Section 1.1 Study Guide: Solving Simple Equations

Core Learning Targets and Vocabulary

• Learning Objectives:

  • Write and solve one-step linear equations in one variable.

  • Apply properties of equality to produce equivalent equations.

  • Solve linear equations using addition, subtraction, multiplication, or division.

  • Model and solve real-life situations using linear equations.

• Essential Vocabulary:

  • Equation: A mathematical statement asserting that two algebraic expressions are equal.

  • Linear Equation in One Variable: An equation that can be written in the form $ax + b = 0$, where $a$ and $b$ are constants and $a \neq 0$.

  • Solution: A specific value substituted for a variable that makes an equation a true statement.

  • Equivalent Equations: Two or more equations that possess identical solution sets.

  • Inverse Operations: Operations that reverse or undo the effect of one another. Addition and subtraction are inverse operations; multiplication and division are inverse operations.

Properties of Equality

• Addition, Subtraction, and Substitution Properties:

  • Addition Property of Equality: Adding the same real number to both sides of an equation maintains equality and produces an equivalent equation.     $\text{If } a = b, \text{ then } a + c = b + c$

  • Subtraction Property of Equality: Subtracting the same real number from both sides of an equation maintains equality and produces an equivalent equation.     $\text{If } a = b, \text{ then } a - c = b - c$

  • Substitution Property of Equality: If two quantities are equal, one may replace the other in any expression or equation without changing the truth value.     $\text{If } a = b, \text{ then } a \text{ can be substituted for } b \text{ (or } b \text{ for } a \text{) in any equation or expression}$

• Multiplication and Division Properties:

  • Multiplication Property of Equality: Multiplying both sides of an equation by the same non-zero real number produces an equivalent equation.     $\text{If } a = b, \text{ then } a \cdot c = b \cdot c, \text{ provided } c \neq 0$

  • Division Property of Equality: Dividing both sides of an equation by the same non-zero real number produces an equivalent equation.     $\text{If } a = b, \text{ then } \frac{a}{c} = \frac{b}{c}, \text{ provided } c \neq 0$

Solving Linear Equations Step-by-Step

• Procedure for Addition and Subtraction Equations:

  • Identify the operation acting on the variable.

  • Apply the inverse operation to both sides of the equation to isolate the variable.

  • Simplify both sides.

  • Verify the solution by substituting it back into the original equation.

• Worked Example 1a: Equation with Subtraction

  • Solve: $x - 3 = -5$

  • Apply Addition Property of Equality (add $3$ to both sides):     $x - 3 + 3 = -5 + 3$

  • Simplify:     $x = -2$

  • Verification:     $-2 - 3 = -5 \implies -5 = -5 \quad \checkmark$

• Worked Example 1b: Equation with Addition

  • Solve: $0.9 = y + 2.8$

  • Apply Subtraction Property of Equality (subtract $2.8$ from both sides):     $0.9 - 2.8 = y + 2.8 - 2.8$

  • Simplify:     $-1.9 = y$

  • Verification:     $0.9 = -1.9 + 2.8 \implies 0.9 = 0.9 \quad \checkmark$

• Procedure for Multiplication and Division Equations:

  • Identify whether the variable is multiplied or divided by a constant.

  • Apply the corresponding inverse operation (multiply to undo division, or divide to undo multiplication).

  • Simplify and verify.

• Worked Example 2a: Equation with Division

  • Solve: $-\frac{n}{5} = -3$

  • Apply Multiplication Property of Equality (multiply both sides by $-5$):     $-5 \cdot \left(-\frac{n}{5}\right) = -5 \cdot (-3)$

  • Simplify:     $n = 15$

  • Verification:     $-\frac{15}{5} = -3 \implies -3 = -3 \quad \checkmark$

• Worked Example 2b: Equation with Constant Factor $\pi$

  • Solve: $\pi x = -2\pi$

  • Apply Division Property of Equality (divide both sides by $\pi$):     $\frac{\pi x}{\pi} = \frac{-2\pi}{\pi}$

  • Simplify:     $x = -2$

  • Verification:     $\pi(-2) = -2\pi \implies -2\pi = -2\pi \quad \checkmark$

• Worked Example 2c: Equation with Decimal Coefficient

  • Solve: $1.3z = 5.2$

  • Apply Division Property of Equality (divide both sides by $1.3$):     $\frac{1.3z}{1.3} = \frac{5.2}{1.3}$

  • Simplify:     $z = 4$

  • Verification:     $1.3(4) = 5.2 \implies 5.2 = 5.2 \quad \checkmark$

Problem-Solving Framework for Real-Life Applications

• Three-Step Problem-Solving Plan:

  1. Understand the Problem: Define the unknown quantity, list given information, and explicitly state what is being calculated.

  2. Make a Plan: Select appropriate strategies such as writing a verbal model, setting up an equation, constructing a table, or sketching a diagram.

  3. Solve and Check: Execute the plan algebraically, examine the result, and verify that the numeric answer is reasonable in context.

• Common Problem-Solving Strategies:

  • Use a verbal model.

  • Draw a diagram or sketch a graph/number line.

  • Write an algebraic equation.

  • Look for patterns or make a structured list.

  • Work backward from a known end state.

  • Break a complex problem into smaller parts.

  • Construct a table or use guess, check, and revise.

• Worked Example 3: Calculating Average Speed (2016 Olympic 200m Dash)

  

  • Scenario: In the 2016 Olympics, Usain Bolt won the 200-meter dash in $19.78\,\text{seconds}$. Find his average speed to the nearest hundredth of a meter per second.

  • Given: Distance $d = 200\,\text{m}$, Time $t = 19.78\,\text{sec}$.

  • Distance Formula: $d = r \cdot t$

  • Substitution:     $200 = r \cdot 19.78$

  • Division Property of Equality:     $\frac{200}{19.78} = \frac{19.78r}{19.78}$

  • Simplification:     $r \approx 10.11\,\text{m/sec}$

  • Check Reasonableness:     Rounding speed to $10\,\text{m/sec}$, time for $200\,\text{m}$ is $200 \div 10 = 20\,\text{seconds}$. Since $20\,\text{sec}$ is extremely close to $19.78\,\text{sec}$, $10.11\,\text{m/sec}$ is reasonable.

• Worked Example 4: Temperature Drop Model

  • Scenario: On January 22, 1943, the temperature in Spearfish, South Dakota, dropped from $54^\circ\text{F}$ at 9:00 a.m. to $-4^\circ\text{F}$ at 9:27 a.m. Determine the total degree drop in temperature.

  • Verbal Model:     $\text{Temperature at 9:27 a.m.} = \text{Temperature at 9:00 a.m.} - \text{Temperature Drop}$

  • Variable Definition: Let $T$ be the number of degrees Fahrenheit the temperature fell.

  • Equation:     $-4 = 54 - T$

  • Subtraction Property of Equality:     $-4 - 54 = 54 - 54 - T \implies -58 = -T$

  • Division by $-1$:     $T = 58$

  • Result: The temperature fell by $58^\circ\text{F}$.

  • Check with Number Line: Distance on a number line from $54$ down to $0$ is $54$, and from $0$ down to $-4$ is $4$. Total distance is $54 + 4 = 58^\circ\text{F}$.

Explorations and Detailed Practice Problems

• Okavango Delta Flow Rate Analysis:

  

  • Context: The Okavango Delta in southern Africa provides freshwater for 1 million people.

  • Peak Flow Rate: Graph shows peak flow rate $f = 800\,\text{m}^3/\text{sec}$.

  • Friend's Calculation:     $f = 800t$     $100,000 = 800t \implies t = \frac{100,000}{800} = \frac{1000}{8} = 125\,\text{seconds}$

  • Dimensional Analysis Check:     $\text{m}^3 \div \frac{\text{m}^3}{\text{sec}} = \text{m}^3 \cdot \frac{\text{sec}}{\text{m}^3} = \text{sec}$

• Detailed Problem Solutions:

  • Self-Assessment Item 1:     $n + 3 = -7 \implies n = -7 - 3 = -10$

  • Self-Assessment Item 2:     $g - \frac{1}{3} = -\frac{2}{3} \implies g = -\frac{2}{3} + \frac{1}{3} = -\frac{1}{3}$

  • Self-Assessment Item 3:     $-6.5 = p + 3.9 \implies p = -6.5 - 3.9 = -10.4$

  • Self-Assessment Item 4 (Equivalence Test):     Compare $6x = -5$ and $-1 = 6x + 4$.     For $6x = -5$, $x = -\frac{5}{6}$.     For $-1 = 6x + 4$, subtract $4$ gives $-5 = 6x \implies x = -\frac{5}{6}$.     Both equations have the exact same solution, so they are equivalent.

  • Self-Assessment Items 5–8:

  • $\frac{y}{3} = -6 \implies y = 3(-6) = -18$

  • $z \div 25 = -4.5 \implies z = -4.5 \times 25 = -112.5$

  • $9\pi = \pi x \implies x = \frac{9\pi}{\pi} = 9$

  • $0.05w = 1.4 \implies w = \frac{1.4}{0.05} = 28$

  • Autonomous Vehicle Travel Distance (Problem 10):

  • Speed $r = 15.7\,\text{mph}$, Time $t = 9\,\text{days} = 9 \times 24 = 216\,\text{hours}$.

  • Distance $d = 15.7 \times 216 = 3391.2\,\text{miles}$.

  • Olympic Comparison (Problem 11):

  • 2012 Bolt 200m average speed $10.35\,\text{m/sec}$. Time $t_{2012} = \frac{200}{10.35} \approx 19.32\,\text{sec}$.

  • 2016 Bolt 200m time $19.78\,\text{sec}$.

  • Bolt was faster in 2012 by $19.78 - 19.32 = 0.46\,\text{seconds}$.

  • Unrecorded Savings Withdrawal (Problem 12):

  • Expected balance: $\$68.33$, Actual balance: $\$26.33$.

  • Equation: $68.33 - w = 26.33 \implies w = 68.33 - 26.33 = \$42.00$.

  • Bluefin Tuna Egg Release (Problem 13):

  • Bluefin tuna releases $300\%$ more eggs than Atlantic sturgeon.

  • Bluefin egg count = $10,000,000$.

  • Let $s$ be sturgeon egg count. $100\% + 300\% = 400\% = 4s$.

  • $4s = 10,000,000 \implies s = 2,500,000\,\text{eggs}$.

  • Discounted Ticket Price (Problem 11, Section Practice):

  • Discount: $\$12.95$ off original price $p$. Sale price: $\$44.00$.

  • Equation: $p - 12.95 = 44.00 \implies p = \$56.95$.

  • Carton Egg Equation (Problem 37):

  • Total eggs = $162$, eggs per carton = $18$.

  • Equation: $18x = 162 \implies x = 9\,\text{cartons}$.

  • Error Analysis:

  • Problem 33 Error: $-0.8 + r = 12.6$. The error was subtracting $-0.8$ instead of adding $0.8$. Correct: $r = 12.6 + 0.8 = 13.4$.

  • Problem 34 Error: $-\frac{m}{3} = -4$. The error was multiplying by $3$ instead of $-3$. Correct: $-3 \cdot \left(-\frac{m}{3}\right) = -3 \cdot (-4) \implies m = 12$.

  • Geometry - Quadrilateral Angles (Problems 35 & 36):

  • Sum of interior angles of a quadrilateral is $360^\circ$.

  • Problem 35: $x + 100 + 120 + 100 = 360 \implies x + 320 = 360 \implies x = 40^\circ$.

  • Problem 36: $x + 150 + 77 + 48 = 360 \implies x + 275 = 360 \implies x = 85^\circ$.

  • Japanese Tatami Mats Layout (Problem 44):

  • Total area = $81\,\text{ft}^2$. Layout comprises 4 identical rectangular mats and 1 square mat.

  • Rectangular mat length $l = 2w$. Area of one rectangular mat A_r = l \cdot w = 2w^2$.\n * Area of square mat A_s = \frac{1}{2} A_r = w^2$.

  • Total area equation:       $4 A_r + A_s = 81 \implies 4(2w^2) + w^2 = 81 \implies 9w^2 = 81 \implies w^2 = 9 \implies w = 3\,\text{ft}$

  • Rectangular mat dimensions: Width $= 3\,\text{ft}$, Length $= 6\,\text{ft}$.