Magnitude Mastery Worksheet
Algebra I & II Expressions & Number Systems
Magnitude Mastery
Evaluating Absolute Value Expressions & Order of Operations
Name:
Date:
Period:
Geometric Meaning \( |x| \) is the non-negative distance from \( 0 \) to \( x \) on a number line. \( |x| \ge 0 \).
Grouping Hierarchy Treat absolute value bars like parentheses: simplify the entire expression inside first!
External Signs A negative sign outside the bars persists: \( -|x| \) is always \( \le 0 \). Never distribute into bars!
A Warm-Up: Core Numerical Values
Simplify each expression completely
- \( | -23.4 | \)
Answer:
- \( -| -58 | \)
Answer:
- \( | 4.15 - 9.75 | \)
Answer:
- \( \left| -\frac{11}{6} \right| - \left| \frac{5}{6} \right| \)
Answer:
B Order of Operations (PEMDAS)
Show each simplification step clearly
- \( 16 - 4 \, | 3 - 8 | \)
Show work
Final Value:
- \( 3 \, | -6 \cdot 2 | - | -5 |^2 \)
Show work
Final Value:
- \( \dfrac{| -24 + 6 |}{| -2 |^3 - 2} \)
Show work
Final Value:
- \( -2 \, | 7 - 19 | + 5 \, | -3 - (-10) | \)
Show work
Final Value:
C Single-Variable Evaluation
Evaluate for \( x = -4 \)
- \( 5 \, | 2x + 3 | - 11 \)
Final Value:
- \( -| x^2 - 25 | + 3x \)
Final Value:
Magnitude Mastery • High School Algebra Page 1 of 2
Magnitude Mastery Part II • Advanced Applications
Student Name:
D Multi-Variable Substitution
Evaluate using \( a = -3 \), \( b = 5 \), \( c = -2 \)
- \( | 2a - 3c | + | b \cdot c | \)
Final Value:
- \( \dfrac{| a^2 - b^2 |}{| a - c |} \)
Final Value:
- \( -3 \, | ab + 4c | - | c |^3 \)
Final Value:
- \( | a - b | - | b - a | \)
Final Value:
E Diagnostic Analysis & Concept Exploration
Critical reasoning
15. Spot the Flaw Error Analysis
A student simplified \( -2 \, | 4 - 9 | \) as follows:
Step 1: \( |-8 + 18| \) → Step 2: \( |10| = 10 \)
Explain the student's conceptual mistake and find the correct value:
Correct Value:
16. Triangle Inequality Check Exploration
Let \( x = -7 \) and \( y = 4 \). Evaluate both expressions:
\( |x + y| \)
\( |x| + |y| \)
Show work for both, then circle which statement is true:
\( |x + y| = \) _____
\( |x| + |y| = \) _____
[ ] \( |x+y| < |x|+|y| \) [ ] \( |x+y| = |x|+|y| \)
F Real-World Application: Engineering Tolerance
Quality Assurance
An aerospace manufacturer specifies that the diameter \( d \) of a precision titanium pin must be exactly \( 15.000\text{ mm} \). The absolute deviation from the target is given by the formula: \( \text{Error} = |d - 15.000| \). A part passes inspection if \( \text{Error} \le 0.025\text{ mm} \).
Pin A: \( d = 15.018\text{ mm} \)
\( \text{Error} = |15.018 - 15| = \) _____
Decision:
PASS REJECT
Pin B: \( d = 14.968\text{ mm} \)
\( \text{Error} = |14.968 - 15| = \) _____
Decision:
PASS REJECT
Pin C: \( d = 15.025\text{ mm} \)
\( \text{Error} = |15.025 - 15| = \) _____
Decision:
PASS REJECT
Magnitude Mastery • High School Algebra Page 2 of 2
Magnitude Mastery Answer Key
Teacher Resource Full Solutions & Grading Guide
Magnitude Mastery • Answer Key
Evaluating Absolute Value Expressions & Order of Operations
Grading Breakdown
Total Points: 32 Pts
Sec A: 4 | Sec B: 8 | Sec C: 6
Sec D: 8 | Sec E: 4 | Sec F: 2
Pedagogical Anchor: Emphasize that absolute value acts as a grouping symbol. External negatives never get neutralized by the bars.
Complete Steps Shown
A Section A: Core Numerical Values (1 pt each)
Verify direct definition of distance
- \( | -23.4 | \)
23.4
- \( -| -58 | \)
-58
- \( | 4.15 - 9.75 | \)
5.6
- \( \left| -\frac{11}{6} \right| - \left| \frac{5}{6} \right| \)
1
B Section B: Order of Operations (2 pts each: 1 pt work, 1 pt answer)
PEMDAS compliance
- \( 16 - 4 \, | 3 - 8 | \)
• Inside: \( 3 - 8 = -5 \implies |-5| = 5 \)
• Multiply: \( 4 \cdot 5 = 20 \)
• Subtract: \( 16 - 20 = -4 \)
Common error: \( 16-4=12 \) first
Ans: -4
- \( 3 \, | -6 \cdot 2 | - | -5 |^2 \)
• First bar: \( |-12| = 12 \implies 3(12) = 36 \)
• Second bar: \( |-5| = 5 \implies 5^2 = 25 \)
• Subtract: \( 36 - 25 = 11 \)
Note: \( | -5 |^2 = 25 \), not \( -25 \)
Ans: 11
- \( \dfrac{| -24 + 6 |}{| -2 |^3 - 2} \)
• Numerator: \( |-18| = 18 \)
• Denominator: \( 2^3 - 2 = 8 - 2 = 6 \)
• Divide: \( \frac{18}{6} = 3 \)
Evaluate top and bottom separately
Ans: 3
- \( -2 \, | 7 - 19 | + 5 \, | -3 - (-10) | \)
• Term 1: \( -2 \, |-12| = -2(12) = -24 \)
• Term 2: \( 5 \, |-3 + 10| = 5|7| = 35 \)
• Combine: \( -24 + 35 = 11 \)
Watch: \( -(-10) = +10 \)
Ans: 11
C Section C: Single-Variable Evaluation (3 pts each)
Substitution value: \( x = -4 \)
- \( 5 \, | 2x + 3 | - 11 \)
• Sub: \( 2(-4) + 3 = -8 + 3 = -5 \)
• Absolute Value: \( |-5| = 5 \)
• Compute: \( 5(5) - 11 = 25 - 11 = 14 \)
Partial: 1 pt sub, 1 pt abs val, 1 pt result
Ans: 14
- \( -| x^2 - 25 | + 3x \)
• Inside: \( (-4)^2 - 25 = 16 - 25 = -9 \)
• Apply sign: \( -|-9| = -(9) = -9 \)
• Second term: \( 3(-4) = -12 \implies -9 + (-12) = -21 \)
Common error: \( -|-9| \neq +9 \)
Ans: -21
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Magnitude Mastery Answer Key Part II • Advanced & Applied
Grading Criteria: Step clarity + Final precision
D Section D: Multi-Variable Substitution (2 pts each)
Given: \( a = -3 \), \( b = 5 \), \( c = -2 \)
- \( | 2a - 3c | + | b \cdot c | \)
• Bar 1: \( |2(-3) - 3(-2)| = |-6 + 6| = |0| = 0 \)
• Bar 2: \( |(5)(-2)| = |-10| = 10 \)
• Sum: \( 0 + 10 = 10 \)
Ans: 10
- \( \dfrac{| a^2 - b^2 |}{| a - c |} \)
• Numerator: \( |(-3)^2 - 5^2| = |9 - 25| = |-16| = 16 \)
• Denominator: \( |-3 - (-2)| = |-1| = 1 \)
• Quotient: \( \frac{16}{1} = 16 \)
Ans: 16
- \( -3 \, | ab + 4c | - | c |^3 \)
• Bar 1: \( -3|(-3)(5) + 4(-2)| = -3|-23| = -3(23) = -69 \)
• Second term: \( |-2|^3 = 2^3 = 8 \)
• Combine: \( -69 - 8 = -77 \)
Ans: -77
- \( | a - b | - | b - a | \)
• \( |a - b| = |-3 - 5| = |-8| = 8 \)
• \( |b - a| = |5 - (-3)| = |8| = 8 \)
• Subtract: \( 8 - 8 = 0 \) (symmetric property)
Ans: 0
E Section E: Diagnostic Analysis & Concept Exploration (2 pts each)
Conceptual mastery
15. Solution: Spot the Flaw 2 Points
Conceptual Flaw: The student distributed \( -2 \) inside the absolute value bars. You cannot distribute into absolute value because \( |-2(x)| = 2|x| \), which incorrectly turns the negative factor positive.
Correct Order: Simplify inside first: \( 4 - 9 = -5 \), then take absolute value: \( |-5| = 5 \), then multiply: \( -2(5) = -10 \).
1 pt explanation, 1 pt correct value
Correct: -10
16. Solution: Triangle Inequality 2 Points
\( | -7 + 4 | = |-3| = \mathbf{3} \)
\( |-7| + |4| = 7 + 4 = \mathbf{11} \)
Pedagogical Insight: Because \( x \) and \( y \) have opposite signs, they partially cancel inside the bars. Thus \( |x+y| < |x|+|y| \).
✓ Checked: \( |x+y| < |x|+|y| \)
3 < 11
F Section F: Engineering Tolerance Solutions (2 pts total)
Pass Rule: \( \text{Error} = |d - 15.000| \le 0.025\text{ mm} \)
Pin A: \( d = 15.018\text{ mm} \)
\( |15.018 - 15.000| = \mathbf{0.018} \)
\( 0.018 \le 0.025 \) → Within tolerance
Decision: PASS
Pin B: \( d = 14.968\text{ mm} \)
\( |14.968 - 15.000| = |-0.032| = \mathbf{0.032} \)
\( 0.032 > 0.025 \) → Exceeds limit!
Decision: REJECT
Pin C: \( d = 15.025\text{ mm} \)
\( |15.025 - 15.000| = \mathbf{0.025} \)
\( 0.025 \le 0.025 \) → Exactly on boundary
Decision: PASS
Magnitude Mastery • Teacher Answer Key Page 2 of 2
