Algebra Blueprint Worksheet
Project File: 01-EQUATIONS
ALGEBRA BLUEPRINT
TOPIC: MULTISTEP EQUATIONS (VARIABLES ON BOTH SIDES & DISTRIBUTIVE PROPERTY)
NAME:
DATE:
PERIOD:
The Builder's Key & Specification Guide
Cyan Boxes = Variable Terms (with x)
Amber Boxes = Constant Terms (numbers only)
STEP 1 (Clear the Frame): Distribute to clear parentheses first.
STEP 2 (Group Materials): Use inverse operations to collect variables on one side, constants on the other.
STEP 3 (Verify & Solve): Combine terms and divide to find the final value of \(x\).
1 Problem #1 (Standard Assembly) Solve for \(x\)
3 ( x - 2 ) = 2 ( x + 1 )
SHOW ALL WORKSTEPS (DISTRIBUTE, ISOLATE, SOLVE) IN THE SPACE BELOW:
Ensure final value is a whole integer.
FINAL ANSWER: x =
2 Problem #2 (Expanded Coefficients) Solve for \(x\)
4 ( 2x + 1 ) = 3 ( x + 8 )
SHOW ALL WORKSTEPS (DISTRIBUTE, ISOLATE, SOLVE) IN THE SPACE BELOW:
Ensure final value is a whole integer.
FINAL ANSWER: x =
ALGEBRA BLUEPRINT WORKPLACE • SPEC 01 PAGE 1 OF 2
THE ALGEBRA WORK SITE
COMPLETE PROBLEMS #3, #4, AND #5 TO FINALIZE THE BLUEPRINT
STAGE: STRUCTURAL ANALYSIS
3 Problem #3 (Complex Frame Addition) Solve for \(x\)
2 ( 3x - 5 ) + 4 = 3 ( x + 1 )
SHOW ALL WORKSTEPS IN THE SPACE BELOW:
Careful! Combine like terms on the left side after distributing.
FINAL ANSWER: x =
4 Problem #4 (The Inverse Balance) Solve for \(x\)
5 ( x - 1 ) = 2 ( 2x + 3 )
SHOW ALL WORKSTEPS IN THE SPACE BELOW:
Ensure variables are collected on one side.
FINAL ANSWER: x =
5 Problem #5 (Negative Distribution Challenge) Solve for \(x\)
-2 ( x - 4 ) = 3 ( x - 1 ) - 9
SHOW ALL WORKSTEPS IN THE SPACE BELOW:
Watch the sign! Multiplying a negative number by a negative number results in a positive.
FINAL ANSWER: x =
Blueprint Inspection Checklist
I distributed correctly. I combined variable terms. I checked my integer solution.
ALGEBRA BLUEPRINT WORKPLACE • SPEC 01 PAGE 2 OF 2
Algebra Blueprint Answer Key
Project File: 01-EQUATIONS • SOLUTION SCHEME
ALGEBRA BLUEPRINT (ANSWER KEY)
TOPIC: MULTISTEP EQUATIONS (VARIABLES ON BOTH SIDES & DISTRIBUTIVE PROPERTY)
TEACHER REVIEW VERIFIED PATHS INTEGER ANSWERS
Grading Specs & Rubric Recommendations
Total Points: 50 (10 pts each)
Distribute step: 3 points
Isolation step: 4 points
Final Solve step: 3 points
Cyan highlight: Variable Terms (with x). Students should combine these on one side first.
Amber highlight: Constant Terms (numbers). Students should isolate these on the opposite side.
Red notation below represents full step-by-step structural paths for grading.
1 Problem #1 (Standard Assembly) Solution Scheme
3 ( x - 2 ) = 2 ( x + 1 )
1. Distribute: \(3x - 6 = 2x + 2\)
2. Move Variable: Subtract \(2x\) from both sides:
\(3x - 2x - 6 = 2 \implies x - 6 = 2\)
3. Move Constant: Add \(6\) to both sides:
\(x = 2 + 6 \implies x = 8\)
Common Error: Multiplying only the first term inside parentheses.
FINAL ANSWER: x = 8
2 Problem #2 (Expanded Coefficients) Solution Scheme
4 ( 2x + 1 ) = 3 ( x + 8 )
1. Distribute: \(8x + 4 = 3x + 24\)
2. Move Variable: Subtract \(3x\) from both sides:
\(8x - 3x + 4 = 24 \implies 5x + 4 = 24\)
3. Move Constant & Solve: Subtract \(4\) from both sides: \(5x = 20\)
Divide both sides by \(5\): \(x = 4\)
Common Error: Forgetting to distribute to the second term inside parentheses (e.g. \(4 \cdot 1\) or \(3 \cdot 8\)).
FINAL ANSWER: x = 4
ALGEBRA BLUEPRINT WORKPLACE • SOLUTION SCHEME PAGE 1 OF 2
THE ALGEBRA WORK SITE (ANSWERS)
COMPLETE SOLUTIONS FOR PROBLEMS #3, #4, AND #5
STAGE: DETAILED DEBRIEF
3 Problem #3 (Complex Frame Addition) Solution Scheme
2 ( 3x - 5 ) + 4 = 3 ( x + 1 )
1. Distribute: \(6x - 10 + 4 = 3x + 3\)
2. Combine constants on Left: \(6x - 6 = 3x + 3\)
3. Group variables (subtract \(3x\)): \(3x - 6 = 3\)
4. Group constants (add \(6\)): \(3x = 9\)
5. Isolate variable (divide by \(3\)): \(x = 3\)
Common Trap: Combining \(-10 + 4\) as \(-14\) instead of \(-6\).
FINAL ANSWER: x = 3
4 Problem #4 (The Inverse Balance) Solution Scheme
5 ( x - 1 ) = 2 ( 2x + 3 )
1. Distribute: \(5x - 5 = 4x + 6\)
2. Move variables (subtract \(4x\) from both sides): \(x - 5 = 6\)
3. Move constants (add \(5\) to both sides): \(x = 6 + 5 \implies x = 11\)
Check: \(5(11-1) = 5(10) = 50\) and \(2(2(11)+3) = 2(25) = 50\). Verified!
FINAL ANSWER: x = 11
5 Problem #5 (Negative Distribution Challenge) Solution Scheme
-2 ( x - 4 ) = 3 ( x - 1 ) - 9
1. Distribute with Negative: \(-2x + 8 = 3x - 3 - 9\)
2. Combine constants on Right: \(-2x + 8 = 3x - 12\)
3. Group variables (add \(2x\) to both sides): \(8 = 5x - 12\)
4. Group constants (add \(12\) to both sides): \(20 = 5x\)
5. Isolate & solve: \(x = 4\)
Common Trap: Forgetting that \(-2 \cdot (-4) = +8\) (sign change).
FINAL ANSWER: x = 4
Diagnostic Assessment Guide
If student answers with \(x = -4\) or fractions, diagnose distributive property sign multiplication or incorrect sign change during isolation operations.
ALGEBRA BLUEPRINT WORKPLACE • SOLUTION SCHEME PAGE 2 OF 2
