Linear Relationships Homework 3 Answer Key

Linear Relationships Homework 3: Answer Key & Comprehensive Guide

Linear relationships are a fundamental concept in algebra, forming the bedrock for understanding more complex mathematical ideas. This comprehensive guide serves as an answer key and detailed explanation for a hypothetical "Linear Relationships Homework 3," covering various aspects of linear functions, including slope, intercepts, graphing, and equation forms. We'll delve into each problem type, providing step-by-step solutions and valuable insights to solidify your understanding. Remember, understanding the why behind the answer is as important as the answer itself.

Understanding Linear Relationships

Before diving into the solutions, let's recap the core components of linear relationships. A linear relationship is characterized by a constant rate of change, meaning the relationship between two variables can be represented by a straight line on a graph. Key elements include:

• Slope (m): Represents the rate of change; it's the steepness of the line. A positive slope indicates a positive relationship (as x increases, y increases), while a negative slope indicates a negative relationship (as x increases, y decreases). The slope is calculated as: m = (y₂ - y₁) / (x₂ - x₁) where (x₁, y₁) and (x₂, y₂) are two points on the line.

• y-intercept (b): The point where the line crosses the y-axis (where x = 0). It represents the initial value or starting point.

• Equation Forms: Linear relationships are commonly expressed in these forms:

  • Slope-intercept form: y = mx + b (most commonly used)
  • Standard form: Ax + By = C
  • Point-slope form: y - y₁ = m(x - x₁)

Homework 3 Problems & Solutions

Let's assume Homework 3 contains the following problems (these are examples; replace with your actual problems):

Problem 1: Finding the Slope and y-intercept

Find the slope (m) and y-intercept (b) of the line passing through points (2, 5) and (4, 9). Then write the equation of the line in slope-intercept form.

Solution:

1. Calculate the slope (m):

   m = (9 - 5) / (4 - 2) = 4 / 2 = 2

2. Find the y-intercept (b): Use the point-slope form with one of the points (let's use (2, 5)) and the slope:

   y - 5 = 2(x - 2) y - 5 = 2x - 4 y = 2x + 1

   Therefore, the y-intercept (b) is 1.

3. Equation in slope-intercept form: y = 2x + 1

Problem 2: Graphing a Linear Equation

Graph the linear equation: 3x - 2y = 6

Solution:

1. Find the x-intercept: Set y = 0 and solve for x:

   3x - 2(0) = 6 3x = 6 x = 2 The x-intercept is (2, 0).

2. Find the y-intercept: Set x = 0 and solve for y:

   3(0) - 2y = 6 -2y = 6 y = -3 The y-intercept is (0, -3).

3. Plot the intercepts: Plot the points (2, 0) and (0, -3) on a coordinate plane.

4. Draw the line: Draw a straight line passing through both points. This line represents the graph of the equation 3x - 2y = 6.

Problem 3: Writing an Equation from a Graph

Write the equation of the line shown in the graph (assume a graph is provided showing a line). Let's assume the graph shows a line passing through points (1, 2) and (3, 6).

Solution:

1. Find the slope (m):

   m = (6 - 2) / (3 - 1) = 4 / 2 = 2

2. Use point-slope form: Use one of the points (let's use (1, 2)) and the slope:

   y - 2 = 2(x - 1) y - 2 = 2x - 2 y = 2x

   The equation of the line is y = 2x.

Problem 4: Parallel and Perpendicular Lines

Find the equation of a line that is parallel to y = 3x + 5 and passes through the point (1, 2).

Solution:

Parallel lines have the same slope. The slope of y = 3x + 5 is 3. Use the point-slope form with the given point (1, 2) and the slope 3:

y - 2 = 3(x - 1) y - 2 = 3x - 3 y = 3x - 1

The equation of the parallel line is y = 3x - 1.

Problem 5: Real-World Application

A taxi charges a $3 initial fee plus $2 per mile. Write a linear equation representing the total cost (y) as a function of the number of miles (x). What is the cost of a 5-mile trip?

Solution:

The initial fee is the y-intercept ($3). The cost per mile is the slope ($2). The equation is:

y = 2x + 3

For a 5-mile trip (x = 5):

y = 2(5) + 3 = 10 + 3 = $13

The cost of a 5-mile trip is $13.

Problem 6: Solving a System of Linear Equations

Solve the following system of equations using either substitution or elimination:

x + y = 5 x - y = 1

Solution: (Using Elimination)

Add the two equations together:

(x + y) + (x - y) = 5 + 1 2x = 6 x = 3

Substitute x = 3 into either equation (let's use the first one):

3 + y = 5 y = 2

The solution is x = 3, y = 2.

Problem 7: Interpreting Slope and Intercept in Context

A company's profit (P) is modeled by the equation P = 100x - 5000, where x is the number of units sold. Interpret the slope and y-intercept in the context of this problem.

Solution:

• Slope (100): This represents the profit increase per unit sold. For each additional unit sold, the profit increases by $100.

• y-intercept (-5000): This represents the initial profit (or loss) when no units are sold. In this case, the company has a loss of $5000 before any sales occur.

Problem 8: Writing Equations from Word Problems

The number of bacteria in a culture doubles every hour. If there are initially 100 bacteria, write an equation that models the number of bacteria (B) after t hours.

Solution:

This is an exponential growth problem, not strictly linear. However, understanding linear functions is crucial for grasping exponential ones. The equation would be:

B = 100 * 2^t

While not linear, the core concepts of representing relationships mathematically are similar.

Further Practice and Resources

This detailed answer key provides a comprehensive understanding of the concepts involved in linear relationships. Remember to practice regularly to solidify your understanding. You can create more practice problems by:

• Changing the numbers: Use different coordinates, intercepts, or equations to test your understanding.
• Working backward: Start with an equation and find points on the line or create word problems from the equation.
• Using online resources: Explore free online resources (many educational websites offer interactive exercises and quizzes on linear equations).

By mastering the fundamentals of linear relationships, you'll build a strong foundation for more advanced mathematical concepts. Continue practicing, and remember that consistent effort leads to success!