Chapter 1: Linear Equations

Solving Single-Variable Linear Equations and Understanding the Properties of Equality (SAT Math)

Introduction

Linear equations are the foundation of algebra, and mastering them is crucial for success in the SAT Math section. This comprehensive guide will delve deep into solving single-variable linear equations, emphasizing the properties of equality that underpin these solutions. We'll cover definitions, detailed explanations, illustrative examples, and case studies to ensure a thorough understanding.

1. What are Linear Equations?

A linear equation is an algebraic equation where the highest power of the variable is 1. It represents a straight line when graphed. The standard form of a linear equation in one variable is:  

ax + b = c

where:

'x' is the variable

'a' and 'b' are coefficients (real numbers)

'c' is a constant (a real number)

2. Properties of Equality

The properties of equality are fundamental rules that allow us to manipulate equations while maintaining balance. These properties are essential for isolating the variable and solving the equation.  

Addition Property of Equality: If a = b, then a + c = b + c. You can add the same value to both sides of an equation.  

Subtraction Property of Equality: If a = b, then a - c = b - c. You can subtract the same value from both sides of an equation.  

Multiplication Property of Equality: If a = b, then a * c = b * c. You can multiply both sides of an equation by the same non-zero value.  

Division Property of Equality: If a = b and c ≠ 0, then a / c = b / c. You can divide both sides of an equation by the same non-zero value.  

3. Solving Single-Variable Linear Equations

The goal in solving a linear equation is to isolate the variable on one side of the equation. This is achieved by applying the properties of equality step-by-step.  

General Steps:

Simplify both sides: Use the distributive property to remove parentheses and combine like terms.

Isolate the variable term: Use the addition or subtraction property of equality to move the variable term to one side and the constant terms to the other side.

Solve for the variable: Use the multiplication or division property of equality to get the variable by itself.

Check your solution: Substitute the solution back into the original equation to verify its correctness.

4. Illustrative Examples

Example 1: Basic Equation

Solve for x: 2x + 5 = 11

Isolate the variable term: Subtract 5 from both sides:

2x + 5 - 5 = 11 - 5

2x = 6

Solve for the variable: Divide both sides by 2:

2x / 2 = 6 / 2

x = 3

Check the solution: Substitute x = 3 back into the original equation:

2(3) + 5 = 11

6 + 5 = 11

11 = 11 (The solution is correct)

Example 2: Equation with Variables on Both Sides

Solve for x: 3x - 7 = x + 5

Isolate the variable term: Subtract x from both sides:

3x - x - 7 = x - x + 5

2x - 7 = 5

Isolate the variable term: Add 7 to both sides:

2x - 7 + 7 = 5 + 7

2x = 12

Solve for the variable: Divide both sides by 2:

2x / 2 = 12 / 2

x = 6

Check the solution: Substitute x = 6 back into the original equation:

3(6) - 7 = 6 + 5

18 - 7 = 11

11 = 11 (The solution is correct)

Example 3: Equation with Parentheses

Solve for x: 2(x + 3) = 4x - 8

Simplify both sides: Use the distributive property:

2x + 6 = 4x - 8

Isolate the variable term: Subtract 2x from both sides:

2x - 2x + 6 = 4x - 2x - 8

6 = 2x - 8

Isolate the variable term: Add 8 to both sides:

6 + 8 = 2x - 8 + 8

14 = 2x

Solve for the variable: Divide both sides by 2:

14 / 2 = 2x / 2

7 = x

Check the solution: Substitute x = 7 back into the original equation:

2(7 + 3) = 4(7) - 8

2(10) = 28 - 8

20 = 20 (The solution is correct)

5. Case Studies

Case Study 1: Real-World Application

A taxi service charges a base fare of $3 plus $2.50 per mile. If a passenger's total fare was $20.50, how many miles did they travel?

Set up the equation: Let 'x' be the number of miles traveled. The equation representing the situation is: 3 + 2.50x = 20.50

Solve for x:

Subtract 3 from both sides: 2.50x = 17.50

Divide both sides by 2.50: x = 7

Answer: The passenger traveled 7 miles.

Case Study 2: Geometry Problem

The perimeter of a rectangle is 30 cm. If the length is 3 cm more than the width, find the dimensions of the rectangle.

Set up the equations: Let 'w' be the width and 'l' be the length. We have two equations:

l = w + 3

2l + 2w = 30

Solve for the dimensions:

Substitute 'l = w + 3' into the second equation: 2(w + 3) + 2w = 30

Simplify and solve for 'w': 2w + 6 + 2w = 30 => 4w = 24 => w = 6

Substitute 'w = 6' into 'l = w + 3' to find 'l': l = 6 + 3 => l = 9

Answer: The width of the rectangle is 6 cm, and the length is 9 cm.

6. Tips for the SAT Math Section

Practice regularly: Consistent practice with a variety of linear equations will improve your speed and accuracy.

Understand the concepts: Don't just memorize procedures; focus on understanding the underlying properties of equality.

Be careful with signs: Pay close attention to positive and negative signs when manipulating equations.

Check your work: Always substitute your solution back into the original equation to ensure it's correct.

Manage your time: Pace yourself during the SAT to ensure you have enough time for all questions.

Conclusion

Solving single-variable linear equations is a fundamental skill in algebra that is frequently tested on the SAT Math section.

By mastering the properties of equality and practicing regularly, you can confidently tackle these problems and achieve your desired score. Remember to focus on understanding the concepts, pay attention to detail, and manage your time effectively during the exam.  

Solving Equations with Fractions and Decimals: A Comprehensive Guide for SAT Math

Introduction

Linear equations are foundational to algebra and appear frequently on the SAT Math section. A solid understanding of how to solve these equations, especially those involving fractions and decimals, is crucial for achieving a high score. This guide provides a detailed exploration of the topic, incorporating the most updated information for the 2024 SAT.  

1. Fundamental Concepts

Before diving into specific techniques, let's review some essential algebraic principles:

Linear Equations: These are equations where the highest power of the variable is 1. They generally take the form ax + b = c, where a, b, and c are constants, and x is the variable.  

Properties of Equality: These properties allow us to manipulate equations without changing their solutions:  

Addition Property: If a = b, then a + c = b + c

Subtraction Property: If a = b, then a - c = b - c

Multiplication Property: If a = b, then ac = bc

Division Property: If a = b and c ≠ 0, then a/c = b/c  

2. Solving Equations with Fractions

Fractions can make equations appear more complex, but the underlying principles remain the same. Here's a step-by-step approach:

Eliminate the Fractions: The most efficient way to deal with fractions is to eliminate them. Find the least common denominator (LCD) of all the fractions in the equation. Multiply both sides of the equation by the LCD. This will cancel out the denominators, leaving you with a simpler equation to solve.  

Solve for the Variable: Use the properties of equality to isolate the variable on one side of the equation.  

Check Your Solution: Substitute the value you found for the variable back into the original equation to ensure it satisfies the equation.

Example:

Solve for x: (2/3)x + 1/4 = 5/6

Find the LCD: The LCD of 3, 4, and 6 is 12.

Multiply by the LCD:

12 * [(2/3)x + 1/4] = 12 * (5/6)

8x + 3 = 10

Solve for x:

8x = 7

x = 7/8

Check:

(2/3)(7/8) + 1/4 = 5/6

7/12 + 1/4 = 5/6

10/12 = 5/6 (This is true, so our solution is correct)

3. Solving Equations with Decimals

Decimals can be handled similarly to fractions. Here's the process:  

Clear the Decimals: Determine the decimal with the most digits after the decimal point. Multiply both sides of the equation by 10 raised to the power of that number of digits. This...