Study Notes: Writing and Solving Equations and Inequalities

Absolute Value Equations

• Key methods
  • If you have an equation of the form |f(x)| = c, then you solve by splitting into two equations: f(x) = c or f(x) = -c. In shorthand: $|f(x)| = c \Rightarrow f(x) = c \text{or} \ f(x) = -c.$
  • If you have |f(x)| ≤ c, then you solve by writing the compound inequality: $|f(x)| \le c \Rightarrow -c \le f(x) \le c.$
  • If you have |f(x)| ≥ c, you solve by splitting: $|f(x)| \ge c \Rightarrow f(x) \le -c \text{or} \ f(x) \ge c.$
  • Be mindful of extraneous solutions when squaring; verify solutions in the original absolute-value equation.

Problem-based examples from the transcript (with standard interpretations)

7) Nail length with tolerance

• Context: A penny-number-2 nail is nominally 1 inch long; actual length allowed to vary by ±0.05 in (assumed from the text).
• Setup (absolute value model): $|x - 1| \le 0.05,$ where x is the nail length in inches.
• Solution: $x \in [0.95,\; 1.05].$
• Part B: If a nail is 1.05 inches long, is it acceptable?
  • Yes, because 1.05 is within the acceptable range [0.95, 1.05].

8) Town survey on garage sale

• Given: 56% in favor, 44% opposed. The actual percent could vary by ±5% from the observed results.
• Part A (in favor): Solve using absolute-value tolerance around the observed proportion 0.56.
  • Inequality: $|p - 0.56| \le 0.05 \Rightarrow p \in [0.51,\; 0.61].$
  • Proportion opposed q = 1 - p, so q ∈ [1 - 0.61, 1 - 0.51] = [0.39,\; 0.49].
  • Therefore, the least and greatest possible opposed percentages are 39% and 49%.
• Part B (conflict check): A friend claims that half the town is opposed (0.50).
  • 0.50 lies outside the possible opposed interval [0.39, 0.49], so the claim conflicts with the survey data given the stated tolerance.

9) Aisha and Carlos bike route (ambiguous in transcription)

• The transcript is unclear about the exact setup for this problem, so the exact absolute-value model cannot be determined from the text provided. If you have the precise statement (distances, home location, etc.), we can formulate the equation for how far Aisha bikes after school using an absolute-value model (e.g., representing deviation from a planned route or a fixed route length). If you can supply the missing details, I can fill this in with a complete setup and solution.

40) Normal body temperature variation

• Statement: Normal body temperature is 98.6 °F with a possible variation of ±0.5 °F.
• Correct absolute-value model: $|\,x - 98.6\,| \le 0.5,$ where x is the actual body temperature in °F.
• Answer choice: This corresponds to option C (represented correctly as an inequality, not an equality).

Inequalities (Solving and set-builder notation)

4) Solve the inequality (from Page 2): 5n - 4 > 7n + 20

• Solve:
  • Subtract 5n from both sides: -4 > 2n + 20
  • Subtract 20: -24 > 2n
  • Divide by 2: -12 > n or equivalently n < -12.
• Set-builder form: { n \mid n < -12 }.

8) Compound inequality with OR

• Given: -13 > q + 2 \quad \text{or} \quad 5q \ge -15.
• Solve each:
  • First part: -13 > q + 2 \Rightarrow q < -15.
  • Second part: $5q \ge -15 \Rightarrow q \ge -3.$
• With OR: the solution set is { q \mid q < -15 \ \text{or} \ q \ge -3 }.

10) Single inequality

• Given: $-2x - 5 \le 10.$
• Solve: $-2x \le 15 \Rightarrow x \ge -\frac{15}{2} = -7.5.$
• Solution set: ${ x \mid x \ge -7.5 }.$

11) Bank minimum balance and $20 withdrawals

• Scenario: Must maintain at least $50; current balance $280; withdrawal is in $20 bills.
• Part A: How many $20 bills can you withdraw without going below minimum?
  • Model: $280 - 20w \ge 50 \Rightarrow 20w \le 230 \Rightarrow w \le 11.5.$
  • Since you can’t withdraw a fraction of a bill, $w \le 11.$ (Thus up to 11 bills.)
• Part B: ATM fee of $2.50 per withdrawal, still with $20 bills.
  • Cost per withdrawal is $20, plus $2.50 fee: total deduction per withdrawal is $22.50.
  • Model: $280 - 22.5w \ge 50 \Rightarrow 22.5w \le 230 \Rightarrow w \le 230/22.5 \approx 10.22.$
  • Therefore, at most 10 withdrawals of $20 can be made under this fee structure.

12) Tuxedo sleeve lengths

• Given: Sleeve lengths available are from 20 in to 40 in. Shop says sleeves should be about 1.2 times arm length.
• Let arm length be A. Requirement for the shop to provide tuxedos: 1.2A must lie within [20, 40].
• Solve for A: $20 \le 1.2A \le 40 \Rightarrow \frac{20}{1.2} \le A \le \frac{40}{1.2} \Rightarrow 16.666\ldots \le A \le 33.333\ldots.$
• Therefore, the arm lengths for which the shop does not provide tuxedos are: $A < \frac{20}{1.2} \quad\text{or}\quad A > \frac{40}{1.2}$ i.e., $A < 16.67 \text{ in or } A > 33.33 \text{ in}.$

13) Andy’s paper boxes (volume and base area)

• Given: Base area = 30 in^2 exactly; Volume V ≤ 240 in^3.
• Volume formula: V = base area × height = 30h.
• Inequality: $30h \le 240 \Rightarrow h \le 8.$
• Height must be positive: 0 < h \le 8.
• Answer: B.

14) Groceries budget with pounds of ground beef

• Prices: tomatoes $1.35/lb, oranges $1.65/lb, ground beef $4.94/lb.
• Joe spends at most $23 on groceries, buying 3 lb tomatoes and 4 lb oranges, plus x lb beef:
• Inequality: $1.35(3) + 1.65(4) + 4.94x \le 23.$
• Compute constants: $4.05 + 6.60 + 4.94x \le 23 \Rightarrow 10.65 + 4.94x \le 23.$
• Solve for x: $4.94x \le 12.35 \Rightarrow x \le \frac{12.35}{4.94} \approx 2.5.$
• Therefore, the possible pounds of ground beef satisfy $x \le 2.5.$

15) An inequality from the transcript (6x + 1 ≤ 5x + 2)

• Solve: $6x + 1 \le 5x + 2 \Rightarrow x \le 1.$
• Answer: x ≤ 1.

One-variable Inequalities (Graphing) – Quick Guidelines

• Solve the inequality, then graph the solution on the number line.
• Watch for strict vs non-strict inequalities: < or ≤; > or ≥ affect whether endpoints are included.
• When inequalities combine with AND (intersection), shade the overlap; with OR (union), shade the larger region defined by either condition.

Linear Equations (Solve and Check)

1) Simple linear equation

• Example: x + 5 = 9
• Solution: x = 4.

2) Linear equation with p

• Example: -43 = 12 - 6p + p
• Solve: -43 = 12 - 5p \Rightarrow -55 = -5p \Rightarrow p = 11.

3) (Transcription ambiguity)

• Some entries in the transcript are unclear; if you provide the exact text, I’ll solve and verify.

4) Another linear equation

• Example: 3(x - 2) + 5(2 - x) = 16
• Solve: 3x - 6 + 10 - 5x = 16 \Rightarrow -2x + 4 = 16 \Rightarrow -2x = 12 \Rightarrow x = -6.

5) Linear equation with t

• Example: 5t + 7 = 3t - 9
• Solve: 2t = -16 \Rightarrow t = -8.

6) Equation with a common factor (x - 4)

• Example: -2(x - 4) = 7(x - 4)
• Solve: -2x + 8 = 7x - 28 \Rightarrow 8 + 28 = 9x \Rightarrow x = 4.

7) Distributive equation with x

• Example: (3 - 6x) = -3(8x - 4)
• Solve: 3 - 6x = -24x + 12 \Rightarrow -6x + 24x = 12 - 3 \Rightarrow 18x = 9 \Rightarrow x = \tfrac{1}{2}.

8) Another multi-step linear equation

• Example: 3(k + 1) + 11k = 2(4 + 5k) + 3
• Solve: 3k + 3 + 11k = 8 + 10k + 3 \Rightarrow 14k + 3 = 11 + 10k \Rightarrow 4k = 8 \Rightarrow k = 2.

9) Landscaping-hours application (linear equation)

• Given: Landscaping bill with labor rate; Total = 720, materials = 375, labor rate = 34.50 per hour.
• Equation: 720 = 375 + 34.50h
• Solve: h = (720 - 375)/34.50 = 345/34.50 = 10 hours.

12) Rope-length problem (solving a setup with pieces)

• Problem: Eight inches are trimmed from the end of a rope. The remaining rope is cut into 5 equal pieces, each of which is 40 inches less than the original length of the rope.
• Let original length be L. After trimming: L - 8. Then five equal pieces: (L - 8)/5. Each piece is L - 40.
• Equation: $\frac{L - 8}{5} = L - 40.$
• Solve: L - 8 = 5L - 200 \Rightarrow -8 + 200 = 5L - L \Rightarrow 192 = 4L \Rightarrow L = 48 \text{ inches}.$

13) Product and sum problem (algebraic expression)

• Statement (interpreted): "The opposite of the product of 4 and the sum of twice x and 3 is equal to one fourth of the quantity 5 less than x."
• Correct equation form: $-4\,(2x + 3) = \frac{x - 5}{4}.$
• Solve: Multiply both sides by 4: $-16x - 12 = x - 5 \Rightarrow -17x = 7 \Rightarrow x = -\frac{7}{17}.$
• Note: The transcript’s multiple-choice options appear to be misprinted; the correctly derived equation uses the division by 4 on the right-hand side.

20) Perimeter comparison – square vs. triangles (application)

• Given: The perimeter of a square is 4 units greater than the combined perimeter of two congruent equilateral triangles. Square side s = 10.
• Perimeter of square: $P<em>{square} = 4s = 40.$ Perimeter of one equilateral triangle with side t: $P</em>{triangle} = 3t.$ Two triangles have total perimeter $P<em>{triangles} = 2\cdot 3t = 6t.$ The relation says: $P</em>{square} = P_{triangles} + 4.$
• Solve: $40 = 6t + 4 \Rightarrow 6t = 36 \Rightarrow t = 6.$

28) Original length of a rope – quick check

• (Covered above under 12) L = 48 inches.

Quick notes on formatting and checking answers

• Always verify by substituting solutions back into the original equation to avoid extraneous results from squaring or algebraic manipulation.
• When dealing with budget/price problems, isolate the unknown variable on one side and simplify constants first to reduce arithmetic error.
• In word problems, translate the scenario into a clean algebraic model (variables for unknown quantities, constants for known values), then solve step by step.

Small glossary of common patterns

• Absolute value equations: |A| = B or |A| ≤ B or |A| ≥ B, leading to one or two linear equations or to a compound inequality.
• Compound inequalities use AND for intersection (both conditions) or OR for union (either condition).
• Set-builder notation: { x | condition } describes the solution set.
• Unit-consistency checks: especially in word problems (money, length, volume) to ensure units align across terms.

If you can provide the exact missing pieces from the transcript (especially item 9 and the ambiguous linear equation fragments), I can fill in the remaining solutions with full step-by-step derivations and check-work.