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
