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Edexcel GCSE Maths (Higher) Study Notes

2.4.2 Solving Linear Equations

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=103x + 4 = 10.

Solution:

Isolate xx.

3x+4=103x + 4 = 10

3x=1043x = 10 - 4

3x=63x = 6

x=63x = \frac{6}{3}

x=2x = 2

Example 2: Solve 52x=3(x+7)5 - 2x = 3(x + 7).

Solution:

Find xx.

52x=3(x+7)5 - 2x = 3(x + 7)

52x=3x+215 - 2x = 3x + 21

2x3x=215-2x - 3x = 21 - 5

5x=16-5x = 16

x=165x = \frac{16}{-5}

x=3.2x = -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 2x7=132x - 7 = 13.

Solution:

Solve for xx.

2x7=132x - 7 = 13

2x=13+72x = 13 + 7

2x=202x = 20

x=202x = \frac{20}{2}

x=10x = 10

Problem 2: Solve x5+3=8\frac{x}{5} + 3 = 8.

Solution:

Determine xx.

x5+3=8\frac{x}{5} + 3 = 8

x5=83 \frac{x}{5} = 8 - 3

x5=5\frac{x}{5} = 5

x=5×5x = 5 \times 5

x=25x = 25

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