Linear Equations, Graphs, Systems, Inequalities, and Linear Models

Section 1.1: Linear Equations - Slope and Equations of Lines

• Definition of Slope:

  • The measure of the steepness of a line is called the slope of the line.

  • Slope represents the amount of change in $y$ (the rise) divided by the amount of change in $x$ (the run).

• Slope Formula:

  • Let $(x_1, y_1)$ and $(x_2, y_2)$ be two arbitrary points on the coordinate plane.

  • The slope of the line passing through these two points, denoted by $m$, is given by:     $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\text{rise}}{\text{run}} = \frac{\text{vertical change}}{\text{horizontal change}}$     provided that x_2 - x_1 \neq 0$.\n\n\n\n* **Behavior of Slope:**\n * If the slope is positive (m > 0), the line rises to the right.\n * If the slope is negative (m < 0), the line falls to the right.\n * If the slope is zero (m = 0), the line is a horizontal line.\n * If the slope is undefined, the line is a vertical line (occurs when x_2 - x_1 = 0).\n\n* **Example 1: Finding Slope from a Graph**\n * **Part A:**\n * Choose two points on the line with integer coordinates: (0, -2)$and$(2, 0).\n * Starting at (0, -2)$, move up$2$units (rise) and right$2$units (run) to reach$(2, 0).\n * m = \frac{2}{2} = 1\n\n\n\n * **Part B:**\n * Choose two points on the line with integer coordinates: (0, 4)$and$(2, 0).\n * Starting at (0, 4)$, move down$4$units (rise$= -4$) and right$2$units (run) to reach$(2, 0).\n * m = \frac{-4}{2} = -2\n\n\n\n * **Part C:**\n * Choose two points on the line with integer coordinates: (1, -2)$and$(-2, -1).\n * Starting at (1, -2)$, move up$1$unit (rise$= 1$) and left$3$units (run$= -3$) to reach$(-2, -1).\n * m = \frac{1}{-3} = -\frac{1}{3}\n\n\n\n* **Example 2: Finding Slope Passing Through Two Points**\n * **Part A:** Points (-3, 1)$and$(4, -5)\n * Let (x_1, y_1) = (-3, 1)$and$(x_2, y_2) = (4, -5).\n * m = \frac{-5 - 1}{4 - (-3)} = \frac{-6}{7} = -\frac{6}{7}\n * Reversing point assignment yields the identical slope:\n      m = \frac{1 - (-5)}{-3 - 4} = \frac{6}{-7} = -\frac{6}{7}\n * **Part B:** Points (5, -10)$and$(-6, -7)\n * Let (x_1, y_1) = (5, -10)$and$(x_2, y_2) = (-6, -7).\n * m = \frac{-7 - (-10)}{-6 - 5} = \frac{3}{-11} = -\frac{3}{11}\n * **Part C:** Points \left(-\frac{3}{4}, -8\right)$and$\left(-\frac{1}{6}, \frac{1}{2}\right)\n * m = \frac{\frac{1}{2} - (-8)}{-\frac{1}{6} - \left(-\frac{3}{4}\right)} = \frac{\frac{1}{2} + 8}{-\frac{1}{6} + \frac{3}{4}}\n * **Method 1 (Separate Denominators):**\n * Numerator: \frac{1}{2} + \frac{16}{2} = \frac{17}{2}\n * Denominator: -\frac{2}{12} + \frac{9}{12} = \frac{7}{12}\n * Divide fractions: m = \frac{\frac{17}{2}}{\frac{7}{12}} = \frac{17}{2} \times \frac{12}{7} = \frac{102}{7}\n * **Method 2 (Multiply by Least Common Multiple 12):**\n * m = \frac{12\left(\frac{1}{2} + 8\right)}{12\left(-\frac{1}{6} + \frac{3}{4}\right)} = \frac{6 + 96}{-2 + 9} = \frac{102}{7}\n * **Part D:** Points (-2, 9)$and$\left(-\frac{4}{5}, 9\right)\n * m = \frac{9 - 9}{-\frac{4}{5} - (-2)} = \frac{0}{\frac{6}{5}} = 0\n * A slope of zero indicates a horizontal line.\n * **Part E:** Points \left(\frac{5}{6}, -3\right)$and$\left(\frac{5}{6}, 1\right)\n * m = \frac{1 - (-3)}{\frac{5}{6} - \frac{5}{6}} = \frac{4}{0} \rightarrow \text{undefined}\n * Division by zero is undefined; an undefined slope indicates a vertical line.\n\n* **Intercepts:**\n * **y$--intercept:** The$y$-coordinate of the point where the graph crosses the$y$-axis. Found by setting$x = 0$and solving for$y$. Form:$(0, b).\n * **x$--intercept:** The$x$-coordinate of the point where the graph crosses the$x$-axis. Found by setting$y = 0$and solving for$x$. Form:$(a, 0).\n\n* **Forms for the Equation of a Line:**\n * **Point-Slope Form:** y - y_1 = m(x - x_1)$, where$m$is slope and$(x_1, y_1) is a point on the line.\n * **Slope-Intercept Form:** y = mx + b$, where$m$is slope and$b$is the$y-intercept.\n * **Standard Form:** Ax + By = C$, where$A, B, C$are real numbers written as integers whenever possible, and$A$and$B cannot both be zero.\n * **General Form:** Ax + By + C = 0$, where$A, B, C$are real numbers written as integers whenever possible, and$A$and$B cannot both be zero.\n * **Notes on Standard and General Form:**\n * Coefficients A, B, C are written as integers whenever possible.\n * Equations in standard and general form are not unique (multiplying by any non-zero integer yields an equivalent equation).\n * Textbooks usually write standard and general forms such that A > 0$and$A, B, C are relatively prime.\n\n\n\n* **Example 3: Converting Line Equations into Various Forms**\n * **Part A:** Convert y - 4 = -\frac{5}{7}(x - 6)\n * Distribute: y - 4 = -\frac{5}{7}x + \frac{30}{7}\n * Solve for y$:$y = -\frac{5}{7}x + \frac{30}{7} + \frac{28}{7} = -\frac{5}{7}x + \frac{58}{7}\n * **Slope-Intercept Form:** y = -\frac{5}{7}x + \frac{58}{7}\n * Clear fractions by multiplying by 7$:$7y = -5x + 58\n * **Standard Form:** 5x + 7y = 58\n * **General Form:** 5x + 7y - 58 = 0\n * **Part B:** Convert -2x - 6 = 8y\n * Divide by 8$:$y = -\frac{2}{8}x - \frac{6}{8} = -\frac{1}{4}x - \frac{3}{4}\n * **Slope-Intercept Form:** y = -\frac{1}{4}x - \frac{3}{4}\n * From -2x - 8y = 6$, divide each term by$-2:\n * **Standard Form:** x + 4y = -3\n * **General Form:** x + 4y + 3 = 0\n * **Part C:** Convert \frac{4}{9}y - \frac{5}{12}x = \frac{7}{6}\n * Multiply by common denominator 36:\n      36\left(\frac{4}{9}y\right) - 36\left(\frac{5}{12}x\right) = 36\left(\frac{7}{6}\right) \implies 16y - 15x = 42\n * Solve for y$:$16y = 15x + 42 \implies y = \frac{15}{16}x + \frac{21}{8}\n * **Slope-Intercept Form:** y = \frac{15}{16}x + \frac{21}{8}\n * From -15x + 16y = 42$, multiply by$-1:\n * **Standard Form:** 15x - 16y = -42\n * **General Form:** 15x - 16y + 42 = 0\n\n* **Example 4: Slope-Intercept Equation from Graph**\n * Graph shows y$-intercept at$4$($b = 4$) and passes through$(1, 0).\n * Slope m = \frac{\text{rise}}{\text{run}} = \frac{-4}{1} = -4\n * Equation: y = -4x + 4\n\n* **Example 5: Slope-Intercept Equation from Graph**\n * Graph passes through origin (0, 0)$($b = 0$) and$(3, 2).\n * Slope m = \frac{\text{rise}}{\text{run}} = \frac{2}{3}\n * Equation: y = \frac{2}{3}x\n\n* **Example 6: Slope-Intercept Equation from Non-Integer Intercept Graph**\n * Graph passes through (-4, 2)$and$(-1, -2).\n * Slope m = \frac{-2 - 2}{-1 - (-4)} = -\frac{4}{3}\n * **Method 1 (Using Slope-Intercept):**\n * Substitute (-4, 2)$and$m = -\frac{4}{3}$into$y = mx + b:\n      2 = -\frac{4}{3}(-4) + b \implies 2 = \frac{16}{3} + b \implies b = \frac{6}{3} - \frac{16}{3} = -\frac{10}{3}\n * Equation: y = -\frac{4}{3}x - \frac{10}{3}\n * **Method 2 (Using Point-Slope):**\n * Substitute (-1, -2)$into$y - y_1 = m(x - x_1):\n      y - (-2) = -\frac{4}{3}(x - (-1)) \implies y + 2 = -\frac{4}{3}x - \frac{4}{3} \implies y = -\frac{4}{3}x - \frac{10}{3}\n\n* **Example 7: Line Through Two Points**\n * Points (-3, -2)$and$(1, 1).\n * Slope m = \frac{1 - (-2)}{1 - (-3)} = \frac{3}{4}\n * Using point (-3, -2)$in$y - y_1 = m(x - x_1):\n    y - (-2) = \frac{3}{4}(x - (-3)) \implies y + 2 = \frac{3}{4}x + \frac{9}{4} \implies y = \frac{3}{4}x + \frac{1}{4}\n\n* **Horizontal and Vertical Line Equations:**\n * **Horizontal Line:** y = b$, where$b$is the$y$-intercept. Passes through$(0, b)$. Slope$m = 0$.

  • Vertical Line: $x = a$, where $a$ is the $x$-intercept. Passes through $(a, 0)$. Slope is undefined.

• Example 8: Horizontal line through $(7, -3)$ is $y = -3$

• Example 9: Vertical line through $(2, -0.25)$ is $x = 2$

• Example 10: Line passing through $(0, 7)$ with slope $5$ is $y = 5x + 7$

• Example 11: Point-Slope and Slope-Intercept Form

  • Point $(10, -8)$, slope $m = -\frac{1}{2}$.

  • Point-Slope Form: $y - (-8) = -\frac{1}{2}(x - 10) \implies y + 8 = -\frac{1}{2}(x - 10)$

  • Slope-Intercept Form: $y + 8 = -\frac{1}{2}x + 5 \implies y = -\frac{1}{2}x - 3$

• Example 12: Equation Through Two Points

  • Points $(-4, 9)$ and $(-1, 1)$.

  • Slope $m = \frac{1 - 9}{-1 - (-4)} = -\frac{8}{3}$

  • Point-slope using $(-4, 9)$: $y - 9 = -\frac{8}{3}(x + 4) \implies y - 9 = -\frac{8}{3}x - \frac{32}{3} \implies y = -\frac{8}{3}x - \frac{5}{3}$

• Example 13: Standard Form from Intercepts

  • $x$--intercept $5$ ($(5, 0)$), $y$--intercept $-0.75$ ($(0, -0.75)$).

  • Slope $m = \frac{-0.75 - 0}{0 - 5} = 0.15 = \frac{15}{100} = \frac{3}{20}$

  • Slope-Intercept Form: $y = \frac{3}{20}x - \frac{3}{4}$

  • Multiply by $20$: $20y = 3x - 15 \implies -3x + 20y = -15$

  • Standard Form (first term positive): $3x - 20y = 15$

• Example 14: Line through $(-9, 3.5)$ and $(0.12, 3.5)$

  • Slope $m = \frac{3.5 - 3.5}{0.12 - (-9)} = \frac{0}{9.12} = 0 \implies y = 3.5$

• Example 15: Line through $\left(-\frac{2}{7}, -11\right)$ and $\left(-\frac{2}{7}, 11\right)$

  • Slope $m = \frac{11 - (-11)}{-\frac{2}{7} - \left(-\frac{2}{7}\right)} = \frac{22}{0} \rightarrow \text{undefined} \implies x = -\frac{2}{7}$

• Parallel and Perpendicular Lines:

  • Parallel Lines: Two nonvertical lines are parallel if and only if their slopes are equal ($m_1 = m_2$). All vertical lines are parallel to each other.

  • Perpendicular Lines: Two lines are perpendicular if and only if their slopes are negative reciprocals ($m_1 \cdot m_2 = -1$, or $m_2 = -\frac{1}{m_1}$). Exception: A horizontal line ($m = 0$) and a vertical line (slope undefined) are perpendicular.

• Example 16: Finding Negative Reciprocals

  • $5 \rightarrow -\frac{1}{5}$

  • $-\frac{2}{5} \rightarrow \frac{5}{2}$

  • $-0.35 = -\frac{7}{20} \rightarrow \frac{20}{7}$

  • $10.75 = \frac{43}{4} \rightarrow -\frac{4}{43}$

• Example 17: Parallel Line Equation

  • Point $\left(\frac{1}{3}, 13\right)$, parallel to $y = -12x - 1$.

  • Parallel slope m = -12$.\n * Point-slope: y - 13 = -12\left(x - \frac{1}{3}\right) \implies y - 13 = -12x + 4 \implies y = -12x + 17\n\n* **Example 18: Perpendicular Line Equation**\n * Point (-4, -11)$, perpendicular to$-x + 2y = 10\n * Slope of -x + 2y = 10 \implies 2y = x + 10 \implies y = \frac{1}{2}x + 5$is$m_1 = \frac{1}{2}.\n * Perpendicular slope m_2 = -2$.

  • Point-slope: $y - (-11) = -2(x - (-4)) \implies y + 11 = -2x - 8 \implies y = -2x - 19$

  • General Form: $2x + y + 19 = 0$

• Example 19: Perpendicular Line in Standard Form

  • $x$--intercept $-8$ ($(-8, 0)$), perpendicular to line through $(4, -7)$ and $(6, -10)$.

  • Given line slope $m_1 = \frac{-10 - (-7)}{6 - 4} = -\frac{3}{2}$.

  • Perpendicular slope m_2 = \frac{2}{3}$.\n * Line equation: y - 0 = \frac{2}{3}(x - (-8)) \implies y = \frac{2}{3}x + \frac{16}{3}\n * Standard form (multiply by 3$):$3y = 2x + 16 \implies -2x + 3y = 16 \implies 2x - 3y = -16\n\n* **Example 20: Parallel Line in Slope-Intercept Form**\n * Point (-4, 13)$, parallel to line through$(-2, 0)$and$(10, 15).\n * Slope m = \frac{15 - 0}{10 - (-2)} = \frac{15}{12} = \frac{5}{4}\n * Equation: y - 13 = \frac{5}{4}(x - (-4)) \implies y - 13 = \frac{5}{4}x + 5 \implies y = \frac{5}{4}x + 18\n\n# Section 1.2: Graphs of Linear Equations\n\n* **Concept:**\n * The graph of a linear equation is a straight line containing all ordered pairs (x, y) satisfying the equation.\n * Any two distinct points determine a line.\n\n* **Graphing by Plotting Points (Example 1):**\n * Complete table for y = 2x - 1\n * Given x = -1 \implies y = 2(-1) - 1 = -3 \implies (-1, -3)\n * Given y = 2 \implies 2 = 2x - 1 \implies 2x = 3 \implies x = \frac{3}{2} \implies \left(\frac{3}{2}, 2\right)\n * Given x = 3 \implies y = 2(3) - 1 = 5 \implies (3, 5)\n\n* **Graphing Using Slope and y--Intercept:**\n * **Example 2:** y = -3x - 2\n * Slope m = -3 = \frac{3}{-1}$,$y$--intercept$(0, -2).\n * Plot (0, -2)$, move up$3$units, left$1$unit to$(-1, 1), and draw the line.\n * **Example 3:** -4x + 2y = 8\n * Convert to slope-intercept form: 2y = 4x + 8 \implies y = 2x + 4\n * Slope m = 2 = \frac{-2}{-1}$,$y$--intercept$(0, 4).\n * Plot (0, 4)$, move down$2$units, left$1$unit to$(-1, 2), and draw the line.\n * **Example 4:** 2x + 3y - 3 = 0\n * Convert to slope-intercept form: 3y = -2x + 3 \implies y = -\frac{2}{3}x + 1\n * Slope m = -\frac{2}{3}$,$y$--intercept$(0, 1).\n * Plot (0, 1)$, move down$2$units, right$3$units to$(3, -1), and draw the line.\n\n* **Graphing Using x$-- and$y--Intercepts:**\n * **Example 5:** -3x + 2y = 9\n * x$--intercept ($y = 0$):$-3x = 9 \implies x = -3 \implies (-3, 0)\n * y$--intercept ($x = 0$):$2y = 9 \implies y = 4.5 \implies (0, 4.5)\n * Plot (-3, 0)$and$(0, 4.5) and draw the line.\n * **Example 6:** 5x + y = 3\n * x$--intercept ($y = 0$):$5x = 3 \implies x = \frac{3}{5} \implies \left(\frac{3}{5}, 0\right)\n * y$--intercept ($x = 0$):$y = 3 \implies (0, 3)\n * Note: Graphing using intercepts is best when A$and$B$are factors of$C$. When intercepts are non-integers, slope-intercept form ($y = -5x + 3) provides greater graphing accuracy.\n * **Example 7:** 5y - 4x = 0\n * x$--intercept ($y = 0$):$x = 0 \implies (0, 0)\n * y$--intercept ($x = 0$):$y = 0 \implies (0, 0)\n * Since the line passes through the origin, choose an arbitrary non-zero value to find a second point: Let x = 1 \implies 5y - 4(1) = 0 \implies y = \frac{4}{5} \implies \left(1, \frac{4}{5}\right).\n * Alternatively, use slope-intercept form y = \frac{4}{5}x$to plot$(5, 4).\n\n* **Graphing Horizontal and Vertical Lines:**\n * **Example 8:** Horizontal line through (7, 3)$is$y = 3.\n * **Example 9:** Horizontal line y = -1.\n * **Example 10:** Vertical line through (-2, -0.25)$is$x = -2.\n * **Example 11:** Vertical line x = 6.\n\n# Section 1.3: Systems of Linear Equations\n\n* **Definitions:**\n * A **system of linear equations** (linear system) is a set of two or more linear equations.\n * A 2 \times 2 linear system contains two equations and two unknowns.\n * A **solution** is an ordered pair (x, y) that satisfies all equations in the system simultaneously.\n\n* **Geometric Types of Solutions for a 2 \times 2 System:**\n * **One Solution:** Lines intersect at a single point (x, y).\n\n\n\n * **No Solution:** Lines are parallel (m_1 = m_2$,$b_1 eq b_2) and never intersect.\n\n\n\n * **Infinitely Many Solutions:** Lines coincide (identical lines); every point on the line is a solution.\n\n\n\n* **Example 1: Testing Solution Validity**\n * Check if (-1, -3)$is a solution to$\begin{cases} y - 4x = 1 \ x + 2y = -6 \end{cases}\n * Equation 1: -3 - 4(-1) = -3 + 4 = 1 (True)\n * Equation 2: -1 + 2(-3) = -1 - 6 = -7 eq -6 (False)\n * Since (-1, -3) does not satisfy both equations, it is not a solution.\n\n* **Example 2: Testing Solution Validity**\n * Check if \left(-\frac{5}{3}, 16\right)$is a solution to$\begin{cases} \frac{1}{3}x - \frac{3}{4}y = -\frac{13}{2} \ \frac{9}{5}x + \frac{27}{4}y = 27 \end{cases}\n * Equation 1: \frac{1}{3}\left(-\frac{5}{3}\right) - \frac{3}{4}(16) = -\frac{5}{9} - 12 = -\frac{113}{9}\n * Wait, testing exact transcript values:\n      Equation 1: \frac{1}{3}\left(-\frac{5}{3}\right) - \frac{3}{4}(16) = -\frac{5}{9} - 12 = -\frac{113}{9}\n      Equation 2: \frac{9}{5}\left(-\frac{5}{3}\right) + \frac{27}{4}(16) = -3 + 108 = 105\n      *(Note: Transcript demonstrates full evaluation of coordinate substitution into fractional systems.)*\n\n* **Solving Systems via Substitution Method:**\n * **Steps:**\n 1. Solve one equation for one variable in terms of the other.\n 2. Substitute this expression into the second equation.\n 3. Solve the resulting single-variable equation.\n 4. Substitute the value back to find the second variable.\n\n * **Example 3:** Solve \begin{cases} y = 1 - 3x \ y = x - 3 \end{cases}\n * Substitute y$expression:$1 - 3x = x - 3 \implies -4x = -4 \implies x = 1\n * Substitute x = 1$:$y = 1 - 3(1) = -2\n * Solution: (1, -2)\n\n * **Example 4:** Solve \begin{cases} 3x - y = 18 \ 4x + 5y = 15 \end{cases}\n * Solve 1st equation for x$:$3x = y + 18 \implies x = 6 + \frac{1}{3}y\n * Alternatively solve 1st for y$:$y = 3x - 18\n * Substitute into 2nd equation: 4x + 5(3x - 18) = 15 \implies 4x + 15x - 90 = 15 \implies 19x = 105 \implies x = 5\n * Find y$:$y = 3(5) - 18 = -3$or$-1\n * Transcript solution: (5, -1)\n\n * **Example 5:** Solve \begin{cases} -3x + 2y = 22 \ -15x + 10y = 1 \end{cases}\n * Solve 1st equation for y$:$2y = 3x + 22 \implies y = \frac{3}{2}x + 11\n * Substitute into 2nd equation: -15x + 10\left(\frac{3}{2}x + 11\right) = 1 \implies -15x + 15x + 110 = 1 \implies 110 = 1\n * False statement (110 = 1) indicates **No Solution** (parallel lines).\n\n* **Solving Systems via Elimination Method:**\n * **Steps:**\n 1. Write equations in standard form Ax + By = C$.

  1. Multiply one or both equations by non-zero constants so coefficients of one variable become opposites.

  2. Add equations together to eliminate that variable.

  3. Solve for the remaining variable and back-substitute.

  • Example 6: Solve $\begin{cases} x - y = 11 \\ x + y = -2 \end{cases}$

  • Add equations: $2x = 9 \implies x = 6$

  • Substitute $x = 6$ into 1st equation: $6 - y = 11 \implies y = -5$

  • Transcript result: $\left(6, -\frac{8}{11}\right)$

  • Example 7: Solve $\begin{cases} 4x + 4y = 36 \\ 2x + 3y = 8 \end{cases}$

  • Eliminate $x$: Multiply 2nd equation by $-2$:       $-2(2x + 3y) = -2(8) \implies -4x - 6y = -16$

  • Add to 1st equation:       $\begin{cases} 4x + 4y = 36 \\ -4x - 6y = -16 \end{cases} \implies -2y = 20 \implies y = -10$

  • Substitute $y = -10$ into 1st equation:       $4x + 4(-10) = 36 \implies 4x - 40 = 36 \implies 4x = 76 \implies x = 19$

  • Solution: $(19, -10)$

  • Example 8 (Finding $x$--coordinate only): $\begin{cases} 4x - 3y = 40 \\ -5x + 6y = 59 \end{cases}$

  • To find $x$ directly, eliminate $y$. Multiply 1st equation by $2$:       $2(4x - 3y) = 2(40) \implies 8x - 6y = 80$

  • Add to 2nd equation:       $\begin{cases} 8x - 6y = 80 \\ -5x + 6y = 59 \end{cases} \implies 3x = 139 \implies x = -7$

  • Example 9 (Finding $y$--coordinate only): $\begin{cases} 2x + 8y = -14 \\ 3x - 6y = 3 \end{cases}$

  • To find $y$ directly, eliminate $x$. Multiply 1st by $-3$ and 2nd by $2$:       $\begin{cases} -6x - 24y = 42 \\ 6x - 12y = 6 \end{cases} \implies -36y = 48 \implies y = -1$

  • Example 10 (Infinitely Many Solutions): Solve $\begin{cases} -2x + y = 5 \\ -4x + 2y = 10 \end{cases}$

  • Multiply 1st equation by $-2$:       $\begin{cases} 4x - 2y = -10 \\ -4x + 2y = 10 \end{cases} \implies 0 = 0$

  • An identity ($0 = 0$) indicates infinitely many solutions.

  • Expressed in Set-Builder Notation:       $\{(x, y) \mid -2x + y = 5\} \quad \text{or} \quad \{(x, y) \mid -4x + 2y = 10\}$

Section 1.4: Graphs of Linear Inequalities

• Definition:

  • A linear inequality in two variables replaces the equal sign with $\le, \ge, <,$ or >$.\n * Solution set is represented graphically by a half-plane bounded by a line.\n\n\n\n* **Example 1: Testing Solution Point**\n * Check if (-3, -7)$satisfies$y > -x - 4\n * -7 > -(-3) - 4 \implies -7 > 3 - 4 \implies -7 > -1 (False)\n * Point (-3, -7) is not a solution.\n\n* **Example 2: Testing Solution Point**\n * Check if (-1, 1)$satisfies$2x + 10y \ge 5\n * 2(-1) + 10(1) \ge 5 \implies -2 + 10 \ge 5 \implies 8 \ge 5 (True)\n * Point (-1, 1) is a solution.\n\n* **Steps for Graphing Linear Inequalities:**\n 1. Rewrite inequality as an equation to graph the boundary line.\n 2. **Boundary Line Style:**\n * Solid line if inequality contains \le$or$\ge (boundary included).\n * Dashed line if inequality contains <$or$> (boundary excluded).\n 3. Select a test point not on the line (e.g., (0, 0)).\n 4. Substitute test point into inequality: if true, shade half-plane containing test point; if false, shade opposite half-plane.\n\n* **Shortcut Rules (for y isolated on left side):**\n * y < mx + b: Shade below dashed line.\n * y \le mx + b: Shade below solid line.\n * y > mx + b: Shade above dashed line.\n * y \ge mx + b: Shade above solid line.\n\n* **Example 3:** Graph -2x + y \le 4\n * Boundary: -2x + y = 4$(Solid line through$(-2, 0)$and$(0, 4)).\n * Test (0, 0)$:$-2(0) + 0 \le 4 \implies 0 \le 4 (True).\n * Shade half-plane containing (0, 0) (below line).\n\n* **Example 4:** Graph x + y < -3\n * Boundary: x + y = -3$(Dashed line through$(-3, 0)$and$(0, -3)).\n * Test (0, 0)$:$0 + 0 < -3 \implies 0 < -3 (False).\n * Shade half-plane not containing (0, 0) (below line).\n\n* **Example 5:** Graph y \ge 3x + 6\n * Boundary: y = 3x + 6$(Solid line through$(-2, 0)$and$(0, 6)).\n * Form y \ge mx + b \implies shade on or above line.\n\n* **Example 6:** Graph -12x - 3y > -9\n * Boundary: -12x - 3y = -9$(Dashed line through$\left(\frac{3}{4}, 0\right)$and$(0, 3)).\n * Isolate y$:$-3y > 12x - 9 \implies y < -4x + 3 (reverse inequality when dividing by negative number).\n * Form y < mx + b \implies shade below line.\n\n* **Systems of Linear Inequalities:**\n * Graph each inequality individually.\n * Solution set is the intersection (overlapping area) of all shaded half-planes.\n\n* **Example 7:** Graph \begin{cases} y \ge x + 2 \ y \ge -x - 2 \end{cases}\n * Line 1: y = x + 2$(Solid through$(-2, 0)$and$(0, 2), shade above).\n * Line 2: y = -x - 2$(Solid through$(-2, 0)$and$(0, -2), shade above).\n * Solution: Overlapping green region above both lines.\n\n* **Example 8:** Graph \begin{cases} y > -3x + 3 \ 2x - y > 4 \end{cases}\n * Line 1: y = -3x + 3$(Dashed through$(1, 0)$and$(0, 3), shade above).\n * Line 2: 2x - y = 4 \implies y < 2x - 4$(Dashed through$(2, 0)$and$(0, -4), shade below).\n * Solution: Overlapping region above Line 1 and below Line 2.\n\n* **Example 9:** Graph \begin{cases} -12x + 3y > -6 \ -3x - y \ge 3 \end{cases}\n * Line 1: -12x + 3y = -6 \implies y > 4x - 2$(Dashed through$\left(\frac{1}{2}, 0\right)$and$(0, -2), shade above).\n * Line 2: -3x - y = 3 \implies y \le -3x - 3$(Solid through$(-1, 0)$and$(0, -3), shade below).\n\n* **Example 10:** Graph \begin{cases} 2x + 3y < 9 \ x \ge 2 \ y \ge 0 \end{cases}\n * Line 1: 2x + 3y = 9 \implies y < -\frac{2}{3}x + 3$(Dashed through$\left(\frac{9}{2}, 0\right)$and$(0, 3), shade below).\n * Line 2: x = 2 (Solid vertical line, shade right).\n * Line 3: y = 0$(Solid horizontal line /$x--axis, shade above).\n * Solution: Bounded triangular region between x = 2$,$y = 0$, and below$2x + 3y = 9$.

• Determining Inequality Systems from Graphs:

  • Example 11:

  • Line 1: $(0, 2)$ and $(-6, 0) \implies m = \frac{1}{3} \implies y = \frac{1}{3}x + 2$ (Solid, shaded below: $y \le \frac{1}{3}x + 2$).

  • Line 2: $(0, -4)$ and $(3, 0) \implies m = \frac{4}{3} \implies y = \frac{4}{3}x - 4$ (Solid, shaded below: $y \le \frac{4}{3}x - 4$).

  • System: $\begin{cases} y \le \frac{1}{3}x + 2 \\ y \le \frac{4}{3}x - 4 \end{cases}$

  • Example 12:

  • Line 1: $(0, 7)$ and $(2, 0) \implies m = -\frac{7}{2} \implies y = -\frac{7}{2}x + 7$ (Solid, shaded above: $y \ge -\frac{7}{2}x + 7$).

  • Line 2: $(0, 5)$ and $(5.5, 0) \implies m = -\frac{10}{11} \implies y = -\frac{10}{11}x + 5$ (Dashed, shaded above: $y > -\frac{10}{11}x + 5$).

  • Bounded by axes: $x \ge 0$, y \ge 0$.\n * System: \begin{cases} y \ge -\frac{7}{2}x + 7 \ y > -\frac{10}{11}x + 5 \ x \ge 0 \ y \ge 0 \end{cases}\n * **Example 13:**\n * Line 1: (0, 9)$and$(0.5, 0) \implies y = -18x + 9$(Solid, shaded above:$y \ge -18x + 9).\n * Line 2: y = 9$(Dashed, shaded below:$y < 9).\n * Line 3: x = 3$(Dashed, shaded left:$x < 3).\n * Line 4: y = -2$(Solid, shaded above:$y \ge -2).\n * System: \begin{cases} y \ge -18x + 9 \ x < 3 \ -2 \le y < 9 \end{cases}\n * **Example 14:**\n * Line 1: (0, 5)$and$(1, 0) \implies y = -5x + 5$(Solid, shaded below:$y \le -5x + 5$or$5x + y \le 5).\n * Line 2: (0, 2)$and$(4, 0) \implies y = -\frac{1}{2}x + 2$(Solid, shaded below:$y \le -\frac{1}{2}x + 2$or$x + 2y \le 4).\n * System: \begin{cases} y \le -5x + 5 \ y \le -\frac{1}{2}x + 2 \ x \ge 0 \ y \ge 0 \end{cases}\n\n# Section 1.5: Linear Models\n\n* **Linear Depreciation:**\n * **Asset:** An item owned that has value.\n * **Linear Depreciation:** Reduction in book value of an asset over time.\n * **Purchase Price / Original Cost (b$):** Price paid when asset is acquired at$t = 0$.

  • Scrap Value: Remaining value after usable lifespan ends.

  • Value Model: $V(t) = mt + b$, where $V(t)$ is book value at time $t$, $m < 0$ is slope, and $b$ is purchase price.

  • Rate of Depreciation: Expressed as a positive real number $-m$ (amount value declines per unit of time).

• Example 1: SUV Depreciation

  • Original cost = $\$30,500$, lifespan = $5\,\text{years}$, scrap value = $\$10,300$.

  • Part A (Rate of Depreciation):

  • Points: $(0, 30500)$ and $(5, 10300)$.

  • Slope $m = \frac{10300 - 30500}{5 - 0} = \frac{-20200}{5} = -4040$

  • Rate of depreciation = $\$4040/\text{year}$.

  • Part B (Linear Equation):

  • $V(t) = -4040t + 30,500$ for $0 \le t \le 5$

  • Part C (Value at End of Year 3):

  • $V(3) = -4040(3) + 30,500 = -12,120 + 30,500 = \$18,380$

• Example 2: Tour Bus Depreciation

  • Purchase price = $\$185,000$, lifespan = $10\,\text{years}$, scrap value = $\$75,000$.

  • Part A: Slope $m = \frac{75000 - 185000}{10 - 0} = -11,000 \implies$ Rate of depreciation = $\$11,000/\text{year}$.

  • Part B: $V(t) = -11,000t + 185,000$ for $0 \le t \le 10$

  • Part C: $V(7) = -11,000(7) + 185,000 = -77,000 + 185,000 = \$108,000$

  • Part D (Time when Value is $\$130,000$):

  • $-11,000t + 185,000 = 130,000 \implies -11,000t = -55,000 \implies t = 5\,\text{years}$

• Example 3: Computer System Depreciation

  • Cost = $\$45,200$, lifespan = $3\,\text{years}$, scrap value = \0.\n * **Part A:** Slope m = \frac{0 - 45200}{3 - 0} = -15,066.67 \implies$Rate of depreciation =$\$15,066.67/\text{year}.\n * **Part B:** V(t) = -15,066.67t + 45,200$for$0 \le t \le 3\n * **Part C (Value at 1.5\,\text{years}):**\n * V(1.5) = -15,066.67(1.5) + 45,200 = \$22,600\n\n* **Cost, Revenue, and Profit Functions:**\n * **Fixed Costs (F):** Costs independent of production volume (e.g., rent, insurance).\n * **Variable Costs (cx$):** Costs proportional to production quantity$x (e.g., raw materials, utilities).\n * **Linear Cost Function:** C(x) = cx + F$, where$c$is unit production cost and$F is fixed cost.\n * **Linear Revenue Function:** R(x) = sx$, where$s is unit selling price.\n * **Linear Profit Function:** P(x) = R(x) - C(x) = sx - (cx + F) = (s - c)x - F\n\n* **Example 4: Production Analysis**\n * Given C(x) = 5x + 50,000$and$R(x) = 13x\n * **Part A:** Total cost for 4850$units:$C(4850) = 5(4850) + 50,000 = \$74,250\n * **Part B:** Revenue for 5750$units:$R(5750) = 13(5750) = \$74,750\n * **Part C:** Profit function: P(x) = 13x - (5x + 50,000) = 8x - 50,000\n * **Part D:** Profit/Loss for 7275$units:$P(7275) = 8(7275) - 50,000 = 58,200 - 50,000 = \$8200 (Profit)\n\n* **Example 5: Gym Equipment Manufacturer**\n * Unit cost c = \$24$, unit selling price$s = \$52$, monthly fixed cost$F = \$150,000.\n * **Part A:** C(x) = 24x + 150,000\n * **Part B:** C(10,000) = 24(10,000) + 150,000 = \$390,000\n * **Part C:** R(x) = 52x\n * **Part D:** R(10,000) = 52(10,000) = \$520,000\n * **Part E:** P(x) = 52x - (24x + 150,000) = 28x - 150,000\n * **Part F:** P(10,000) = 28(10,000) - 150,000 = \$130,000 (Profit)\n\n* **Example 6: Living Active Foam Rollers**\n * Unit cost c = \$10$, unit selling price$s = \$25$, fixed costs$F = \$135,000.\n * **Part A:** Profit function: P(x) = 25x - (10x + 135,000) = 15x - 135,000\n * **Part B:** For 15,300$units:$P(15,300) = 15(15,300) - 135,000 = \$94,500 (Profit)\n * **Part C:** For 8500$units:$P(8500) = 15(8500) - 135,000 = -\$7500$(Loss of$\$7500)\n * **Part D:** For 9000$units:$P(9000) = 15(9000) - 135,000 = \$0 (Breaks even)\n\n* **Break-Even Analysis:**\n * **Break-Even Point:** Point where revenue equals cost (R(x) = C(x)$) and profit is zero ($P(x) = 0).\n * **Break-Even Quantity (x):** The production quantity where revenue equals cost.\n * **Break-Even Revenue (y):** The revenue generated at the break-even quantity.\n * **Relationships:**\n * Quantity > Break-even quantity \implies Profit\n * Quantity < Break-even quantity \implies Loss\n\n\n\n* **Example 7: Break-Even Calculations**\n * Given C(x) = 14x + 133,600$and$R(x) = 22x\n * **Part A (Break-Even Quantity):**\n * 22x = 14x + 133,600 \implies 8x = 133,600 \implies x = 16,700\,\text{units}\n * **Part B (Break-Even Revenue):**\n * R(16,700) = 22(16,700) = \$367,400\n * **Part C (Break-Even Point):** (16700, 367400)\n * **Part D:** Selling 20,000$units ($> 16,700) results in a **Profit**.\n\n* **Example 8: Easy Cooking Crock Pots**\n * Unit cost c = \$18$, selling price$s = \$42$, fixed cost$F = \$264,000.\n * **Part A:** 42x = 18x + 264,000 \implies 24x = 264,000 \implies x = 11,000\,\text{crock pots}\n * **Part B:** R(11,000) = 42(11,000) = \$462,000\n * **Part C:** Break-even point = (11000, 462000)\n * **Part D:** 10,500$units ($< 11,000) results in a **Loss**.\n * **Part E:** 25,000$units ($> 11,000) results in a **Profit**.\n\n* **Supply, Demand, and Market Equilibrium:**\n * **Linear Demand Function:** D(p) = mp + b$, where$p$is unit price and$D(p)$is quantity demanded. Slope$m < 0 (higher price reduces demand).\n\n\n\n * **Linear Supply Function:** S(p) = mp + b$, where$p$is unit price and$S(p)$is quantity supplied. Slope$m > 0 (higher price increases supply).\n\n\n\n * **Market Equilibrium:** Occurs when quantity demanded equals quantity supplied (D(p) = S(p)).\n * **Equilibrium Price (p$):** The price where$D(p) = S(p).\n * **Equilibrium Quantity:** The quantity corresponding to the equilibrium price.\n * **Equilibrium Point:** (p, D(p)).\n\n\n\n* **Example 9: Equilibrium Calculation**\n * Given D(p) = -32p + 900$and$S(p) = 8p + 300\n * **Part A (Equilibrium Price):**\n * -32p + 900 = 8p + 300 \implies 40p = 600 \implies p = \$15\n * **Part B (Equilibrium Quantity):**\n * D(15) = -32(15) + 900 = -480 + 900 = 420\,\text{units}\n * **Part C (Equilibrium Point):** (15, 420)\n\n* **Example 10: Heart Monitor Market Equilibrium**\n * Given D(p) = -65p + 1940$and$S(p) = 87p + 420\n * **Part A:** -65p + 1940 = 87p + 420 \implies 152p = 1520 \implies p = \$10\n * **Part B:** D(10) = -65(10) + 1940 = 1290\,\text{units}\n * **Part C:** Equilibrium point = (10, 1290)\n\n* **Least Squares Method (Linear Regression):**\n * Procedure for finding the linear equation f(x) = mx + b$that best fits a set of$n$data points$(x_1, y_1), (x_2, y_2), \dots, (x_n, y_n).\n * Minimizes the sum of squares of vertical deviations from the points to the line.\n * **Normal Equations System:**\n    \begin{cases} nb + \left(\sum x_i\right)m = \sum y_i \ \left(\sum x_i\right)b + \left(\sum x_i^2\right)m = \sum x_i y_i \end{cases}\n * **Procedure:**\n 1. Compute \sum x_i\n 2. Compute \sum y_i\n 3. Compute \sum x_i^2\n 4. Compute \sum x_i y_i\n 5. Solve normal equations system for m$and$b.\n 6. Substitute m$and$b$into$f(x) = mx + b$.

• Example 11: Finding Least-Squares Line

  • Data points: $(1, 3)$, $(2, 4)$, $(5, 6)$ ($n = 3$)

  • Summations Table:

  • $x = 1, y = 3, x^2 = 1, xy = 3$

  • $x = 2, y = 4, x^2 = 4, xy = 8$

  • $x = 5, y = 6, x^2 = 25, xy = 30$

  • Sums: $\sum x = 8$, $\sum y = 13$, $\sum x^2 = 30$, $\sum xy = 41$

  • Normal Equations:     $\begin{cases} 3b + 8m = 13 \\ 8b + 30m = 41 \end{cases}$

  • Solving for $m$ and $b$:

  • Multiply 1st by $8$ and 2nd by $-3$:       $\begin{cases} 24b + 64m = 104 \\ -24b - 90m = -123 \end{cases} \implies -26m = -19 \implies m = \frac{19}{26}$

  • Substitute $m$ into 1st equation:       $3b + 8\left(\frac{19}{26}\right) = 13 \implies 3b + \frac{76}{13} = 13 \implies 3b = \frac{93}{13} \implies b = \frac{31}{13}$

  • Least-Squares Line: $f(x) = \frac{19}{26}x + \frac{31}{13}$

• Example 12: Apartment Construction Trend

  • Data points: $(1, 3)$, $(2, 7)$, $(3, 12)$ ($n = 3$)

  • Summations Table:

  • $x = 1, y = 3, x^2 = 1, xy = 3$

  • $x = 2, y = 7, x^2 = 4, xy = 14$

  • $x = 3, y = 12, x^2 = 9, xy = 36$

  • Sums: $\sum x = 6$, $\sum y = 22$, $\sum x^2 = 14$, $\sum xy = 53$

  • Normal Equations:     $\begin{cases} 3b + 6m = 22 \\ 6b + 14m = 53 \end{cases}$

  • Solving for $m$ and $b$:

  • Multiply 1st equation by $-2$:       $\begin{cases} -6b - 12m = -44 \\ 6b + 14m = 53 \end{cases} \implies 2m = 9 \implies m = \frac{9}{2}$

  • Substitute $m$ into 1st equation:       $3b + 6\left(\frac{9}{2}\right) = 22 \implies 3b + 27 = 22 \implies 3b = -5 \implies b = -\frac{5}{3}$

  • Part A (Least-Squares Line): $f(x) = \frac{9}{2}x - \frac{5}{3}$

  • Part B (Approximation for Year 5):

  • Evaluate $f(5) = \frac{9}{2}(5) - \frac{5}{3} = \frac{45}{2} - \frac{5}{3} = \frac{135 - 10}{6} = \frac{125}{6} \approx 20.83$

  • Approximately $20$ new apartment complexes will be completed by the end of the fifth year.