Simplifying Difference Quotients and Solving Systems of Linear Equations B5-B7

Simplifying Difference Quotients (Average Rate of Change)

• Mathematical setup for the difference quotient representing average rate of change using input expressions $x$ and $x+h$:   $\frac{f(x+h) - f(x)}{(x+h) - x}$

• Complete algebraic expansion of the numerator $f(x+h) - f(x)$ for a quadratic function $f(x) = x^2 + x - 4$:   $(x^2 + 2xh + h^2 + x + h - 4) - (x^2 + x - 4)$

• Application of the distributive property to the subtracted term $-f(x)$:   $x^2 + 2xh + h^2 + x + h - 4 - x^2 - x + 4$

• Simplification of the denominator:   $(x+h) - x = h$

• Combining like terms in the numerator to cancel terms without $h$:

  • $x^2 - x^2 = 0$

  • $-4 + 4 = 0$

  • $x - x = 0$

• Resulting simplified numerator containing exclusively terms with $h$:   $2xh + h^2 + h$

• Factoring out the common factor $h$ from the numerator:   $h(2x + h + 1)$

• Dividing out $h$ from both numerator and denominator:   $\frac{h(2x + h + 1)}{h} = 2x + h + 1$

• Fundamental properties of difference quotients with inputs $x$ and $x+h$:

  • Every term in the numerator that does not contain an $h$ will cancel out completely.

  • The factor $h$ in the denominator will always cancel with a factored $h$ from the numerator.

  • The simplified output remains an algebraic expression containing variables ($2x + h + 1$), rather than a single numerical value, due to using algebraic inputs rather than specific constants.

Systems of Linear Equations and Real-World Applications

• A system of linear equations consists of two or more linear equations evaluated simultaneously to find common solution points $(x, y)$.

• Common structural forms for linear equations:

  • Standard Form: $Ax + By = C$

  • Slope-Intercept Form: $y = mx + b$

• Case Study: Car Rental Agency Comparison over a 3-day period:

  • Fun Time Rentals:

  • Daily rate: $\$40/\text{day}$

  • Per-mile rate: $\$0.10/\text{mile}$

  • Total fixed cost for 3 days: $3 \times \$40 = \$120$

  • Cost function: $f(x) = 0.10x + 120$

  • Good Time Rentals:

  • Daily rate: $\$50/\text{day}$

  • Per-mile rate: $\$0.08/\text{mile}$

  • Total fixed cost for 3 days: $3 \times \$50 = \$150$

  • Cost function: $g(x) = 0.08x + 150$

• Determining the break-even mileage where rental costs are equal:

  • Set cost functions equal: $f(x) = g(x)$

  • Equation setup: $0.10x + 120 = 0.08x + 150$

  • Subtract $0.08x$ from both sides: $0.02x + 120 = 150$

  • Subtract $120$ from both sides: $0.02x = 30$

  • Divide by $0.02$: $x = 1500\,\text{miles}$

• Economic interpretation of the break-even mileage:

  • At $1500\,\text{miles}$ total (an average of $500\,\text{miles/day}$ over 3 days), both rental agencies cost the exact same amount.

  • For driving distances below $1500\,\text{miles}$, Fun Time Rentals ($f(x)$) is cheaper due to its lower fixed daily cost.

  • For driving distances above $1500\,\text{miles}$, Good Time Rentals ($g(x)$) is cheaper due to its lower per-mile rate.

Graphical Method and Solution Classification for Linear Systems

• Solving systems by graphing involves converting equations to slope-intercept form, plotting the lines, and identifying intersection points.

• Example of solving a system graphically:

  • System:     $\text{Line 1: } x - y = 3 \implies y = x - 3$     $\text{Line 2: } x + 2y = 6 \implies y = -\frac{1}{2}x + 3$

  • Line 1 properties: Y-intercept at $(0, -3)$, slope $m = 1$

  • Line 2 properties: Y-intercept at $(0, 3)$, slope $m = -\frac{1}{2}$

  • Point of intersection / Solution: $(4, 1)$

  • Algebraic verification of solution $(4, 1)$:

  • Equation 1 check: $4 - 1 = 3$

  • Equation 2 check: $4 + 2(1) = 6$

• Practical drawbacks and limitations of the graphical method:

  • Graph scaling challenges when dealing with very large numerical values.

  • Inaccuracy when Y-intercepts are non-integers or decimals (e.g., $2.64$, $3.12$).

  • Inability to read precise coordinates if the intersection point lies off grid intersections (e.g., $(4.7, 1.34)$).

• The Three Universal Solution Types for Linear Systems:

  • Intersecting Lines (Exactly One Solution):

  • Occurs when lines have different slopes ($m_1 \neq m_2$).

  • Lines cross at a single distinct ordered pair $(x, y)$.

  • Parallel Lines (No Solution):

  • Occurs when lines have equal slopes ($m_1 = m_2$) but different Y-intercepts ($b_1 \neq b_2$).

  • Lines never meet.

  • Algebraic solving yields an impossible statement or contradiction (e.g., $0 = -24$).

  • Identical / Coincident Lines (Infinitely Many Solutions):

  • Occurs when lines share both equal slopes ($m_1 = m_2$) and equal Y-intercepts ($b_1 = b_2$).

  • Lines overlap completely.

  • Algebraic solving yields a universally true identity (e.g., $0 = 0$).

Algebraic Solution Methods: Substitution Method

• Mechanics of the Substitution Method:

  • Isolate one variable ($x$ or $y$) in either equation.

  • Substitute the isolated variable expression into the remaining equation, creating a single-variable equation.

  • Solve for the single variable.

  • Back-substitute the numerical value into the isolated variable expression to find the second variable value.

  • State the final answer as an ordered pair $(x, y)$.

• Step-by-Step Example using Substitution:

  • System of equations:     $\text{Equation 1: } 4x - y = 8$     $\text{Equation 2: } -2x + 3y = 6$

  • Step 1: Isolate variable $y$ in Equation 1:     $-y = -4x + 8 \implies y = 4x - 8$

  • Step 2: Substitute $(4x - 8)$ for $y$ in Equation 2:     $-2x + 3(4x - 8) = 6$

  • Step 3: Expand and solve for $x$:     $-2x + 12x - 24 = 6$     $10x - 24 = 6$     $10x = 30 \implies x = 3$

  • Step 4: Substitute $x = 3$ into the isolated $y$ equation:     $y = 4(3) - 8 = 12 - 8 = 4$

  • Final Solution: $(3, 4)$

  • Verification:

  • Equation 1 check: $4(3) - 4 = 12 - 4 = 8$

  • Equation 2 check: $-2(3) + 3(4) = -6 + 12 = 6$

Algebraic Solution Methods: Addition/Elimination Method and Special Cases

• Mechanics of the Addition/Elimination Method:

  • Write both equations in Standard Form ($Ax + By = C$).

  • Multiply one or both equations by non-zero constants so that coefficients for one chosen variable are equal in magnitude but opposite in sign.

  • Add the equations together to eliminate that variable.

  • Solve the resulting single-variable equation and back-substitute to find the remaining variable.

• Example 1: Solving a Parallel System (No Solution Case):

  • System setup:     $\text{Equation 1: } 3x + 2y = 6$     $\text{Equation 2: } 6x + 4y = -12$

  • Multiply Equation 1 by $-2$:     $-2(3x + 2y = 6) \implies -6x - 4y = -12$

  • Add modified Equation 1 to Equation 2:     $(-6x + 6x) + (-4y + 4y) = -12 + (-12)$     $0x + 0y = -24 \implies 0 = -24$

  • Conclusion: $0 = -24$ is a mathematical contradiction; both variables cancel out simultaneously, indicating the system has no solution (the lines are parallel with slope $m = -\frac{3}{2}$).

• Example 2: Solving a Coincident System (Infinitely Many Solutions Case):

  • System setup:     $\text{Equation 1: } 3x - 6y = 12$     $\text{Equation 2: } 4x - 8y = 16$

  • Multiply Equation 1 by $4$ and Equation 2 by $-3$:     $4(3x - 6y = 12) \implies 12x - 24y = 48$     $-3(4x - 8y = 16) \implies -12x + 24y = -48$

  • Add modified equations together:     $(12x - 12x) + (-24y + 24y) = 48 - 48$     $0 = 0$

  • Conclusion: $0 = 0$ is a true identity statement, indicating infinitely many solutions.

  • Formulating the formal solution set for infinite solutions:

  • Convert an equation to slope-intercept form ($y = mx + b$):       $-6y = -3x + 12 \implies y = \frac{1}{2}x - 2$

  • State the solution set as a parameterized ordered pair:       $\left(x, \frac{1}{2}x - 2\right)$

Setting Up Systems of Linear Equations for Word Problems

• Procedure for modeling narrative problems using linear systems:

  • Define two distinct variables representing the unknown quantities.

  • Extract linear relationships from problem statements to form two independent equations.

• Standardized Test Problem Example:

  • Scenario context: A test contains $125$ total questions worth a combined total of $1300\,\text{points}$.

  • Problem types: True/False questions worth $8\,\text{points}$ each, and Multiple Choice questions worth $14\,\text{points}$ each.

  • Variable definitions:

  • Let $t$ equal the total number of True/False questions.

  • Let $m$ equal the total number of Multiple Choice questions.

  • System formulation:

  • Equation based on total question count:       $t + m = 125$

  • Equation based on total point values:       $8t + 14m = 1300$