Equations and inequalities: linear, simultaneous, quadratic

📐 Algebra: Equations & Inequalities

Linear Equations

Think of a linear equation as a straight road. The equation tells you where you are on that road.

General form: $ax + b = 0$ where $a eq 0$.

  1. Isolate the variable: $ax = -b$
  2. Divide by the coefficient: $x = -\dfrac{b}{a}$

Example: Solve $3x - 9 = 0$.

Step Equation Result
1 $3x - 9 = 0$ Add 9 to both sides
2 $3x = 9$ Divide by 3
3 $x = 3$ Answer
Exam Tip: Always check your answer by substituting it back into the original equation. If it satisfies the equation, you’re likely correct! 🎯

Simultaneous Equations

Imagine two roads crossing. The point where they intersect is the solution.

Two common methods:

  • Substitution: Solve one equation for a variable, then substitute into the other.
  • Elimination: Add or subtract equations to cancel a variable.

Example: Solve the system

$\begin{cases} 2x + 3y = 12 \\ 5x - y = 9 \end{cases}$

Method Steps
Substitution 1️⃣ Solve first for $y$: $y = \dfrac{12-2x}{3}$
2️⃣ Substitute into second: $5x - \dfrac{12-2x}{3} = 9$
3️⃣ Solve for $x$, then back‑substitute for $y$.
Elimination 1️⃣ Multiply second equation by 3: $15x - 3y = 27$
2️⃣ Add to first: $(2x+3y)+(15x-3y)=12+27$ → $17x = 39$ → $x = \dfrac{39}{17}$
3️⃣ Substitute back to find $y$.
Exam Tip: Keep your work neat. Label each step clearly; examiners appreciate a tidy solution! 🧮

Inequalities

Inequalities are like “not equal” roads. They let you know where you can go.

Key rules:

  1. Adding/subtracting the same number keeps the direction.
  2. Multiplying/dividing by a positive number keeps the direction.
  3. Multiplying/dividing by a negative number reverses the direction.

Example: Solve $-2x + 5 > 1$.

Solution:

  1. $-2x > -4$ (subtract 5)
  2. $x < 2$ (divide by -2, reverse sign)

Graphically, on a number line, shade to the left of 2.

Exam Tip: Always check the sign change when multiplying/dividing by a negative. A quick sign test can save time! 🔍

Quadratic Equations

Quadratics are like roller‑coasters: they go up and down. The standard form is

$ax^2 + bx + c = 0$, $a eq 0$.

Three main solution methods:

  • Factoring (when possible)
  • Completing the square
  • Quadratic formula: $x = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$

Example: Solve $x^2 - 5x + 6 = 0$.

  1. Factor: $(x-2)(x-3)=0$
  2. Set each factor to zero: $x-2=0 \Rightarrow x=2$, $x-3=0 \Rightarrow x=3$

When factoring isn’t obvious, use the quadratic formula.

Example: Solve $2x^2 + 3x - 2 = 0$.

Compute discriminant: $b^2-4ac = 9 - 4(2)(-2) = 9 + 16 = 25$.

Apply formula:

$x = \dfrac{-3 \pm \sqrt{25}}{4} = \dfrac{-3 \pm 5}{4}$.

Thus $x = \dfrac{2}{4} = \dfrac{1}{2}$ or $x = \dfrac{-8}{4} = -2$.

Exam Tip: Memorise the discriminant ($\Delta = b^2-4ac$). It tells you whether you’ll get two real solutions ($\Delta > 0$), one real solution ($\Delta = 0$), or no real solutions ($\Delta < 0$). 📊

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