# Linear equations in one variable

Linear equations in one variable are a fundamental topic in algebra that students must master for success on the Digital SAT exam. These equations involve finding the value of a single variable that makes the equation true. Understanding linear equations in one variable is crucial as it forms the basis for more advanced algebraic concepts and problem-solving skills. Mastery of this topic enables students to solve real-world problems and perform well in standardized tests like the Digital SAT.

### Learning Objectives

By the end of this topic, you should be able to understand the structure and components of linear equations in one variable, solve these equations efficiently, apply various methods such as isolation and balancing to find the solution, and interpret the solutions in the context of real-world problems. You will also develop the ability to verify solutions and understand how to handle special cases and different forms of linear equations in one variable.

Linear equations in one variable are equations that can be expressed in the form \[ax + b = 0\] , where a and b are constants, and x is the variable. These equations have a unique solution.

### General Form

The standard form of a linear equation in one variable is: \[

ax + b = 0

\]

Where:

- a is the coefficient of x
- b is a constant term
- x is the variable

#### Structure and Components

**Variable**: The unknown quantity we need to solve for, represented by x.**Coefficient**: The numerical factor multiplied by the variable, represented by aaa.**Constant**: The fixed value added to or subtracted from the variable term, represented by b.

#### Solving Linear Equations in One Variable

To solve a linear equation in one variable, follow these steps:

##### Step 1: Simplify Both Sides

Combine like terms and simplify both sides of the equation if necessary. For example:

\[

3x + 2 – x = 10

\]

Simplify to:

\[

2x + 2 = 10

\]

##### Step 2: Isolate the Variable Term

Move all terms containing the variable to one side of the equation and constants to the other side. This often involves adding or subtracting terms on both sides. For instance:

\[

2x + 2 = 10

\]

Subtract 2 from both sides:

\[

2x = 8

\]

##### Step 3: Solve for the Variable

Divide both sides by the coefficient of the variable to solve for x. In this example:

\[

x = \frac{8}{2}

\]

\[

x = 4

\]

### Example Problems

#### Example 1

Solve the equation:

\[

5x – 7 = 3x + 5

\]

**Step-by-Step Solution**:

- Subtract 3x3x3x from both sides: \[5x−3x−7=5\]
- Add 7 to both sides:\[2x=12\]
- Divide by 2: x=6

#### Example 2

Solve the equation:

4(2x−1) = 3(x+5)

**Step-by-Step Solution**:

- Expand both sides: 8x−4 = 3x+15
- Subtract 3x from both sides: 8x−3x−4 = 15
- Add 4 to both sides: 5x = 19
- Divide by 5: \[x = \frac{19}{5}\]

### Special Cases

**No Solution**: If simplifying the equation results in a contradiction, such as 0 = 5, then the equation has no solution.

Example: 2x+3 = 2x−5

Simplify: 3 = −5 This is a contradiction, so there is no solution.**Infinite Solutions**: If simplifying the equation results in a tautology, such as 0 = 0, then the equation has infinitely many solutions.

Example: 3(x−1) = 3x−3

Simplify: 3x−3 = 3x−3 This is always true, so there are infinitely many solutions.

### Properties of Linear Equations in One Variable

**Unique Solution**: Linear equations in one variable always have a unique solution.**No Solution**: If the equation simplifies to a false statement (e.g., 0=5), then it has no solution.**Infinite Solutions**: If the equation simplifies to a true statement (e.g., 0=0), then it has infinitely many solutions.

### Application in Real-World Problems

Linear equations in one variable can be used to solve real-world problems. For instance, finding the cost of items, determining distances, and calculating time.

#### Example

A company charges a fixed cost of $50 plus $10 per hour for consulting services. If the total bill is $150, how many hours of consulting were provided?

**Solution**:

Let xxx be the number of hours of consulting.

50+10x = 150

Subtract 50 from both sides:

10x = 100

Divide by 10:

x=10

So, 10 hours of consulting were provided.

### Verifying Solutions

To verify a solution, substitute the value of x back into the original equation and check if both sides are equal.

#### Example

Verify the solution x=6 for the equation 5x – 7 = 3x + 5.

Substitute x = 6:

5(6) – 7 = 3(6) + 5

30 – 7 = 18 + 5

23 = 23

Both sides are equal, so the solution is verified.

### Practice Problems

- Solve 7x + 3 = 2x + 18.
- Solve 3(4x – 2) = 2(6x + 1).
- Solve -2x + 5 = 3x – 15.
- A car rental company charges a flat fee of $30 plus $0.25 per mile driven. If the total cost is $55, how many miles were driven?

### Summary

Mastering linear equations in one variable is essential for algebra proficiency and success on the Digital SAT exam. This topic involves understanding the structure of equations, using various methods to solve them, and applying these skills to real-world problems. Practice solving different types of linear equations to build confidence and improve problem-solving abilities.