Module 7: General First-Degree Equations and Inequalities

Overview of Equations within the Syllabus

Module 7 focuses specifically on first-order equations and inequalities in one variable. This is part of a broader curriculum containing several types of equations:

1. First-order equations and inequalities in one variable (Module 7).
2. Second-order equations in one variable, also known as quadratic equations (Module 13 and Module 16).
3. Rational equations (Module 14), which simplify into either first-order equations or second-order quadratic equations.
4. Radical equations (Module 15), which also simplify into either first-order or second-order quadratic equations.
5. First-order equations in two variables (Module 9 and Module 10).
6. Systems of first-order equations in two variables (Module 11).
7. Descriptions of second-order equations/functions in two variables, such as quadratic functions (Module 17).
8. Descriptions of second-order equations in two variables that are non-functions, such as circles.

Foundations of Solving First-Degree Equations of the Form $ax + b = c$

In solving an equation of the form $ax + b = c$, the primary goal is to rewrite the given equation into a equivalent form expressed as $\text{variable} = ext{constant}$. This process requires applying both the Addition and Multiplication Properties of Equations.

• The Addition Property of Equations: This property is used to remove a term from one side of the equation by adding the opposite (additive inverse) of that term to both sides. The inverse property of addition states that $(-a) + (a) = 0$. This ensures that the equations remain equivalent throughout the process.
• The Multiplication Property of Equations: This property is used to remove a coefficient from one side of the equation by multiplying each side by the reciprocal of that coefficient. This is based on the inverse property of multiplication: (a) imes (rac{1}{a}) = 1. Because division can be defined in terms of multiplication, one may also divide both sides of an equation by the same nonzero number without altering the solution.

Example 1: Solve $3x - 7 = -5$.

1. Add 7 to each side: $3x - 7 + 7 = -5 + 7$.
2. Simplify: $3x = 2$.
3. Divide each side by 3 to isolate the variable.

Example 2: Solve $x - 2 = -11$.

1. Add 2 to each side: $x - 2 + 2 = -11 + 2$.
2. Simplify: $x = -9$.
   (Note: The transcript also lists a solution where $x = -12$ for an unspecified variant of this form involving multiplication by a constant.)

Clearing Denominators in Equations with Fractions

Solving equations containing two or more fractions is often simplified by multiplying each side of the equation by the least common multiple (LCM) or least common denominator (LCD) of all denominators involved. This process, known as clearing denominators, results in an equivalent equation that does not contain fractions. The LCM is defined as the smallest number into which all the denominators will divide evenly.

Example: Solve \frac{4}{5}x - rac{1}{2} = rac{3}{4}.

1. Identify the LCM of 5, 2, and 4, which is 20.
2. Multiply every term on both sides by 20: 20(rac{4}{5}x) - 20(rac{1}{2}) = 20(rac{3}{4}).
3. Simplify to remove fractions: $16x - 10 = 15$.
4. Add 10 to each side: $16x = 25$.
5. Divide by 16: x = rac{25}{16}. The solution is $\frac{25}{16}$.

Solving Equations of the Form $ax + b = cx + d$

When variables appear on both sides of the equation, the objective remains to reach the form $\text{variable} = ext{constant}$. The specific steps include:

1. Rewrite the equation so there is only one variable term by collecting variable terms to one side. (It is often preferable to move the smaller of the two terms to avoid negative coefficients.)
2. Rewrite the equation so there is only one constant term by collecting constant terms to the opposite side.
3. Apply the Addition and Multiplication Properties of Equations as necessary.

Example: Solve $4x - 5 = 8x - 7$.

1. Subtract $8x$ from each side: $4x - 8x - 5 = 8x - 8x - 7$, resulting in $-4x - 5 = -7$.
2. Add 5 to each side: $-4x - 5 + 5 = -7 + 5$, resulting in $-4x = -2$.
3. Divide by $-4$: x = rac{-2}{-4} = rac{1}{2}.

Solving Equations Containing Parentheses

When grouping symbols are present, the Distributive Property must be utilized to simplify variable expressions. The property states that if $a$, $b$, and $c$ are real numbers, then $a(b + c) = ab + ac$. Conversely, it can be used to combine terms: $ba + ca = (b + c)a$ (e.g., $2x + 3x = (2 + 3)x = 5x$).

Example 1: Solve $4 + 5(2x - 3) = 3(4x - 1)$.

1. Apply Distributive Property: $4 + 10x - 15 = 12x - 3$.
2. Combine like terms: $10x - 11 = 12x - 3$.
3. Subtract $12x$ from each side: $-2x - 11 = -3$.
4. Add 11 to each side: $-2x = 8$.
5. Divide by $-2$: $x = -4$.

Example 2: Solve $3x - 4(2 - x) = 3(x - 2) - 4$.

1. Distributive Property: $3x - 8 + 4x = 3x - 6 - 4$.
2. Combine terms: $7x - 8 = 3x - 10$.
3. Subtract $3x$: $4x - 8 = -10$.
4. Add 8: $4x = -2$.
5. Divide by 4: x = -rac{2}{4} = -rac{1}{2}.

Note: In the case of nested grouping symbols, always begin the solving process by removing the innermost symbols first.

Solving Literal Equations and Formulas

A literal equation is an equation that contains more than one variable. Formulas are specific types of literal equations used to express relationships between physical or business quantities. Common examples include:

• Physics formulas.
• Linear equations: $Ax + By = C$ or $y = mx + b$.
• Business interest: $M = P + Prt$.

The goal when solving a literal equation for a specific variable is to isolate that variable on one side. This may involve factoring out a common factor using the Distributive Property. For example, to solve $S = C - rC$ for $C$:

1. Identify $C$ as a common factor: $S = C(1 - r)$. (Note: Recall that $C = C imes 1$).
2. Divide both sides by the expression $(1 - r)$.
3. Result: C = rac{S}{1 - r}.

Value Mixture Problems

Value mixture problems involve combining ingredients with different prices into a specific blend. These problems utilize the value mixture equation $AC = V$, where $A$ is the amount, $C$ is the unit cost, and $V$ is the total value of the ingredient.

Strategy for Value Mixtures:

1. Create a table recording the amount, unit cost, and value for each ingredient and the final mixture.
2. Apply the principles:
       * The sum of values of all ingredients equals the value of the blend ($V_1 + V_2 = V_{\text{total}}$).
       * The sum of the amounts of all ingredients equals the amount of the blend ($A_1 + A_2 = A_{\text{total}}$).

Example: How many pounds of strawberry ($4/lb$) mixed with $10 ext{ lb}$ avocado ($6/lb$) makes a salad costing $4.32/lb$?

• Strawberry: amount $x$, cost $4$, value $4x$.
• Avocado: amount $10$, cost $6$, value $60$.
• Mixture: amount $10 + x$, cost $4.32$, value $4.32(10 + x)$.
• Equation: $4x + 60 = 4.32(10 + x)$.
• Solve: $4x + 60 = 43.2 + 4.32x \rightarrow -0.32x = -16.8 \rightarrow x = 52.5 ext{ lb}$.

Additional Practice Cases:

• Herbalist: $30 ext{ oz}$ at $2 mixed with $x ext{ oz}$ at $1 to get $1.60/oz$. Answer: $20 ext{ lb}$.
• Grain: $500 ext{ lb}$ @ $1.20 mixed with meal @ $0.80 to get $1.05/lb$. Answer: $300 ext{ lb}$.
• Coffee: $12 ext{ lb}$ @ $8.00 and $30 ext{ lb}$ @ $4.50. Resulting cost: $5.50/lb$.
• Dye: Blue dye @ $1.60 and $18 ext{ L}$ anil @ $2.50 to get $1.90/L$. Answer: $36 ext{ L}$.

Percent Mixture Problems

Percent mixture problems involve combining solutions with different concentrations of a substance. These rely on the substance amount equation $Ar = Q$, where $A$ is the solution amount, $r$ is the concentration percentage, and $Q$ is the quantity of the substance within the solution.

Strategy for Percent Mixtures:

1. Maintain a table for amount, percentage (as a decimal), and quantity ($Ar$).
2. Principle: The sum of the quantities of ingredients equals the quantity in the blend.

Example: Making $2 ext{ L}$ of 8% acid using 10% and 5% solutions.

• 10% solution: amount $x$, quantity $0.10x$.
• 5% solution: amount $2 - x$, quantity $0.05(2 - x)$.
• 8% mixture: amount $2$, quantity $0.16$.
• Equation: $0.10x + 0.05(2 - x) = 0.08(2)$.
• Solve: $0.10x + 0.10 - 0.05x = 0.16 \rightarrow 0.05x = 0.06 \rightarrow x = 1.2 ext{ L}$ (10% solution) and $0.8 ext{ L}$ (5% solution).

Additional Practice Cases:

• Plant food: 9% nitrogen combined with 25% for $10 ext{ gal}$ of 15%. Answer: $6.25 ext{ ml}$.
• Chemist: $50 ext{ ml}$ of 16% using 13% and 18%. Answer: $20 ext{ ml}$ of 13% and $30 ext{ ml}$ of 18%.
• Hair dye: 7% and 4% solutions to make $300 ext{ ml}$ of 5%. Answer: $100 ext{ ml}$ of 7% and $200 ext{ ml}$ of 4%.
• Shampoo: $12 ext{ oz}$ of 20% conditioner + $8 ext{ oz}$ pure shampoo. Answer: 12% concentration.

Uniform Motion Problems

Uniform motion occurs when an object travels at a constant speed in a straight line. The basis of these problems is the equation $rt = d$ (rate $imes$ time = distance).

Strategy for Uniform Motion:

1. Identify the rate, time, and distance for two moving objects/scenarios.
2. Determine the relationship between distances (e.g., total distance known, or distances are equal if they catch up).
3. It is helpful to draw a sketch of the situation.

Example 1: Car A at $40 ext{ mph}$. Two hours later, Car B at $60 ext{ mph}$ follows. When will Car B pass Car A?

• Car B time: $t$. Car A time: $t + 2$.
• Distance A: $40(t + 2)$. Distance B: $60t$.
• Set distances equal: $40(t + 2) = 60t \rightarrow 40t + 80 = 60t \rightarrow 80 = 20t \rightarrow t = 4 ext{ h}$.

Example 2: Two cars move in opposite directions, one $10 ext{ mph}$ faster than the other. In $3 ext{ h}$, they are $300 ext{ mi}$ apart.

• Rate A: $r$. Rate B: $r + 10$.
• Equation: $3r + 3(r + 10) = 300$.
• Solve: $3r + 3r + 30 = 300 \rightarrow 6r = 270 \rightarrow r = 45 ext{ mph}$.
• Rates are $45 ext{ mph}$ and $55 ext{ mph}$.

First-Degree Inequalities and Set Notation

Unlike equations, inequalities have multiple solutions forming a solution set. Symbols include:

• a > b: $a$ is greater than $b$.
• $a \ge b$: $a$ is greater than or equal to $b$.
• a < b: $a$ is less than $b$.
• $a \le b$: $a$ is less than or equal to $b$.

Methods of Writing Solution Sets:

• Roster Method: Encloses a list of elements in braces ${ }$. Not used if elements are continuous. Elements are never duplicated.
• Empty Set (Null Set): A set with no elements, represented by $\emptyset$ or ${ }$.
• Set-Builder Notation: Descriptive notation representing a set. For example, {x | x < 10, x \in \text{positive integers}}.
• Interval Notation: Used for continuous sets.
      * Ex: $[-3, 2)$ includes $-3$ (bracket) but excludes $2$ (parenthesis).
      * Infinity ($\infty$) and negative infinity ($-\infty$) always use parentheses because infinity is not a real number.
• Graphing: Represented on a number line. Parentheses or open circles denote exclusion; brackets or closed dots denote inclusion.

Properties for Solving Inequalities

Solving inequalities follows similar steps to equations, with one critical rule change:

• Addition Property of Inequalities: The same term can be added to or subtracted from each side without changing the solution set. If a < b, then a + c < b + c.
• Multiplication Property of Inequalities:
      * Rule 1: Each side can be multiplied or divided by the same positive constant without changing the symbol.
      * Rule 2: If each side is multiplied or divided by a negative constant, the inequality symbol must be reversed to maintain the same solution set.

Example: Solve \frac{1}{6} - rac{3}{4}x > rac{11}{12}.

1. Clear fractions by multiplying by the LCM (12): 2 - 9x > 11.
2. Subtract 2: -9x > 9.
3. Divide by $-9$ and reverse the symbol: x < -1.
4. The solution set is {x | x < -1} in set-builder notation.