Comprehensive Study Notes on Linear Equations and Properties of Equality

Logistics and Administrative Information

• Homework Assignment Coverage: The current assigned homework covers Sections 2.1 and 2.2.

• Due Date: The assignment must be completed by the end of the week, with a final deadline on Sunday night at the latest.

• Course Materials Access:

  • Shared Notebook: The lecture notes correspond to Section 8 in the shared notebook link provided via email immediately prior to class.

  • Brightspace Integration: Homework is accessed directly through Brightspace under the October 1 homework link. Registration must first be completed by navigating to "Start Here" and selecting "Pearson course materials".

  • Homework Attempt Limits: The homework software provides unlimited attempts per problem.

• Class Schedule: Section 2.1 was initiated on Monday and finalized during this session. Section 2.2 is scheduled to begin on Friday.

Addition Property of Equality

• Fundamental Rule: Any mathematical operation (addition or subtraction) performed on one side of an equation sign must be performed on the opposite side to keep the equation balanced.

• Primary Objective in Solving Equations: Isolate the variable on one side of the equal sign by applying inverse (opposite) operations to any surrounding numbers.

• Step-by-Step Example 1:

  • Equation: $x - 9 = 22$

  • Isolation Step: Since $9$ is subtracted from $x$, apply the inverse operation by adding $9$ to both sides of the equal sign:

  • $x - 9 + 9 = 22 + 9$

  • $x = 31$

  • Double-Checking Method (Verification via Substitution):

  • Substitute $x = 31$ into the original equation $x - 9 = 22$:

  • $31 - 9 = 22$

  • $22 = 22$

  • Obtaining identical values on both sides establishes a true statement, confirming that $x = 31$ is the correct solution.

Real-World Application: Price Before Tax Calculation

• Word Problem Scenario:

  • Total purchase cost for a fitness tracker (including tax): $\$144.36$

  • Sales tax included in purchase: $\$6.87$

  • Objective: Calculate the original price $p$ of the fitness tracker prior to tax addition.

• Algebraic Equation:

  • $p + 6.87 = 144.36$

• Solution Steps:

  • Isolate the variable $p$ by subtracting the positive tax value $6.87$ from both sides of the equation:

  • $p + 6.87 - 6.87 = 144.36 - 6.87$

  • $p = 137.49$

  • Practical Calculation Advice: Using a basic calculator prevents simple arithmetic errors when performing decimal subtraction.

  • Final Result: The pre-tax sticker price of the tracker is $\$137.49$.

• Double-Checking Procedure:

  • Substitute $p = 137.49$ into $p + 6.87 = 144.36$:

  • $137.49 + 6.87 = 144.36$

  • $144.36 = 144.36$

  • Yields a true statement, confirming accuracy.

Multiplication Property of Equality

• Formal Definition: For any real numbers $a$, $b$, and non-zero $c$, if $a = b$, then $a \times c = b \times c$.

• Inverse Relationship: Multiplication and division are equal and opposite operations. The multiplication property extends equally to division:

  • If $a = b$, then $\frac{a}{c} = \frac{b}{c}$ (where $c \neq 0$).

• Implicit Multiplication Notation: Writing a numerical coefficient directly adjacent to a variable (such as $3x$) signifies implicit multiplication.

Solving Equations Using Multiplication and Division

• Example 1 (Fractions and Multiplication):

  • Equation: $\frac{1}{7} \times x = 4$

  • Step 1: Write $x$ as a fraction by placing it over $1$: $\frac{x}{1}$.

  • Step 2: Multiply numerators and denominators across:

  • $\frac{1 \times x}{7 \times 1} = 4$

  • $\frac{x}{7} = 4$

  • Step 3: Apply the inverse of dividing by $7$ by multiplying both sides by $7$:

  • 7 \times \n    \left(\frac{x}{7}\right) = 7 \times 4

  • $x = 28$

  • Verification Step:

  • Substitute $x = 28$ into $\frac{1}{7} \times x = 4$:

  • $\frac{1}{7} \times \frac{28}{1} = \frac{28}{7} = 4$

  • $4 = 4$ (True Statement).

• Example 2 (Direct Division):

  • Equation: $3x = 81$

  • Step 1: Isolate $x$ by dividing both sides by $3$:

  • $\frac{3x}{3} = \frac{81}{3}$

  • $x = 27$

  • Verification Step:

  • Substitute $x = 27$ into $3x = 81$:

  • $3(27) = 81$

  • $81 = 81$ (True Statement).

• Example 3 (Using Reciprocals):

  • Equation: $12 = \frac{2}{3}x$

  • Definition of Reciprocal: Flipping a fraction's numerator and denominator upside down creates its reciprocal. The reciprocal of $\frac{2}{3}$ is $\frac{3}{2}$.

  • Reciprocal Strategy (One-Step Isolation):

  • Multiply both sides by the reciprocal $\frac{3}{2}$:

  • $\left(\frac{3}{2}\right) \times 12 = \left(\frac{3}{2}\right) \times \left(\frac{2}{3}x\right)$

  • Convert $12$ to a fraction: $\frac{3}{2} \times \frac{12}{1} = \frac{36}{2} = 18$

  • On the right side, products simplify to $1$: $\left(\frac{3 \times 2}{2 \times 3}\right)x = 1x = x$

  • Solution: $x = 18$

  • Step-by-Step Alternative:

  • Step A: Multiply both sides by $3$ to clear the denominator ($36 = 2x$).

  • Step B: Divide both sides by $2$ to isolate $x$ ($18 = x$).

  • Verification Step:

  • Substitute $x = 18$ into $12 = \frac{2}{3}x$:

  • $12 = \frac{2}{3} \times \frac{18}{1} = \frac{36}{3} = 12$

  • $12 = 12$ (True Statement).

Section 2.1 Homework Problem Solutions

• Problem 1:

  • Problem Prompt: Determine whether $x = 3$ is a solution to $6x - 2 = 16$.

  • Substitution Process: $6(3) - 2 = 18 - 2 = 16$.

  • Result: $16 = 16$. Both sides equal $16$.

  • Answer: Yes.

• Problem 2:

  • Problem Prompt: Determine whether $k = \frac{1}{5}$ is a solution to $10k - 1 = 1$.

  • Substitution Process: $10\left(\frac{1}{5}\right) - 1 = \frac{10}{1} \times \frac{1}{5} - 1 = \frac{10}{5} - 1 = 2 - 1 = 1$.

  • Result: $1 = 1$. Both sides equal $1$.

  • Answer: Yes.

• Problem 3:

  • Equation: $x + 3 = -14$

  • Solution Procedure:

  • Subtract $3$ from both sides: $x = -14 - 3$

  • Combining terms: $x = -17$

• Problem 4:

  • Equation: $x - \frac{7}{5} = \frac{3}{5}$

  • Solution Procedure:

  • Add $\frac{7}{5}$ to both sides: $x = \frac{3}{5} + \frac{7}{5}$

  • Combine common denominators: $x = \frac{3 + 7}{5} = \frac{10}{5} = 2$

• Problem 5:

  • Equation: $\frac{5}{6} = x - \frac{1}{8}$

  • Solution Procedure:

  • Add $\frac{1}{8}$ to both sides: $x = \frac{5}{6} + \frac{1}{8}$

  • Find Least Common Multiple (LCM) for denominators $6$ and $8$:

    • Multiples of $6$: $6, 12, 18, 24, 30, \dots$

    • Multiples of $8$: $8, 16, 24, 32, \dots$

    • Least Common Multiple = $24$

  • Convert fractions to equivalent forms with denominator $24$:

    • $\frac{5}{6} = \frac{5 \times 4}{6 \times 4} = \frac{20}{24}$

    • $\frac{1}{8} = \frac{1 \times 3}{8 \times 3} = \frac{3}{24}$

  • Add converted numerators: $x = \frac{20}{24} + \frac{3}{24} = \frac{20 + 3}{24} = \frac{23}{24}$

  • Simplification Check: Since $23$ is prime, $\frac{23}{24}$ cannot be reduced further.

• Problem 6:

  • Equation: $x + 1.1 = -7.5$

  • Solution Procedure:

  • Subtract $1.1$ from both sides: $x = -7.5 - 1.1$

  • Combining negative decimals: $x = -8.6$

• Problem 7:

  • Equation: $9c = 18$

  • Solution Procedure:

  • Divide both sides by $9$: $c = \frac{18}{9} = 2$

  • Critical Warning on Common Mistakes: Do not divide by $18$. Always divide by the coefficient attached to the variable ($9$) to achieve isolation.

• Problem 8:

  • Equation: $-8n = 40$

  • Solution Procedure:

  • Divide both sides by $-8$: $n = \frac{40}{-8} = -5$

  • Sign Rule: Dividing a positive real number by a negative real number produces a negative result.

• Problem 9:

  • Equation: $\frac{5}{4}a = 10$

  • Solution Procedure:

  • Identify reciprocal of $\frac{5}{4}$, which is $\frac{4}{5}$.

  • Multiply both sides by $\frac{4}{5}$:

    • $\left(\frac{4}{5}\right) \times \left(\frac{5}{4}a\right) = \left(\frac{4}{5}\right) \times 10$

    • $a = \frac{4}{5} \times \frac{10}{1} = \frac{40}{5} = 8$

• Problem 10:

  • Equation: $-\frac{2}{3}z = -4$

  • Solution Procedure:

  • Move the negative sign explicitly to the numerator to eliminate confusion: $\frac{-2}{3}z = -4$

  • Multiply both sides by the reciprocal $\frac{3}{-2}$:

    • $z = \left(\frac{3}{-2}\right) \times (-4) = \frac{3}{-2} \times \frac{-4}{1} = \frac{-12}{-2} = 6$

  • Double negatives cancel to yield a positive product.

• Problem 11:

  • Equation: $-\frac{3}{8}t = \frac{1}{6}$

  • Solution Procedure:

  • Rewrite negative sign in the numerator: $\frac{-3}{8}t = \frac{1}{6}$

  • Multiply both sides by the reciprocal $\frac{8}{-3}$:

    • $t = \left(\frac{8}{-3}\right) \times \left(\frac{1}{6}\right) = \frac{8 \times 1}{-3 \times 6} = \frac{8}{-18}$

  • Reduce fraction by dividing top and bottom by common factor $2$:

    • $t = \frac{8 \div 2}{-18 \div 2} = \frac{4}{-9} = -\frac{4}{9}$