y_1\)):
\(6 - 2 = \) \(\underline{\mathbf{4}}\)
STEP 4 & 5: Square Your Differences & Add Them Multiply by itself, then sum
Square them:
Horizontal: \(3^2 = \) \(\underline{\mathbf{9}}\)
Vertical: \(4^2 = \) \(\underline{\mathbf{16}}\)
Add them together:
\(\underline{\mathbf{9}} + \underline{\mathbf{16}} = \underline{\mathbf{25}}\)
STEP 6: Take the Square Root Solve for final segment length
\(\text{Distance} = \sqrt{\underline{\mathbf{25}}} \rightarrow d = \underline{\mathbf{5}}\)
TEACHER FACILITATION COMPASS PAGE 3 FOCUS
What to Do Directions: Lead direct instruction. Model the step-by-step math map carefully. Emphasize that we break down the giant formula into 6 tiny sub-steps. Have students physically write down each number inside their boxes on Page 3.
Script: "Scholars, let's walk down our roadmap. First, we label our coordinate values. Second, we find the horizontal change: what is \(4 - 1\)? Yes, 3! Third, what is vertical change \(6 - 2\)? Yes, 4! Now, let's square them both. What is 3 times itself? 9! What is 4 times itself? 16! Add them together to get 25. What is the square root of 25? It is 5!"
Break It Down Question:
"If you are walking horizontal from X = 1 to X = 4, how many steps is that? (3). If you are walking vertical from Y = 2 to Y = 6, how many steps is that? (4). Those are the horizontal and vertical legs of our right triangle!"
Check for Understanding (CFU):
"Why do we square the numbers in step 4? (Because the formula says $(x_2-x_1)^2$ and squaring removes negative values and helps us solve diagonal paths using the Pythagorean theorem!)."
Segment Length & Distance Formula (TEACHER EDITION) Page 3
Teacher Facilitation Guide & Key — Replica Format
Time: 0:25 - 0:35 (10 min)
PROBLEM 1: Segment connecting \(E(3, 2)\) and \(F(9, 10)\) Find Length
Fill-in Setup:
\(x_1 = 3\), \(y_1 = 2\)
\(x_2 = 9\), \(y_2 = 10\)
Horizontal: \(9 - 3 = \) \(\underline{\mathbf{6}}\)
Vertical: \(10 - 2 = \) \(\underline{\mathbf{8}}\)
Square, Sum, & Square Root:
Squares: \(6^2 = 36\) and \(8^2 = 64\)
Sum of Squares: \(36 + 64 = 100\)
\(d = \sqrt{100} \rightarrow d = 10\)
PROBLEM 2: Segment connecting \(G(1, 1)\) and \(H(6, 13)\) Find Length
Fill-in Setup:
\(x_1 = 1\), \(y_1 = 1\)
\(x_2 = 6\), \(y_2 = 13\)
Horizontal: \(6 - 1 = \) \(\underline{\mathbf{5}}\)
Vertical: \(13 - 1 = \) \(\underline{\mathbf{12}}\)
Square, Sum, & Square Root:
Squares: \(5^2 = 25\) and \(12^2 = 144\)
Sum of Squares: \(25 + 144 = 169\)
\(d = \sqrt{169} \rightarrow d = 13\)
TEACHER MONITORING & GUIDANCE PAGE 4 FOCUS
Monitoring Laps (What to Do):
Break It Down Scaffold:
"If students are stuck on Problem 1 squaring, ask: 'What does \(6^2\) mean? Does it mean \(6 \times 2\)? No, it means \(6 \times 6\), which is 36! Write that down inside your square box.'"
Script: "Scholars, as you work through these problems, fill out the coordinate box on the left first. That gives you your horizontal and vertical sides! Once you have those two numbers, squaring them and adding them gives you the square root target."
Segment Length & Distance Formula (TEACHER EDITION) Page 4
Teacher Facilitation Guide & Key — Replica Format
Time: 0:35 - 0:47 (12 min)
Problem 1 (Bronze): Points \(A(1, 1)\) and \(B(4, 5)\) X-Change = 3, Y-Change = 4
\(d = \sqrt{3^2 + 4^2} \rightarrow d = \sqrt{9 + 16} \rightarrow d = \sqrt{25}\)
\(\text{Distance } d = \underline{\mathbf{5}}\)
Problem 2 (Silver): Points \(P(2, 3)\) and \(Q(8, 11)\) Find horizontal & vertical changes first!
\(x\text{-change} = 8 - 2 = 6\), \(y\text{-change} = 11 - 3 = 8\)
\(d = \sqrt{6^2 + 8^2} \rightarrow d = \sqrt{36 + 64} \rightarrow d = \sqrt{100}\)
\(\text{Distance } d = \underline{\mathbf{10}}\)
Problem 3 (Gold): Points \(R(1, 2)\) and \(S(13, 7)\) Find segment length \(RS\)
\(x\text{-change} = 13 - 1 = 12\), \(y\text{-change} = 7 - 2 = 5\)
\(d = \sqrt{12^2 + 5^2} \rightarrow d = \sqrt{144 + 25} \rightarrow d = \sqrt{169}\)
\(\text{Distance } d = \underline{\mathbf{13}}\)
My Distance Self-Check (All Checked):
[✓] Labeled Points
[✓] Subtracted Correctly
[✓] Squared Values First
[✓] Solved Root
TEACHER MONITORING TRACKS PAGE 5 FOCUS
Anticipated Errors (Highlights):
Prob 2 Error: Students might write \(6^2 = 12\) instead of 36. Remind them: "A square is a number times *itself*, not times 2! \(6 \times 6 = 36\)."
Prob 3 Error: Students may get intimidated by \(12^2 = 144\) and \(5^2 = 25\). Encourage them: "You know these! Write \(12 \times 12\) on your margins."
CFU Checkups:
"Why is the distance in Problem 1 5, and not 7? (Because we take the root of 25. \(5 \times 5 = 25\))."
"Why can we never get a negative number inside the square root? (Because squaring any number always results in a positive!)"
Script: "Scholars, this is where you show your power as Distance Detectives! Go ahead and start. Label, subtract, square, sum, and root. Check your squares twice!"
Segment Length & Distance Formula (TEACHER EDITION) Page 5
Teacher Facilitation Guide & Key — Replica Format
Time: 0:47 - 0:50 (3 min)
Your Goal: Find length of segment \(XY\) \(X(3, 4)\) and \(Y(9, 12)\)
Step 1: Label: \(x_1=3, y_1=4\), \(x_2=9, y_2=12\)
Step 2: Subtraction: \(x\text{-diff} = 9-3 = 6\), \(y\text{-diff} = 12-4 = 8\)
Step 3: Square: \(6^2 = 36\), \(8^2 = 64\)
Step 4 & 5: Sum & Root: \(36 + 64 = 100 \rightarrow d = \sqrt{100} \rightarrow \underline{\mathbf{d = 10}}\)
Scholars rate their self-confidence level. Goal is 😊! [✓] Smiling Emoji Selected
Self-Reflection Blanks Answers:
1. To keep my steps straight, I find it most helpful to:
label the coordinates on my endpoints first before subtracting.
2. One coordinate trick I have to watch out for is:
matching X-values with X-values, and Y-values with Y-values when subtracting.
EXIT TICKET STRATEGY PAGE 6 FOCUS
What to Do Directions: Direct students to begin the Exit Ticket individually and in absolute silence. Scan coordinates setups. Check that students are writing $d = 10$. Collect tickets at 3 minutes.
Script: "Distance Detectives, final mission! Show me your coordinate-solving skills on the Exit Ticket. Remember your steps: subtract, square, add, and root. Absolute silence starts now."
Anticipated Error:
Students may mix X and Y subtraction (e.g. subtracting \(9 - 4\)). Guide them: "Coordinates are always \((x, y)\). Don't mix up your fruit cups! Keep X with X."
CFU Validation Check:
If student writes \(d = 10\), verify: \(XY = \sqrt{6^2 + 8^2} = \sqrt{100} = 10\). Fist bump and collect ticket.
Segment Length & Distance Formula (TEACHER EDITION) Page 6
Teacher Facilitation Guide & Key — Replica Format
Time: 0:50 - 0:55 (5 min)
Problem 1 Answers
Point B: (4, 6) Point A: (1, 2) Length AB = ?
\(x\Delta = 4 - 1 = 3\), \(y\Delta = 6 - 2 = 4\)
\(AB = \sqrt{3^2 + 4^2} \rightarrow \sqrt{9 + 16} = \sqrt{25}\)
\(\text{Answer: } \underline{\mathbf{AB = 5}}\)
Problem 2 Answers
Point Q: (8, 9) Point P: (2, 1) Length PQ = ?
\(x\Delta = 8 - 2 = 6\), \(y\Delta = 9 - 1 = 8\)
\(PQ = \sqrt{6^2 + 8^2} \rightarrow \sqrt{36 + 64} = \sqrt{100}\)
\(\text{Answer: } \underline{\mathbf{PQ = 10}}\)
Problem 3 Answers
Point S: (5, 12) Point R: (0, 0) Length RS = ?
\(x\Delta = 5 - 0 = 5\), \(y\Delta = 12 - 0 = 12\)
\(RS = \sqrt{5^2 + 12^2} \rightarrow \sqrt{25 + 144} = \sqrt{169}\)
\(\text{Answer: } \underline{\mathbf{RS = 13}}\)
HOMEWORK ROUTINE & WRAP-UP PAGE 7 FOCUS
What to Do Directions: Direct students to write down today's Homework. Point out the coordinate visual helpers next to each question on Page 7. Wrap up the coordinate unit with a solid high-five.
Script: "Scholars, your homework tonight features three segment mapping problems. Each one has its own coordinate helper box to help you visualize length, width, and distance. You are all completely equipped to crush this tonight! Keep your skills sharp, and I will see you tomorrow morning."
Key Homework Message:
"Homework is your shield. Bring these 3 problems completed tomorrow. They will be scanned at the door for 100% homework completion marks."
Daily Goal Met Check:
"We determined segment length using the distance formula! Scholars proved that we can find diagonal lengths on grids as easily as flat roads!"
Solving Area Problems with Equations (TEACHER EDITION) Page 7
Add X values (\(x_1 + x_2\)):
\(2 + 8 = \)
Divide sum by 2:
\(\frac{\text{sum}}{2} = \)
STEP 3: Find the Y-Coordinate Midpoint Average Y values
Add Y values (\(y_1 + y_2\)):
\(1 + 7 = \)
Divide sum by 2:
\(\frac{\text{sum}}{2} = \)
FINAL STEP: Write as a Coordinate Pair! M = (X-midpoint, Y-midpoint)
Midpoint \(M = \) ( , )
Segment Midpoints & Midpoint Formula Page 3
Time: 10 Mins
Let's solve these midpoint equations with our partner. Use the step boxes to keep coordinates organized!
PROBLEM 1: Midpoint of segment \(CD\) connecting \(C(1, 3)\) and \(D(5, 9)\) Find Midpoint
Fill-in Setup:
\(x_1 = 1\), \(y_1 = 3\)
\(x_2 = 5\), \(y_2 = 9\)
Sum of X: \(1 + 5 = \)
Sum of Y: \(3 + 9 = \)
Divide sums by 2 & Assemble:
PROBLEM 2: Midpoint of segment \(EF\) connecting \(E(2, 4)\) and \(F(10, 8)\) Find Midpoint
Fill-in Setup:
\(x_1 = 2\), \(y_1 = 4\)
\(x_2 = 10\), \(y_2 = 8\)
Sum of X: \(2 + 10 = \)
Sum of Y: \(4 + 8 = \)
Divide sums by 2 & Assemble:
Segment Midpoints & Midpoint Formula Page 4
Time: 12 Mins
Solve these segment midpoint problems on your own. Remember: add first coordinates and divide, then add second coordinates and divide!
Problem 1 (Bronze): Points \(A(1, 5)\) and \(B(7, 9)\) Find Midpoint \(M\)
Problem 2 (Silver): Points \(P(3, 2)\) and \(Q(9, 12)\) Hint: Sum and average X values first, then do the same for Y!
Problem 3 (Gold): Points \(R(0, 4)\) and \(S(12, 14)\) Hint: Write your final answer in coordinate format: \((x, y)\).
Did I correctly label first and second coordinates? Did I add coordinates instead of subtracting? Did I divide both X and Y sums by 2? Did I write my final midpoint as a coordinate pair \((x, y)\)?
Segment Midpoints & Midpoint Formula Page 5
Time: 3 Mins
Complete this final coordinate midpoint problem independently. Show all steps in your map clearly!
Your Goal: Find midpoint of segment \(XY\) \(X(2, 6)\) and \(Y(8, 12)\)
How do you feel about applying the Midpoint Formula today?
😊 😐 ☹️
Midpoint Formula Reflection:
1. To avoid mixing up distance and midpoint calculations, I remember that midpoint:
2. The most common mistake I have to watch out for is:
Segment Midpoints & Midpoint Formula Page 6
Week 2 Day 6
Complete these three segment midpoint questions tonight. Use the helper diagrams on the left to set up coordinates correctly!
Problem 1
Point B: (7, 9) Point A: (1, 3) Midpoint AB = ?
Your Step-by-Step Solving:
Problem 2
Point Q: (8, 11) Point P: (2, 5) Midpoint PQ = ?
Your Step-by-Step Solving:
Problem 3
Point S: (12, 8) Point R: (4, 2) Midpoint RS = ?
Your Step-by-Step Solving:
Segment Midpoints & Midpoint Formula Page 7
The Midpoint Formula:
\(M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)\)
Midpoint
Exact middle point of a line segment.
Coord Pair
An ordered pair \((x, y)\) showing a position.
Bisect
To cut a segment into exactly two equal parts.
Average
Add values together and divide by 2.
TEACHER FACILITATION COMPASS PAGE 2 FOCUS
What to Do Directions: Project Page 2. Explain that a midpoint is a single coordinate that acts as the physical middle. Emphasize that coordinates are dual-natured (X and Y), so we must average the X-coordinates and Y-coordinates separately.
Script: "Scholars, look at this. We have Point A and Point B on our grid. The midpoint, \(M\), is the exact balance point in the center. Because coordinates are made of X and Y, we must find the middle of the X's, and then the middle of the Y's! That's why the formula is written as one big coordinate pair."
Check for Understanding (CFU):
"If endpoints are $(2, 4)$ and $(8, 8)$, how do we find the X-midpoint? (Add X's: $2 + 8 = 10$, then divide by 2 to get 5!)."
Vocabulary Focus:
"To *bisect* means to cut in half. The midpoint bisects the segment into two congruent halves!"
Segment Midpoints & Midpoint Formula (TEACHER EDITION) Page 2
Teacher Facilitation Guide & Key — Replica Format
Time: 0:15 - 0:25 (10 min)
We are solving: Midpoint of segment \(AB\) with endpoints \(A(2, 1)\) and \(B(8, 7)\)
STEP 1: Label Your Endpoints Identify x and y values
\(x_1\) 2
\(y_1\) 1
\(x_2\) 8
\(y_2\) 7
STEP 2: Find the X-Coordinate Midpoint Average X values
Add X values (\(x_1 + x_2\)):
\(2 + 8 = \) \(\underline{\mathbf{10}}\)
Divide sum by 2:
\(\frac{10}{2} = \) \(\underline{\mathbf{5}}\)
STEP 3: Find the Y-Coordinate Midpoint Average Y values
Add Y values (\(y_1 + y_2\)):
\(1 + 7 = \) \(\underline{\mathbf{8}}\)
Divide sum by 2:
\(\frac{8}{2} = \) \(\underline{\mathbf{4}}\)
FINAL STEP: Write as a Coordinate Pair! M = (X-midpoint, Y-midpoint)
Midpoint \(M = \) \(\mathbf{(}\) \(\underline{\mathbf{5}}\) \(\mathbf{,}\) \(\underline{\mathbf{4}}\) \(\mathbf{)}\)
TEACHER FACILITATION COMPASS PAGE 3 FOCUS
What to Do Directions: Lead direct instruction. Focus heavily on how we find X-midpoint and Y-midpoint separately. Walk through adding the coordinates together (not subtracting!), and then dividing by 2. Fill in the blanks exactly as shown.
Script: "Scholars, let's solve this step-by-step. First, label your values. Second, we add X coordinates: \(2 + 8 = 10\). Now, divide 10 by 2. What is half of 10? Yes, 5! That is our X-midpoint! Write 5 in your template. Now, let's do Y: \(1 + 7 = 8\). Half of 8 is 4! That's our Y-midpoint. So our final midpoint coordinates are \((5, 4)\)!"
Break It Down Question:
"If you have 2 dollars, and your friend has 8 dollars, and you pool your money and share it equally... how many dollars does each of you get? (5!). That is taking the average!"
Check for Understanding (CFU):
"What is the biggest difference between Distance Formula and Midpoint Formula? (Distance formula uses *subtraction* and squaring. Midpoint formula uses *addition* and division by 2!)."
Segment Midpoints & Midpoint Formula (TEACHER EDITION) Page 3
Teacher Facilitation Guide & Key — Replica Format
Time: 0:25 - 0:35 (10 min)
PROBLEM 1: Midpoint of segment \(CD\) connecting \(C(1, 3)\) and \(D(5, 9)\) Find Midpoint
Fill-in Setup:
\(x_1 = 1\), \(y_1 = 3\)
\(x_2 = 5\), \(y_2 = 9\)
Sum of X: \(1 + 5 = \) \(\underline{\mathbf{6}}\)
Sum of Y: \(3 + 9 = \) \(\underline{\mathbf{12}}\)
Divide sums by 2 & Assemble (Key):
\(X\text{-midpoint} = \frac{6}{2} = 3\)
\(Y\text{-midpoint} = \frac{12}{2} = 6\)
Midpoint \(M = (3, 6)\)
PROBLEM 2: Midpoint of segment \(EF\) connecting \(E(2, 4)\) and \(F(10, 8)\) Find Midpoint
Fill-in Setup:
\(x_1 = 2\), \(y_1 = 4\)
\(x_2 = 10\), \(y_2 = 8\)
Sum of X: \(2 + 10 = \) \(\underline{\mathbf{12}}\)
Sum of Y: \(4 + 8 = \) \(\underline{\mathbf{12}}\)
Divide sums by 2 & Assemble (Key):
\(X\text{-midpoint} = \frac{12}{2} = 6\)
\(Y\text{-midpoint} = \frac{12}{2} = 6\)
Midpoint \(M = (6, 6)\)
TEACHER MONITORING & GUIDANCE PAGE 4 FOCUS
Monitoring Laps (What to Do):
Break It Down Scaffold:
"If students subtract coordinates (e.g. \(5 - 1 = 4\) and divide to get 2), say: 'No! Distance uses subtraction, but Midpoint is an *average*, so we must *add* the values first! \(1 + 5 = 6\). What is half of 6?'"
Script: "Scholars, as you solve, make sure you are checkmarking your checklists. Watch out for the operation sign! Midpoint is addition! We add coordinates and divide by 2."
Segment Midpoints & Midpoint Formula (TEACHER EDITION) Page 4
Teacher Facilitation Guide & Key — Replica Format
Time: 0:35 - 0:47 (12 min)
Problem 1 (Bronze): Points \(A(1, 5)\) and \(B(7, 9)\) Find Midpoint \(M\)
\(X_{\text{mid}} = \frac{1+7}{2} = \frac{8}{2} = 4\), \(Y_{\text{mid}} = \frac{5+9}{2} = \frac{14}{2} = 7\)
\(\text{Midpoint } M = \underline{\mathbf{(4, 7)}}\)
Problem 2 (Silver): Points \(P(3, 2)\) and \(Q(9, 12)\) Average X and Y values separately!
\(X_{\text{mid}} = \frac{3+9}{2} = \frac{12}{2} = 6\), \(Y_{\text{mid}} = \frac{2+12}{2} = \frac{14}{2} = 7\)
\(\text{Midpoint } M = \underline{\mathbf{(6, 7)}}\)
Problem 3 (Gold): Points \(R(0, 4)\) and \(S(12, 14)\) Find Midpoint \(RS\)
\(X_{\text{mid}} = \frac{0+12}{2} = \frac{12}{2} = 6\), \(Y_{\text{mid}} = \frac{4+14}{2} = \frac{18}{2} = 9\)
\(\text{Midpoint } M = \underline{\mathbf{(6, 9)}}\)
My Midpoint Self-Check (All Checked):
[✓] Labeled Points
[✓] Added Coordinates
[✓] Divided Sums by 2
[✓] Wrote as Coordinate
TEACHER MONITORING TRACKS PAGE 5 FOCUS
Anticipated Errors (Highlights):
Prob 2 Error: Students might divide the sum of X by 3 because there are three points in the segment label. Guide them: "Average of two points is always dividing by 2! We are only averaging two numbers at a time."
Prob 3 Error: Students may write \(M = (12, 18)\) forgetting to divide by 2 entirely. Prompt them: "Look at your sums! Half of 12 is 6, and half of 18 is 9."
CFU Checkups:
"For Problem 1, does the point \((4, 7)\) look like it is exactly halfway between \((1, 5)\) and \((7, 9)\)? (Yes, 4 is halfway between 1 and 7, and 7 is halfway between 5 and 9!)."
"Why do we always divide by 2? (Because we are finding the middle of *two* points)."
Script: "Midpoint Matchers, show your individual power! Complete each level. Check that you are writing your answers in parentheses. Make sure you add coordinates, do not subtract!"
Segment Midpoints & Midpoint Formula (TEACHER EDITION) Page 5
Teacher Facilitation Guide & Key — Replica Format
Time: 0:47 - 0:50 (3 min)
Your Goal: Find midpoint of segment \(XY\) \(X(2, 6)\) and \(Y(8, 12)\)
Step 1: Label: \(x_1=2, y_1=6\), \(x_2=8, y_2=12\)
Step 2: Sums: \(x\text{-sum} = 2+8 = 10\), \(y\text{-sum} = 6+12 = 18\)
Step 3: Average: \(X_{\text{mid}} = \frac{10}{2} = 5\), \(Y_{\text{mid}} = \frac{18}{2} = 9\)
Step 4: Coordinate Pair: \(M = (\underline{\mathbf{5}}, \underline{\mathbf{9}})\)
Scholars rate their self-confidence level. Goal is 😊! [✓] Smiling Emoji Selected
Self-Reflection Blanks Answers:
1. To avoid mixing up distance and midpoint calculations, I remember that midpoint:
uses addition to find averages, while distance uses subtraction and squares.
2. The most common mistake I have to watch out for is:
forgetting to divide X and Y sums by 2 at the very end.
EXIT TICKET STRATEGY PAGE 6 FOCUS
What to Do Directions: Direct students to begin the Exit Ticket individually and in absolute silence. Scan coordinate averages. Check that students are writing $M = (5, 9)$. Collect tickets at 3 minutes.
Script: "Midpoint Matchers, final stretch! Pencils up. Show me how you find the halfway coordinates between \((2, 6)\) and \((8, 12)\). Absolute silent work starts now."
Anticipated Error:
Students may write the midpoint as \((10, 18)\). Write on board: "An endpoint sum of 10 and 18 means we must slice them both in half to find the actual midpoint! Divide by 2!"
CFU Validation Check:
If student writes \(M = (5, 9)\), verify: \(X = \frac{10}{2} = 5, Y = \frac{18}{2} = 9\). Quiet high-five and collect ticket.
Segment Midpoints & Midpoint Formula (TEACHER EDITION) Page 6
Teacher Facilitation Guide & Key — Replica Format
Time: 0:50 - 0:55 (5 min)
Problem 1 Answers
Point B: (7, 9) Point A: (1, 3) Midpoint AB = ?
\(X_{\text{mid}} = \frac{1+7}{2} = 4\)
\(Y_{\text{mid}} = \frac{3+9}{2} = 6\)
\(\text{Answer: } \underline{\mathbf{M = (4, 6)}}\)
Problem 2 Answers
Point Q: (8, 11) Point P: (2, 5) Midpoint PQ = ?
\(X_{\text{mid}} = \frac{2+8}{2} = 5\)
\(Y_{\text{mid}} = \frac{5+11}{2} = 8\)
\(\text{Answer: } \underline{\mathbf{M = (5, 8)}}\)
Problem 3 Answers
Point S: (12, 8) Point R: (4, 2) Midpoint RS = ?
\(X_{\text{mid}} = \frac{4+12}{2} = 8\)
\(Y_{\text{mid}} = \frac{2+8}{2} = 5\)
\(\text{Answer: } \underline{\mathbf{M = (8, 5)}}\)
HOMEWORK ROUTINE & WRAP-UP PAGE 7 FOCUS
What to Do Directions: Direct students to write down today's Homework. Explain that each question features its own visual coordinate helper diagram to keep coordinates aligned. Praise their incredible coordinate geometry work today!
Script: "Scholars, your homework tonight features three midpoint mapping problems. Each one has its own coordinate helper box to help you visualize X and Y coordinates. You've mastered area, segment length, and midpoints in just three days! Keep up this incredible momentum."
Key Homework Message:
"Homework is your seal. Do these 3 tonight and bring them completed tomorrow morning. Excellent job!"
Daily Goal Met Check:
"We found segment midpoints using the midpoint formula! Scholars proved they can find the exact center coordinate between any two points!"
Segment Midpoints & Midpoint Formula (TEACHER EDITION) Page 7
Part A: Model the Area with a Grid Model Area = Width × Length
\(5 \cdot x\)
\(5 \cdot 2\)
\(x\) \(2\) 5
How do we find area?
Area is computed by multiplying width times length. We write it with parentheses to represent the distributive property.
\(\text{Area} = \text{width} \cdot (\text{length expression})\)
Part B: Set Up Your Linear Equation and Solve for \(x\) Distribute width first!
Area Solving Steps: 1. Write equation: \(5(x + 2) = 30\) 2. Distribute: multiply width to BOTH terms inside parentheses 3. Subtract constant, then divide to find final \(x\)
Your Active Solving Area:
Mid-Unit Quiz Assessment Page 3
Part A: Label All Four Sides of the Rectangle (Completed Key)
\(2x + 2\) \(2x + 2\) 4 4 P = 24
How do we find perimeter?
Perimeter is the sum of **all four sides**: add length + width + length + width.
\(P = \text{side 1} + \text{side 2} + \text{side 3} + \text{side 4}\)
Part B: Set Up Your Linear Equation and Solve for \(x\) (Completed Key)
Perimeter Steps (Completed):
[✓] 1. Set up sum equation
[✓] 2. Combine like terms
[✓] 3. Isolate and solve
Your Active Solving Area:
\(4 + 4 + (2x + 2) + (2x + 2) = 24\)
Combine: \(4x + 12 = 24\)
Subtract 12: \(4x = 12\)
Divide by 4: \(\frac{4x}{4} = \frac{12}{4}\)
\(x = 3\)
GOAL 2 MONITORING & GRADING PAGE 2 KEY
What to Do Directions: Walk around the room as students attempt Page 2. Ensure they are labeling all four sides of the rectangle model in Part A before solving in Part B. Check that students combined the X values to get \(4x\), and the numerical sides to get \(12\).
Script: "Scholars, remember! A rectangle has four sides. If the left side is 4, what is the right side? Yes, it must be 4 too! Write that down. If the top side is \(2x+2\), what is the bottom side? \(2x+2\)! Now add all four sides together and make them equal 24."
Anticipated Error Highlight:
Students may only add two sides: \(4 + (2x + 2) = 24\), leading to \(x = 9\). Remind them immediately: "Is a rectangle made of only two sides? No, four! Walk all the way around the yard!"
Grading Rubric:
Give 2 points for correct rectangle side labeling in Part A, 1 point for setting up the equation, 1 point for combining like terms, and 1 point for getting \(x = 3\). Total: 5 points.
Mid-Unit Quiz Assessment (TEACHER EDITION) Page 2
Teacher Facilitation Guide & Key — Replica Format
Time: 15 Mins
A rectangle has a total Area of 30 square inches. Width is 5 inches, and length is \(x + 2\).
Part A: Model the Area with a Grid Model (Completed Key)
\(5x\)
\(10\)
\(x\) \(2\) 5
How do we find area?
Area is computed by multiplying width times length. We write it with parentheses to represent the distributive property.
\(\text{Area} = \text{width} \cdot (\text{length expression})\)
Part B: Set Up Your Linear Equation and Solve for \(x\) (Completed Key)
Area Steps (Completed):
[✓] 1. Set up formula equation
[✓] 2. Distribute outer width
[✓] 3. Subtract and divide
Your Active Solving Area:
\(5(x + 2) = 30\)
Distribute 5: \(5x + 10 = 30\)
Subtract 10: \(5x = 20\)
Divide by 5: \(\frac{5x}{5} = \frac{20}{5}\)
\(x = 4\)
GOAL 3 MONITORING & GRADING PAGE 3 KEY
What to Do Directions: Administer Page 3. Ensure students multiply the outer 5 times BOTH \(x\) and 2 inside parentheses. Check that they do not add 2 and 30. Collect all quizzes at the 45-minute mark and offer reassurance.
Script: "Last page, scholars! On Page 3, you are multiplying length times width to find the area of the rectangle. Draw your rainbow arrows to multiply the 5 by \(x\), and then the 5 by 2. That breaks open the parentheses so you can solve!"
Anticipated Error (Highlight):
Students may write \(5x + 2 = 30\) by forgetting to multiply 5 by the 2. Remind them: "The width 5 must scale *both* rooms in the apartment! \(5 \times 2 = 10\)."
Grading Rubric:
Give 1 point for setting up \(5(x+2) = 30\), 2 points for correct distribution (\(5x + 10 = 30\)), and 2 points for isolating and solving for \(x = 4\). Total: 5 points. Quiz Grand Total: 16 points.
Mid-Unit Quiz Assessment (TEACHER EDITION) Page 3
5 × 2x = 10x
5 × 2 = 10
2x 2 5
How to solve Area:
Multiply the width outside by BOTH values inside the length parentheses: \(5(2x + 2) = 40\).
Step 2: Fill in the Blanks to solve the Equation! Just get through each line!
1. Set up formula: 5 ( 2x + 2 ) = 40
2. Distribute 5: x + = 40
3. Subtract constant: - -
4. Simple equation: x =
5. Divide & solve: ÷ ÷
6. Final Answer: x =
Unit Test Assessment Page 3
Coordinate Distance
Our Distance Mission: Find the length of the segment connecting point \(A(2, 3)\) and point \(B(8, 11)\). Complete the 3 easy arithmetic steps below to solve!
Step 1: Horizontal & Vertical Differences
\(x\)-coordinates are: \(2\) and \(8\)
Horizontal change = \(8 - 2 = \)
\(y\)-coordinates are: \(3\) and \(11\)
Vertical change = \(11 - 3 = \)
Roadmap Helper:
Once you find horizontal and vertical changes, put them in Step 2!
Step 2: Solve the 3 Arithmetic Questions below! Calculators are completely allowed!
1. Multiply both numbers by themselves (Square them):
Horizontal: \(6 \times 6 =\)
Vertical: \(8 \times 8 =\)
2. Add your two squared answers together:
3. Take the final Square Root of the sum:
d = \(\sqrt{\vphantom{1}}\) =
Unit Test Assessment Page 4
Part A: Label All Four Sides of the Rectangle (Completed Key)
\(3x + 3\) \(3x + 3\) Width: 6 Width: 6 P = 36
How do we find perimeter?
Perimeter is the sum of **all four sides**: add length + width + length + width.
\(P = \text{side 1} + \text{side 2} + \text{side 3} + \text{side 4}\)
Part B: Set Up Your Linear Equation and Solve for \(x\) (Completed Key)
1. Set up sum: 6 + 6 + ( 3x + 3 ) + ( 3x + 3 ) = 36
2. Combine terms: 6 x + 18 = 36
3. Subtract constant: - 18 - 18
4. Simple equation: 6 x = 18
5. Divide & solve: ÷ 6 ÷ 6
6. Final Answer: x = 3
GOAL 2 MONITORING & GRADING PAGE 2 KEY
What to Do Directions: Walk around the room. For 9th-grade SPED independent comprehension, the algebra equation has been laid out as a linear template. Scan to make sure students are filling in the boxes chronologically.
Script: "Scholars, look at Page 2. The formula steps are pre-written as blank blanks! Follow the roadmap on Part B. Line 2 combines the \(x\)'s together and the plain numbers together."
Anticipated Error Highlight:
In step 2, students might combine \(6+6+3+3 = 18\) and add the \(3x + 3x = 6x\) correctly, but misalign the subtraction. Point to the minus signs: "Subtract 18 from 36 first!"
Grading Rubric:
Give 1 point per correctly filled-in line in Part B (Lines 2, 3, 4, 5, and 6 are worth 1 pt each). Total: 5 points.
Unit Test Assessment (TEACHER EDITION) Page 2
Teacher Facilitation Guide & Key — Replica Format
Time: 15 Mins
Our Area Mission: A rectangle has a total Area of 40 square inches.
• The **Width** of the rectangle is **5 inches**.
• The **Length** is split into two parts: **\(2x\)** and **\(2\)**. Find \(x\).
Part A: Model the Area with a Grid Model (Completed Key)
5 × 2x = 10x
5 × 2 = 10
2x 2 5
How do we find area?
Area is computed by multiplying width times length. We write it with parentheses to represent the distributive property.
\(\text{Area} = \text{width} \cdot (\text{length expression})\)
Part B: Set Up Your Linear Equation and Solve for \(x\) (Completed Key)
1. Set up formula: 5 ( 2x + 2 ) = 40
2. Distribute 5: 10 x + 10 = 40
3. Subtract constant: - 10 - 10
4. Simple equation: 10 x = 30
5. Divide & solve: ÷ 10 ÷ 10
6. Final Answer: x = 3
GOAL 3 MONITORING & GRADING PAGE 3 FOCUS
What to Do Directions: Walk around the room. Check if students are using the pre-calculated areas shown inside the visual model to fill out Line 2 of the algebraic template on Page 3.
Script: "Scholars, look at Step 1 area model. The areas of the split boxes are pre-calculated for you! It shows that \(5 \times 2x = 10x\) and \(5 \times 2 = 10\). Write those two numbers directly on Line 2 of your equation solver!"
Anticipated Error Highlight:
Students may struggle with dividing by 10. Assure them: "If \(10x = 30\), we divide 30 by 10. Think: how many 10s are inside 30? (3!)."
Grading Rubric:
Give 1 point per correctly filled-in line in Part B (Lines 2, 3, 4, 5, and 6 are worth 1 pt each). Total: 5 points.
Unit Test Assessment (TEACHER EDITION) Page 3
Teacher Facilitation Guide & Key — Replica Format
Time: 15 Mins
Points: \(A(2, 3)\) and point \(B(8, 11)\).
Step 1: Horizontal & Vertical Differences (Completed Key):
\(x\)-coordinates: 2 and 8
Horizontal change: \(8 - 2 = \) \(\underline{\mathbf{6}}\)
\(y\)-coordinates: 3 and 11
Vertical change: \(11 - 3 = \) \(\underline{\mathbf{8}}\)
Roadmap Helper:
Once you find horizontal and vertical changes, put them in Step 2!
Step 2: Solve the 3 Arithmetic Questions below! (Completed Key)
1. Multiply both numbers by themselves (Square them):
Horizontal: \(6 \times 6 =\)
36
Vertical: \(8 \times 8 =\)
64
2. Add your two squared answers together:
36 + 64 = 100
3. Take the final Square Root of the sum:
d = \(\sqrt{\underline{\mathbf{100}}}\) = 10
GOAL 4 MONITORING & GRADING PAGE 4 FOCUS
What to Do Directions: Administer Page 4. Check horizontal and vertical subtraction setups. Assist students with arithmetic calculations. Instruct them to perform the steps in order: Step 1, then Step 2, then Step 3.
Script: "Distance Detectives, final step! Page 4 has three simple arithmetic questions to find diagonal segment length. Step 1: find horizontal change and vertical change. Step 2: multiply them by themselves! Step 3: add, and Step 4: take square root!"
Anticipated Error Highlight:
Students may add 6 and 8 in step 1, getting 14, and then squaring 14. Prompt them: "Square them first separately! \(6 \times 6 = 36\) and \(8 \times 8 = 64\)."
Grading Rubric:
Give 2 points for Part A coordinate subtractions, 1 point for squaring, 1 point for sum, and 1 point for final root of 10. Page total: 5 points. Exam Total: 21 points.
Unit Test Assessment (TEACHER EDITION) Page 4
