Algebraic Equations

Linear Equation Solution

\(ax + b = 0 \quad \Rightarrow \quad x = -\frac{b}{a}\)

Quadratic Formula

\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

For \(ax^2 + bx + c = 0\), where \(a \neq 0\)

Important Algebraic Identities

Method 1: Balancing (Isolation) Method

Apply the same operation to both sides to isolate the variable.

Key Principle: Whatever you do to one side, do to the other!

Method 2: Transposition Method

Move terms from one side to another by changing their sign.

Rule: + becomes -, × becomes ÷, and vice versa

Method 3: Substitution Method

For systems of equations: solve one equation for a variable, then substitute into the other.

Best for: Systems where one variable is already isolated

Method 4: Elimination Method

For systems of equations: add or subtract equations to eliminate one variable.

Best for: Systems with coefficients that easily cancel

Example 1: Solving a Simple Linear Equation

Problem: Solve \(2x + 7 = 19\)

Step 1: Subtract 7 from both sides

\(2x + 7 - 7 = 19 - 7\)

\(2x = 12\)

Step 2: Divide both sides by 2

\(\frac{2x}{2} = \frac{12}{2}\)

\(x = 6\)

✓ Solution: \(x = 6\)

Example 2: Variables on Both Sides

Problem: Solve \(5x - 3 = 2x + 12\)

Step 1: Subtract 2x from both sides

\(5x - 2x - 3 = 2x - 2x + 12\)

\(3x - 3 = 12\)

Step 2: Add 3 to both sides

\(3x = 15\)

Step 3: Divide by 3

\(x = 5\)

✓ Solution: \(x = 5\)

Example 3: Solving by Factoring

Problem: Solve \(x^2 - 5x + 6 = 0\)

Step 1: Factor the quadratic

\((x - 2)(x - 3) = 0\)

Step 2: Set each factor equal to zero

\(x - 2 = 0\) or \(x - 3 = 0\)

Step 3: Solve each equation

\(x = 2\) or \(x = 3\)

✓ Solutions: \(x = 2\) or \(x = 3\)

Example 4: Using the Quadratic Formula

Problem: Solve \(2x^2 + 5x - 3 = 0\)

Step 1: Identify a, b, c

\(a = 2, \quad b = 5, \quad c = -3\)

Step 2: Apply the quadratic formula

\(x = \frac{-5 \pm \sqrt{5^2 - 4(2)(-3)}}{2(2)}\)

Step 3: Simplify

\(x = \frac{-5 \pm \sqrt{25 + 24}}{4}\)

\(x = \frac{-5 \pm \sqrt{49}}{4}\)

\(x = \frac{-5 \pm 7}{4}\)

Step 4: Find both solutions

\(x = \frac{-5 + 7}{4} = \frac{2}{4} = \frac{1}{2}\)

\(x = \frac{-5 - 7}{4} = \frac{-12}{4} = -3\)

✓ Solutions: \(x = \frac{1}{2}\) or \(x = -3\)

Example 5: System of Linear Equations (Substitution)

Problem: Solve \(\begin{cases} y = 2x + 1 \\ 3x + y = 11 \end{cases}\)

Step 1: Substitute y from equation 1 into equation 2

\(3x + (2x + 1) = 11\)

Step 2: Solve for x

\(5x + 1 = 11\)

\(5x = 10\), so \(x = 2\)

Step 3: Substitute back to find y

\(y = 2(2) + 1 = 5\)

✓ Solution: \(x = 2, \quad y = 5\)

✅ Trick #1: PEMDAS Reverse

To isolate a variable, undo operations in reverse PEMDAS order: Addition/Subtraction first, then Multiplication/Division, then Exponents.

✅ Trick #2: Keep It Balanced

Think of an equation as a balanced scale. Whatever you do to one side, must be done to the other to maintain balance!

✅ Trick #3: Move & Change

When transposing, remember: + becomes -, × becomes ÷. Moving a term across the equals sign reverses its operation.

✅ Trick #4: Check Your Answer

Always substitute your answer back into the original equation to verify it's correct. If both sides equal, you're right!

✅ Trick #5: Simplify First

Before solving, combine like terms and simplify both sides. This makes the equation much easier to work with!

✅ Trick #6: Factor When Possible

For quadratics, try factoring first! It's usually faster than the quadratic formula when it works.

✅ Trick #7: Zero Product Property

If \(a \times b = 0\), then \(a = 0\) or \(b = 0\). This is crucial for solving factored equations!

✅ Trick #8: Fraction Elimination

Multiply both sides by the LCD (least common denominator) to eliminate fractions early. Much easier to work with whole numbers!

❌ Mistake #1: Not Doing the Same to Both Sides

Wrong: \(x + 5 = 10\) → \(x = 10\) (forgot to subtract 5 from both sides)

Right: \(x + 5 = 10\) → \(x = 5\)

❌ Mistake #2: Sign Errors

Wrong: \(-3x = 12\) → \(x = 4\) (forgot the negative sign)

Right: \(-3x = 12\) → \(x = -4\)

❌ Mistake #3: Distribution Errors

Wrong: \(2(x + 3) = 2x + 3\)

Right: \(2(x + 3) = 2x + 6\) (distribute to BOTH terms)

❌ Mistake #4: Not Checking Solutions

Always substitute your answer back into the original equation to verify!

📍 Note 1: Number of Solutions

An equation of degree \(n\) has at most \(n\) solutions. Linear = 1, Quadratic = 2, Cubic = 3, etc.

📍 Note 2: Discriminant for Quadratics

\(b^2 - 4ac\) tells you: > 0 (two real solutions), = 0 (one solution), < 0 (no real solutions)

📍 Note 3: Division by Zero

Never divide by zero or by a variable that could equal zero without checking!

📍 Note 4: Extraneous Solutions

When squaring both sides, always check for extraneous solutions that don't satisfy the original equation.

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