EQUATIONS · FOUNDATIONS

Equation fundamentals

Before learning formulas, understand what an equation actually says, what a solution is, and which transformations preserve the solution set.

1. Equation, solution and solution set

An equation is a statement that two expressions are equal:

\[L(x)=R(x)\]

A value \(x_0\) is a solution if substituting it makes the statement true. The solution set contains every such value.

Example\(2x+3=11\) has solution set \(\{4\}\).

Identity\(2(x+1)=2x+2\) is true for every real \(x\), so its solution set is \(\mathbb R\).

2. Equivalent transformations

Two equations are equivalent when they have exactly the same solution set. Reliable algebra is built from transformations that preserve equivalence.

Always reversibleAdd or subtract the same expression on both sides. Multiply or divide both sides by the same known non-zero number.

Potentially non-reversibleSquaring, taking roots, multiplying by an expression that can be zero, or applying some nonlinear functions can add or remove candidates.

\[u=v\iff u+k=v+k\qquad\text{and, for }k\neq0,\qquad u=v\iff ku=kv\]

3. Domain comes first

The equation only makes sense where every expression is defined. Record restrictions before transforming.

FractionsFor \(\frac{2}{x-1}=1\), require \(x\neq1\).

Real square rootsFor \(\sqrt{x-3}=x-5\), require \(x\ge3\), and the right side must be non-negative for any solution.

4. A dependable solving workflow

State the domain.
Identify denominator, root, logarithm or other restrictions.

Simplify the structure.
Expand, collect terms or factor only when it helps.

Choose a method.
Linear isolation, factoring, quadratic formula, substitution, numerical methods, and so on.

Track equivalence.
If a step is not reversible, treat resulting values as candidates.

Verify in the original equation.
This removes extraneous candidates and catches arithmetic errors.

5. Worked examples

Linear equation

\[3x-7=2x+5\iff x-7=5\iff x=12\]

Every step is reversible, so \(12\) is the unique solution.

Rational equation

Domain: \(x\neq1\).

\[\frac{2}{x-1}=1\iff 2=x-1\iff x=3\]

The candidate \(3\) lies in the domain and satisfies the original equation.

Squaring can create an extraneous solution

For \(\sqrt{x+4}=x\), the original equation requires \(x\ge0\).

\[x+4=x^2\iff x^2-x-4=0\]

The quadratic has two algebraic candidates, but only the non-negative candidate that satisfies the original square-root equation is valid.

6. Common mistakes

Dividing by an unknown expressionDividing by \(x-2\) may silently discard the case \(x=2\).

Forgetting the original domainA transformed equation can be valid at values where the original expression was undefined.

Treating implication as equivalenceFrom \(x=2\) follows \(x^2=4\), but \(x^2=4\) also permits \(x=-2\).

Skipping verificationChecking a result is especially important after non-reversible operations.

7. Practice

Solve \(5x+4=19\).
Solve \(\frac{6}{x+2}=3\) and state the domain.
Explain why squaring both sides of an equation may enlarge the solution set.

Show answers

1) \(x=3\). 2) Domain \(x\neq-2\); solution \(x=0\). 3) Squaring is not one-to-one on the reals: opposite numbers have the same square.