Understanding how to solve linear equations is crucial in algebra, providing the tools needed to tackle a wide range of mathematical challenges. This section focuses on solving linear equations with one variable, demonstrating techniques through detailed solutions.

**Principles of Solving Linear Equations**

The process of solving linear equations involves isolating the variable on one side of the equation. Key strategies include:

**Maintaining equality:**Perform the same operation on both sides of the equation.**Isolation of the variable:**Achieve an expression where the variable stands alone on one side.

**Worked Examples**

**Example 1: Solve **$3x + 4 = 10$**.**

**Solution: **

Isolate $x$.

$3x + 4 = 10$

$3x = 10 - 4$

$3x = 6$

$x = \frac{6}{3}$

$x = 2$

**Example 2: Solve **$5 - 2x = 3(x + 7)$.

#### Solution:

Find $x$.

$5 - 2x = 3(x + 7)$

$5 - 2x = 3x + 21$

$-2x - 3x = 21 - 5$

$-5x = 16$

$x = \frac{16}{-5}$

$x = -3.2$

**Techniques in Action**

**Balancing Equations**

Key to solving is to keep the equation balanced. When you add, subtract, multiply, or divide, do so equally on both sides.

**Combining Like Terms**

If the same variable appears on both sides, combine them on one side to simplify.

**Clearing Fractions**

Multiply every term by the least common denominator to eliminate fractions.

**Practice Problems**

Let's solve given practice problems with precise calculations.

**Problem 1: Solve **$2x - 7 = 13$.

**Solution: **

Solve for $x$.

$2x - 7 = 13$

$2x = 13 + 7$

$2x = 20$

$x = \frac{20}{2}$

$x = 10$

**Problem 2: Solve **$\frac{x}{5} + 3 = 8$.

#### Solution:

Determine $x$.

$\frac{x}{5} + 3 = 8$

$\frac{x}{5} = 8 - 3$

$\frac{x}{5} = 5$

$x = 5 \times 5$

$x = 25$