14.3 Linear Functions, Slope-Intercept Form & Systems of Linear Equations
Key Takeaways
- A relation is a function if and only if each input element x in the domain corresponds to exactly one output value y in the range, passing the graphical Vertical Line Test.
- Slope measures the constant rate of change: m = (y_2 - y_1)/(x_2 - x_1) = Δy/Δx; slopes are classified as positive (rising), negative (falling), zero (m = 0, horizontal line y = c), or undefined (m = undefined, vertical line x = c).
- Linear equations are expressed in three standard interchangeable formats: Slope-Intercept Form (y = mx + b), Point-Slope Form (y - y_1 = m(x - x_1)), and Standard Form (Ax + By = C, with integer coefficients where A ≥ 0).
- Parallel lines share identical slopes (m_1 = m_2) with different intercepts, while perpendicular lines have negative reciprocal slopes (m_1 · m_2 = -1, or m_2 = -1/m_1).
- Systems of two linear equations in two variables are classified as consistent-independent (1 unique solution, intersecting lines), inconsistent (0 solutions, parallel lines), or consistent-dependent (infinitely many solutions, coincident lines), and can be solved algebraically via Substitution or Elimination.
Linear Functions, Slope-Intercept Form & Systems of Linear Equations
Quick Answer: On the WEST-B Mathematics subtest (Objective 0017), linear functions and systems questions test your coordinate algebra and multi-variable problem-solving. Remember: A relation is a function if every x has exactly one y (passes the Vertical Line Test). Slope is m = (y_2 - y_1)/(x_2 - x_1). Horizontal lines have m = 0 (y = c); vertical lines have undefined slope (x = c). Parallel lines have equal slopes (m_1 = m_2); perpendicular lines have negative reciprocal slopes (m_2 = -1/m_1). For a 2 × 2 system of linear equations, use Substitution when a variable has coefficient ±1, and Elimination when coefficients align nicely. A system has 1 solution (intersecting lines), no solution (parallel lines), or infinite solutions (identical lines).
1. Relations, Functions, Domain, Range & Function Notation
An algebraic relation is any set of ordered pairs (x, y). A function is a specialized relation where each input value x in the domain is paired with exactly one unique output value y in the range.
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| FUNCTION DETERMINATION CRITERIA |
| |
| +-------------------+----------------------------+------------------------------------------+ |
| | REPRESENTATION | IS A FUNCTION (YES) | NOT A FUNCTION (NO) | |
| +-------------------+----------------------------+------------------------------------------+ |
| | Set of Pairs | {(1, 4), (2, 7), (3, 4)} | {(1, 4), (1, 9), (2, 7)} | |
| | | (Outputs can repeat!) | (Input 1 has two different outputs: 4 & 9| |
| +-------------------+----------------------------+------------------------------------------+ |
| | Mapping Diagram | Each x has exactly 1 arrow | An x has 2 or more arrows departing | |
| +-------------------+----------------------------+------------------------------------------+ |
| | Graphical (VLT) | Any vertical line cuts | A vertical line intersects the curve at | |
| | | the curve at AT MOST 1 pt | TWO or more points (fails VLT) | |
| +-------------------+----------------------------+------------------------------------------+ |
| | Equation Form | y = 3x - 5, y = x² + 2 | x² + y² = 25 (Circle), x = 4 (Vertical) | |
| +-------------------+----------------------------+------------------------------------------+ |
+---------------------------------------------------------------------------------------------------+
Function Notation: f(x)
The notation f(x) (read "f of x") represents the output value y generated when input x is processed by function f.
- Evaluating Functions: If f(x) = -2x² + 5x - 3, evaluate f(-3): f(-3) = -2(-3)² + 5(-3) - 3 = -2(9) - 15 - 3 = -18 - 15 - 3 = -36
- Solving for Input: If g(x) = 4x - 7, find x when g(x) = 25: 4x - 7 = 25 => 4x = 32 => x = 8
2. Rate of Change & Slope (m)
The slope (m) of a non-vertical line passing through points (x_1, y_1) and (x_2, y_2) is the ratio of vertical change (rise, Δy) to horizontal change (run, Δx): m = Rise / Run = Δy / Δx = (y_2 - y_1) / (x_2 - x_1)
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| THE FOUR TYPES OF SLOPE |
| |
| Positive Slope (m > 0) Negative Slope (m < 0) |
| / \ |
| / Rises left-to-right \ Falls left-to-right |
| / Δy > 0 as Δx > 0 \ Δy < 0 as Δx > 0 |
| / \ |
| |
| Zero Slope (m = 0) Undefined Slope (m = undefined) |
| ------------------- Horizontal Line | Vertical Line |
| Equation: y = c | Equation: x = c |
| Δy = 0, Δx ≠ 0 | Δx = 0 (division by zero!) |
| Function? YES | Function? NO |
+---------------------------------------------------------------------------------------------------+
Trap Alert: A horizontal line has slope m = 0 and equation y = c. A vertical line has undefined slope and equation x = c. Vertical lines are NOT functions because they fail the Vertical Line Test infinitely.
3. Forms of Linear Equations & Intercepts
Linear equations on the Cartesian coordinate plane can be written in three interchangeable algebraic forms.
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| FORMS OF LINEAR EQUATIONS MATRIX |
| |
| +-----------------------+-----------------------------+-------------------------------------+ |
| | FORM NAME | STANDARD FORMULA | KEY PARAMETERS & UTILITY | |
| +-----------------------+-----------------------------+-------------------------------------+ |
| | Slope-Intercept Form | y = mx + b | m = Slope | |
| | | | b = y-intercept at point (0, b) | |
| | | | (Ideal for graphing & slope reading)| |
| +-----------------------+-----------------------------+-------------------------------------+ |
| | Point-Slope Form | y - y₁ = m(x - x₁) | m = Slope | |
| | | | (x₁, y₁) = Any known point on line | |
| | | | (Ideal for writing line equations) | |
| +-----------------------+-----------------------------+-------------------------------------+ |
| | Standard Form | Ax + By = C | A, B, C are integers, A ≥ 0 | |
| | | | Slope m = -A/B | |
| | | | x-intercept = (C/A, 0) | |
| | | | y-intercept = (0, C/B) | |
| +-----------------------+-----------------------------+-------------------------------------+ |
+---------------------------------------------------------------------------------------------------+
Finding Intercepts
- x-intercept: The point where the graph crosses the x-axis. Set y = 0 and solve for x.
- y-intercept: The point where the graph crosses the y-axis. Set x = 0 and solve for y.
- Example: For 3x - 5y = 30:
- x-intercept: 3x - 5(0) = 30 => 3x = 30 => x = 10 => (10, 0)
- y-intercept: 3(0) - 5y = 30 => -5y = 30 => y = -6 => (0, -6)
4. Parallel & Perpendicular Lines
The geometric relationship between two lines is determined entirely by their slopes.
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| PARALLEL VS. PERPENDICULAR LINE CRITERIA |
| |
| +-------------------+-----------------------------+-----------------------------------------+ |
| | RELATIONSHIP | SLOPE CONDITION | VISUAL / GEOMETRIC EFFECT | |
| +-------------------+----------------------------+-----------------------------------------+ |
| | Parallel Lines | m₁ = m₂ | Same steepness, never intersect | |
| | (L₁ ∥ L₂) | (b₁ ≠ b₂) | (Equal slopes, distinct y-intercepts) | |
| +-------------------+-----------------------------+-----------------------------------------+ |
| | Perpendicular | m₁ · m₂ = -1 | Intersect at right angles (90°) | |
| | Lines (L₁ ⊥ L₂) | m₂ = -1 / m₁ | Slopes are OPPOSITE RECIPROCALS | |
| +-------------------+-----------------------------+-----------------------------------------+ |
| | Horizontal / | m_horiz = 0 | Always perpendicular to each other | |
| | Vertical Lines | m_vert = undefined | y = a is perpendicular to x = b | |
| +-------------------+-----------------------------+-----------------------------------------+ |
+---------------------------------------------------------------------------------------------------+
Finding Opposite Reciprocals
- If m_1 = 2/5 => m_2 = -5/2
- If m_1 = -4 => m_2 = +1/4
- If m_1 = -7/3 => m_2 = +3/7
5. Systems of Linear Equations in Two Variables
A system of linear equations consists of two or more linear equations containing the same variables. Equation 1: a₁x + b₁y = c₁ Equation 2: a₂x + b₂y = c₂
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| SYSTEM CLASSIFICATION & SOLUTIONS |
| |
| +-----------------------+-----------------------+----------------------+--------------------+ |
| | CLASSIFICATION | SLOPES & INTERCEPTS | GEOMETRIC GRAPH | NUMBER OF SOLUTIONS| |
| +-----------------------+-----------------------+----------------------+--------------------+ |
| | Consistent & | Slopes are DIFFERENT | Two lines intersect | Exactly ONE | |
| | Independent | (m₁ ≠ m₂) | at a single point | unique pair (x, y) | |
| +-----------------------+-----------------------+----------------------+--------------------+ |
| | Inconsistent | Slopes are EQUAL, | Parallel lines | NO SOLUTION | |
| | | Intercepts DIFFERENT | (never intersect) | ∅ or { } | |
| | | (m₁ = m₂, b₁ ≠ b₂) | | | |
| +-----------------------+-----------------------+----------------------+--------------------+ |
| | Consistent & | Slopes are EQUAL, | Coincident lines | INFINITELY MANY | |
| | Dependent | Intercepts EQUAL | (same line graphed) | solutions | |
| | | (m₁ = m₂, b₁ = b₂) | | | |
| +-----------------------+-----------------------+----------------------+--------------------+ |
+---------------------------------------------------------------------------------------------------+
Algebraic Methods for Solving Systems
Method 1: Substitution
Best when at least one variable has a coefficient of 1 or -1.
- Isolate one variable in one equation (e.g., y = 3x - 4).
- Substitute that expression into the other equation.
- Solve the resulting single-variable equation.
- Back-substitute to find the second variable.
Method 2: Elimination / Linear Combination
Best when equations are in standard form Ax + By = C with non-unit coefficients.
- Multiply one or both equations by chosen non-zero constants so that the coefficients of one variable become exact opposites (e.g., +6y and -6y).
- Add the two equations vertically to eliminate that variable.
- Solve for the remaining variable.
- Back-substitute the value into either original equation to find the other variable.
6. Real-World Systems Modeling & Applications
WEST-B questions frequently present multi-variable real-world scenarios requiring algebraic systems modeling.
1. Ticket Sales / Quantity-Value Problems
- Quantity Equation: Count of items (x + y = Total Items)
- Value / Revenue Equation: Monetary worth ((Price₁)x + (Price₂)y = Total Revenue)
2. Break-Even Analysis
- Cost Function: C(x) = Fixed Cost + (Variable Cost per unit)x
- Revenue Function: R(x) = (Selling Price per unit)x
- Break-Even Point: Set R(x) = C(x) and solve for x.
3. Mixture and Solution Problems
- Total Volume Equation: v₁ + v₂ = v_total
- Pure Substance Equation: c₁v₁ + c₂v₂ = (c_final)(v_total)
7. Step-by-Step Worked Problems & Exact Derivations
Problem 1: Equation of a Perpendicular Line
Problem: Write the equation in slope-intercept form (y = mx + b) for the line that passes through the point (-3, 4) and is perpendicular to the line 2x - 6y = 15.
Step-by-Step Solution:
- Find the slope of the given line: Convert 2x - 6y = 15 to slope-intercept form: -6y = -2x + 15 => y = (-2/-6)x + (15/-6) => y = (1/3)x - 5/2 The slope of the given line is m₁ = 1/3.
- Determine the perpendicular slope (m₂): Perpendicular lines have negative reciprocal slopes: m₂ = -1 / m₁ = -1 / (1/3) = -3
- Use point-slope form with m = -3 and point (x₁, y₁) = (-3, 4): y - y₁ = m(x - x₁) y - 4 = -3(x - (-3)) y - 4 = -3(x + 3)
- Distribute and simplify to slope-intercept form: y - 4 = -3x - 9 y = -3x - 5
Problem 2: Solving a 2 × 2 System by Elimination
Problem: Solve the system of linear equations: 4x + 3y = -1 3x - 2y = 12 Determine the solution (x, y) and calculate the value of x + y.
Step-by-Step Solution:
- Choose variable to eliminate: Eliminate y by finding the LCM of 3 and 2, which is 6.
- Multiply Equation 1 by 2 and Equation 2 by 3: 2(4x + 3y) = 2(-1) => 8x + 6y = -2 3(3x - 2y) = 3(12) => 9x - 6y = 36
- Add the two equations vertically to eliminate y: (8x + 9x) + (6y - 6y) = -2 + 36 17x = 34 => x = 2
- Substitute x = 2 into Equation 1 to find y: 4(2) + 3y = -1 8 + 3y = -1 3y = -9 => y = -3
- State solution and calculate required value: Solution: (x, y) = (2, -3) x + y = 2 + (-3) = -1
Problem 3: Real-World Ticket Sales Word Problem
Problem: A school auditorium sold 350 total tickets for a musical performance, generating a total revenue of $3,850. Adult tickets cost $14 each and student tickets cost $7 each. How many adult tickets were sold?
Step-by-Step Solution:
- Define the variables: Let A = number of adult tickets, S = number of student tickets.
- Construct the system of equations:
- Ticket count: A + S = 350
- Revenue: 14A + 7S = 3850
- Use substitution: Express S in terms of A: S = 350 - A
- Substitute into revenue equation: 14A + 7(350 - A) = 3850 14A + 2450 - 7A = 3850
- Combine like terms and solve for A: 7A + 2450 = 3850 7A = 1400 => A = 200
- Calculate student tickets and verify: S = 350 - 200 = 150 Check revenue: 14(200) + 7(150) = 2800 + 1050 = 3850. Exact match! Therefore, 200 adult tickets were sold.
What is the equation in slope-intercept form for the line that passes through the point (-3, 4) and is perpendicular to the line 2x - 6y = 15?
Consider the following system of linear equations: 4x + 3y = -1 3x - 2y = 12 What is the value of x + y?
For what value of k will the following system of linear equations have NO solution? 6x - 9y = 15 4x - ky = 8
A theater sold a total of 350 tickets for a performance, collecting $3,850 in total revenue. Adult tickets cost $14 each and student tickets cost $7 each. How many adult tickets were sold?