Section 2.1 Exercise Set: Solving Linear Equations and Verification

Foundations of Linear Equations in One Variable

• Core Definitions and Algebraic Terminology:

  • An algebraic equation is a mathematical statement asserting that two expressions are equal, joined by an equals sign ($=$).

  • A linear equation in one variable (such as $x$ or $y$) is an equation that can be written in the standard form $ax + b = 0$, where $a$ and $b$ are real numbers and $a \neq 0$.

  • The solution (or root) of an equation is the specific value of the variable that makes the equation a true numerical statement.

  • Solving an equation involves finding the solution set through equivalent equations, which are equations that have the exact same solution set.

• Fundamental Properties of Equality:

  • Addition Property of Equality:

  • Formally stated: If $a = b$, then $a + c = b + c$ for any real numbers $a$, $b$, and $c$.

  • Subtraction Corollary: If $a = b$, then $a - c = b - c$. Subtracting a quantity is algebraically equivalent to adding its opposite.

  • Purpose: Used to eliminate constant terms or variable terms on one side of an equation to isolate terms.

  • Multiplication Property of Equality:

  • Formally stated: If $a = b$ and $c \neq 0$, then $a \times c = b \times c$ for any real numbers $a$, $b$, and $c$.

  • Division Corollary: If $a = b$ and $c \neq 0$, then $\frac{a}{c} = \frac{b}{c}$. Dividing by a nonzero quantity is algebraically equivalent to multiplying by its reciprocal.

  • Purpose: Used to reduce the coefficient of the isolated variable to $1$.

  • Symmetric Property of Equality:

  • Formally stated: If $a = b$, then $b = a$.

  • Purpose: Allows the isolated variable to be stated on the left-hand side regardless of which side it was originally isolated on.

• Verification Protocol (The Check Procedure):

  • To verify that a calculated value is the true solution, substitute the value directly into every instance of the variable in the original, unaltered equation.

  • Evaluate both the left-hand side (LHS) and right-hand side (RHS) independently following standard order of operations.

  • If LHS $=$ RHS, the solution is correct and verified. If LHS $\neq$ RHS, an algebraic or arithmetic error has occurred.

One-Step Equations: Multiplication and Division

• Exercise 1:

  • Given Equation:     $-5x = -30$

  • Objective:

  • Isolate the variable $x$, which is currently multiplied by the coefficient $-5$.

  • Algebraic Solution:

  • Apply the Division Property of Equality by dividing both sides of the equation by $-5$:       $\frac{-5x}{-5} = \frac{-30}{-5}$

  • Simplify both sides:

    • Left-hand side: $\frac{-5}{-5} = 1$, yielding $1x = x$.

    • Right-hand side: A negative number divided by a negative number results in a positive quotient: $\frac{-30}{-5} = 6$.

  • Result:       $x = 6$

  • Verification / Check:

  • Substitute $x = 6$ into the original equation $-5x = -30$:       $-5(6) = -30$       $-30 = -30$

  • The statement is true; therefore, the solution set is $\{6\}$.

• Exercise 2:

  • Given Equation:     $-2x = 18$

  • Objective:

  • Isolate the variable $x$, which is currently multiplied by the coefficient $-2$.

  • Algebraic Solution:

  • Apply the Division Property of Equality by dividing both sides of the equation by $-2$:       $\frac{-2x}{-2} = \frac{18}{-2}$

  • Simplify both sides:

    • Left-hand side: $\frac{-2}{-2} = 1$, yielding $1x = x$.

    • Right-hand side: A positive number divided by a negative number results in a negative quotient: $\frac{18}{-2} = -9$.

  • Result:       $x = -9$

  • Verification / Check:

  • Substitute $x = -9$ into the original equation $-2x = 18$:       $-2(-9) = 18$       $18 = 18$

  • The statement is true; therefore, the solution set is $\{-9\}$.

One-Step Equations: Addition and Subtraction

• Exercise 3:

  • Given Equation:     $-10 = x + 12$

  • Objective:

  • Isolate the variable $x$ on the right-hand side of the equation.

  • Algebraic Solution:

  • Apply the Subtraction Property of Equality by subtracting $12$ from both sides of the equation:       $-10 - 12 = x + 12 - 12$

  • Simplify both sides:

    • Left-hand side: Summing two negative values yields $-10 - 12 = -22$.

    • Right-hand side: Additive inverses cancel: $12 - 12 = 0$, yielding $x + 0 = x$.

  • Result:       $-22 = x$

  • By the Symmetric Property of Equality:       $x = -22$

  • Verification / Check:

  • Substitute $x = -22$ into the original equation $-10 = x + 12$:       $-10 = -22 + 12$       $-10 = -10$

  • The statement is true; therefore, the solution set is $\{-22\}$.

• Exercise 4:

  • Given Equation:     $-25 = y + 30$

  • Objective:

  • Isolate the variable $y$ on the right-hand side of the equation.

  • Algebraic Solution:

  • Apply the Subtraction Property of Equality by subtracting $30$ from both sides of the equation:       $-25 - 30 = y + 30 - 30$

  • Simplify both sides:

    • Left-hand side: Combining negative integers yields $-25 - 30 = -55$.

    • Right-hand side: Additive inverses cancel: $30 - 30 = 0$, yielding $y$.

  • Result:       $-55 = y$

  • By the Symmetric Property of Equality:       $y = -55$

  • Verification / Check:

  • Substitute $y = -55$ into the original equation $-25 = y + 30$:       $-25 = -55 + 30$       $-25 = -25$

  • The statement is true; therefore, the solution set is $\{-55\}$.

• Exercise 5:

  • Given Equation:     $x - 2.8 = 1.9$

  • Objective:

  • Isolate the variable $x$ by removing the decimal constant $-2.8$.

  • Algebraic Solution:

  • Apply the Addition Property of Equality by adding $2.8$ to both sides of the equation:       $x - 2.8 + 2.8 = 1.9 + 2.8$

  • Simplify both sides:

    • Left-hand side: $-2.8 + 2.8 = 0$, leaving $x$.

    • Right-hand side: Decimal addition: $1.9 + 2.8 = 4.7$.

  • Result:       $x = 4.7$

  • Verification / Check:

  • Substitute $x = 4.7$ into the original equation $x - 2.8 = 1.9$:       $4.7 - 2.8 = 1.9$       $1.9 = 1.9$

  • The statement is true; therefore, the solution set is $\{4.7\}$.

• Exercise 6:

  • Given Equation:     $y - 8.6 = -6.3$

  • Objective:

  • Isolate the variable $y$ by clearing the subtracted decimal constant $8.6$.

  • Algebraic Solution:

  • Apply the Addition Property of Equality by adding $8.6$ to both sides of the equation:       $y - 8.6 + 8.6 = -6.3 + 8.6$

  • Simplify both sides:

    • Left-hand side: $-8.6 + 8.6 = 0$, leaving $y$.

    • Right-hand side: Adding a positive value of greater magnitude to a negative value: $8.6 - 6.3 = 2.3$.

  • Result:       $y = 2.3$

  • Verification / Check:

  • Substitute $y = 2.3$ into the original equation $y - 8.6 = -6.3$:       $2.3 - 8.6 = -6.3$       $-6.3 = -6.3$

  • The statement is true; therefore, the solution set is $\{2.3\}$.

Multi-Step Equations with Variables on Both Sides

• General Strategy for Linear Equations with Variables on Both Sides:

  • Step 1: Use the Addition or Subtraction Property of Equality to collect all variable terms on one side of the equation.

  • Step 2: Use the Addition or Subtraction Property of Equality to collect all constant numerical terms on the opposite side.

  • Step 3: Use the Multiplication or Division Property of Equality to isolate the variable with a coefficient of $1$.

  • Step 4: Check the solution by evaluating both sides of the initial equation.

• Exercise 7:

  • Given Equation:     $5x - 4 = 26 + 2x$

  • Detailed Algebraic Procedure:

  • Step 1: Eliminate the variable term $2x$ from the right-hand side by subtracting $2x$ from both sides:       $5x - 2x - 4 = 26 + 2x - 2x$       $3x - 4 = 26$

  • Step 2: Eliminate the constant term $-4$ from the left-hand side by adding $4$ to both sides:       $3x - 4 + 4 = 26 + 4$       $3x = 30$

  • Step 3: Isolate the variable $x$ by dividing both sides by the coefficient $3$:       $\frac{3x}{3} = \frac{30}{3}$       $x = 10$

  • Verification / Check:

  • Evaluate Left-Hand Side (LHS) with $x = 10$:       $\text{LHS} = 5(10) - 4 = 50 - 4 = 46$

  • Evaluate Right-Hand Side (RHS) with $x = 10$:       $\text{RHS} = 26 + 2(10) = 26 + 20 = 46$

  • Since $\text{LHS} = \text{RHS} = 46$, the solution is verified.

  • Solution set: $\{10\}$.

• Exercise 8:

  • Given Equation:     $5y - 3 = 11 + 3y$

  • Detailed Algebraic Procedure:

  • Step 1: Eliminate the variable term $3y$ from the right-hand side by subtracting $3y$ from both sides:       $5y - 3y - 3 = 11 + 3y - 3y$       $2y - 3 = 11$

  • Step 2: Eliminate the constant term $-3$ from the left-hand side by adding $3$ to both sides:       $2y - 3 + 3 = 11 + 3$       $2y = 14$

  • Step 3: Isolate the variable $y$ by dividing both sides by the coefficient $2$:       $\frac{2y}{2} = \frac{14}{2}$       $y = 7$

  • Verification / Check:

  • Evaluate Left-Hand Side (LHS) with $y = 7$:       $\text{LHS} = 5(7) - 3 = 35 - 3 = 32$

  • Evaluate Right-Hand Side (RHS) with $y = 7$:       $\text{RHS} = 11 + 3(7) = 11 + 21 = 32$

  • Since $\text{LHS} = \text{RHS} = 32$, the solution is verified.

  • Solution set: $\{7\}$.