Systematic Equation Solving
Master the art of solving equations with clear, step-by-step methods that work for linear, quadratic, and more complex equation types.
Equation Fundamentals
An equation is a mathematical statement that asserts the equality of two expressions. Solving an equation means finding all values of the variable(s) that make the equation true.
The Golden Rule
Whatever you do to one side, do to the other
This maintains balance and preserves the equality. You can add, subtract, multiply, or divide—just do it to both sides.
Addition
Add the same value to both sides to eliminate negative terms
x - 5 + 5 = 10 + 5
x = 15
Multiplication
Multiply both sides to eliminate fractions
x/4 × 4 = 8 × 4
x = 32
Division
Divide both sides to isolate the variable
3x ÷ 3 = 21 ÷ 3
x = 7
Linear Equations
Linear equations have variables raised only to the first power. The general form is ax + b = c, and they always have exactly one solution (unless a = 0).
Standard Form
Example: Solve 3x + 7 = 22
3x + 7 - 7 = 22 - 7
3x = 15
3x ÷ 3 = 15 ÷ 3
x = 5
Example: Solve 2(x - 4) = 3x + 5
2x - 8 = 3x + 5
-8 = x + 5
-13 = x
Quadratic Equations
Quadratic equations contain x² as the highest power. They can have 0, 1, or 2 real solutions depending on the discriminant.
Quadratic Formula
The Discriminant (b² - 4ac)
Example: Solve x² - 5x + 6 = 0
This could also be solved by factoring: (x - 3)(x - 2) = 0
Factoring
Best when equation factors nicely into integers
(x - 2)(x - 3) = 0
x = 2 or x = 3
Completing the Square
Useful for deriving vertex form
(x + 3)² = 16
x = -3 ± 4
Quadratic Formula
Always works for any quadratic
Systems of Equations
When you have multiple equations with multiple unknowns, you need to solve them simultaneously.
Substitution Method
- Solve one equation for one variable
- Substitute into the other equation
- Solve for the remaining variable
- Back-substitute to find the first variable
Elimination Method
- Multiply equations to match coefficients
- Add/subtract to eliminate one variable
- Solve for the remaining variable
- Back-substitute to find the other
Example (Substitution)
Solve: y = 2x + 1 and 3x + y = 11
3x + (2x + 1) = 11
5x + 1 = 11
5x = 10
x = 2
y = 2(2) + 1 = 5
Common Mistakes
Wrong
Forgetting to distribute the negative sign
Right
Distribute to all terms inside
Wrong
Dividing only part of the equation
x + 4 = 5
Right
Divide all terms
x + 2 = 5
Wrong
Forgetting ± in square root
x = 3
Right
Include both solutions
x = ±3