Rational Reckoning Worksheet
Rational Reckoning
Solving Equations & Spotting Imposters
Name:
Date:
Warm-up: Proportion Review
Solve the following proportion for \(x\). Show your steps.
\[ \frac{3}{x} = \frac{9}{12} \]
Reflection: What mathematical property allows you to "cross-multiply" in a proportion?
Evidence Collection: Video Notes
1 Case Study: \(x + \frac{12}{x} = 7\)
Follow along with the video and complete the steps below.
Step 1: Clear the Denominator
What value should we multiply the entire equation by?
Step 2: Factor (PAUSE @ 1:36)
Write the quadratic and factor it before the narrator does:
Discussion Question:
Why do we multiply by the LCD? What is our goal in this first step?
2 Case Study: \(\frac{5}{x} + \frac{8}{x+3} = 7\)
Identify the LCD (PAUSE @ 4:45)
The Least Common Denominator is:
Factoring (PAUSE @ 8:25)
The equation simplifies to \(7x^2 + 8x - 15 = 0\). Show the AC Method here:
Verification Phase (PAUSE @ 11:35)
"Do you think this ugly fraction (\(x = -15/7\)) will work? Why or why not?"
Algebra Forensics: Error Analysis
Three "Student" solutions are provided below. One is perfect, one has an arithmetic error, and one missed an extraneous solution. Your job: **Verify, Identify, and Correct.**
Exhibit A: Solve \(\frac{x}{x-2} = \frac{2}{x-2} + 2\)
Correct
Error
// Student Work 1) Multiply by \((x-2)\):
   &(x) = 2 + 2(x - 2)\)
2) Distribute:
   &(x = 2 + 2x - 4\)
3) Simplify:
   &(x = 2x - 2\)
4) Solve:
   &(-x = -2 \rightarrow x = 2\)
Solution: \(x = 2\)
Your Analysis:
Identify error or confirm correctness...
Correction / Verification:
Exhibit B: Solve \(\frac{5}{x} - \frac{2}{3} = \frac{1}{x}\)
Correct
Error
// Student Work 1) Multiply by LCD \(3x\):
   &(3(5) - x(2) = 1\)
2) Simplify:
   &(15 - 2x = 1\)
3) Subtract 15:
   &(-2x = -14\)
4) Solve:
   &(x = 7\)
Solution: \(x = 7\)
Your Analysis:
Identify error or confirm correctness...
Correction / Verification:
Exhibit C: Solve \(\frac{4}{x} = \frac{12}{x-4}\)
Correct
Error
// Student Work 1) Cross-multiply:
   &(4(x - 4) = 12x\)
2) Distribute:
   &(4x - 16 = 12x\)
3) Subtract 4x:
   &(-16 = 8x\)
4) Solve:
   &(x = -2\)
Solution: \(x = -2\)
Your Analysis:
Identify error or confirm correctness...
Correction / Verification:
Rational Reckoning Presentation Slides
Rational
Reckoning
Solving Rational Equations & Identifying Extraneous Solutions
Algebra 2
"The hunt for imposter solutions"
Mission Objective
Warm-up
Today we will solve equations that contain fractions and identify extraneous solutions—answers that look right but don't actually work.
Focus Keywords
- LCD (Least Common Denominator)
- Extraneous Solution
Solve on your desk:
\[ \frac{3}{x} = \frac{9}{12} \]
"What property allows you to solve this quickly?"
Video Case Study
Embedded media
Get your Guided Notes ready!
Pause & Reflect
Timestamp: 1:36
Factor the quadratic \(x^2 - 7x + 12 = 0\)
Timestamp: 4:45
What is the LCD for Example #2?
Critical Question
"Why do we check our answers in rational equations? Why don't we have to do this for every other equation we've solved?"
Algebra Forensics
Activity
1
Investigate
Analyze the three exhibits on your worksheet. One is a "Perfect Problem," but the others are crimes against math.
2
Categorize
Is it an Arithmetic Error? Or did they miss an Extraneous Solution? Mark your findings.
3
Correct
In the forensics lab space, provide the correct solution and verification for each error you found.
Time Limit: 20 Minutes
\( \frac{1}{x} \)
\( x^2 \)
\( LCD \)
Final Reckoning
Exit Ticket
Solve & Check:
\( x + \frac{6}{x} = 5 \)
1. Clear LCD
2. Solve Quadratic
3. Check!
Rational Reckoning Answer Key
Teacher Answer Key
Rational Reckoning: Solving & Extraneous Solutions
ALGEBRA 2
Warm-up: Proportion Review
\[ \frac{3}{x} = \frac{9}{12} \]
Step 1: Cross-multiply: \( 3(12) = 9x \)
Step 2: \( 36 = 9x \)
Step 3: \( x = 4 \)
Reflection:
The Multiplication Property of Equality allows us to multiply both sides by the denominators to clear them.
Evidence Collection (Video Notes)
Example 1: \(x + 12/x = 7\)
- Multiply by: \(x\)
- Quadratic: \(x^2 - 7x + 12 = 0\)
- Factors: \((x-3)(x-4) = 0\)
- Solutions: \(x=3, x=4\)
- Check: Both valid. No extraneous.
Example 2: \(5/x + 8/(x+3) = 7\)
- LCD: \(x(x+3)\)
- Simplified: \(5(x+3) + 8x = 7x(x+3)\)
- Quadratic: \(7x^2 + 8x - 15 = 0\)
- AC Method: \(7 \cdot -15 = -105\). Factors are \(-7\) and \(15\).
- Solutions: \(x=1, x = -15/7\)
- Check: Both valid.
Algebra Forensics: Error Analysis
Exhibit A: MISSED EXTRANEOUS SOLUTION
The math is correct until the very end. The student found \(x=2\), but plugging \(x=2\) into the original equation causes division by zero (\(2/0\)).
Correct Status: No Solution (or \(x=2\) is extraneous).
Exhibit B: ARITHMETIC ERROR
In Step 1, the student failed to multiply the right side of the equation (\(1/x\)) by the LCD (\(3x\)).
Correct Step 1: \( 3(5) - x(2) = 1(3) \rightarrow 15 - 2x = 3 \)
Correct Final Answer: \( -2x = -12 \rightarrow x = 6 \)
Exhibit C: CORRECT
The cross-multiplication and solving steps are perfectly executed. \(x = -2\) is a valid solution as it does not make any denominator zero.
Exit Ticket: \(x + 6/x = 5\)
- Multiply by \(x\): \(x^2 + 6 = 5x\)
- Set to zero: \(x^2 - 5x + 6 = 0\)
- Factor: \((x - 2)(x - 3) = 0\)
- Solve: \(x = 2, x = 3\)
- Check: Both denominators are just \(x\). Neither 2 nor 3 is zero.
Final Answer: x = 2, 3 (No extraneous)
