IGCSE Mathematics 0580

Algebra: Equations and Inequalities

This section covers linear, simultaneous, and quadratic equations and inequalities. We will explore methods for solving these types of equations and inequalities.

1. Linear Equations

Linear equations are equations where the highest power of the variable is 1. They can be represented in the form $ax + b = 0$ or $y = mx + c$, where $a$, $b$, $m$, and $c$ are constants.

Solving Linear Equations

To solve a linear equation, we aim to isolate the variable on one side of the equation. We use inverse operations to do this.

Example: Solve for $x$: $3x + 5 = 14$

Step 1: Subtract 5 from both sides: $3x + 5 - 5 = 14 - 5 \Rightarrow 3x = 9$

Step 2: Divide both sides by 3: $\frac{3x}{3} = \frac{9}{3} \Rightarrow x = 3$

2. Simultaneous Equations

Simultaneous equations are a set of two or more equations with the same variables. We aim to find the values of the variables that satisfy all equations in the set.

Solving Simultaneous Equations

There are several methods for solving simultaneous equations:

• Substitution: Solve one equation for one variable and substitute that expression into the other equation.
• Elimination: Multiply one or both equations by constants so that the coefficients of one variable are opposites. Then, add the equations together to eliminate that variable.

Example (Substitution): Solve the following equations: $$ \begin{cases} x + y = 5 \\ x - y = 1 \end{cases} $$

Step 1: Solve the first equation for $x$: $x = 5 - y$

Step 2: Substitute this expression for $x$ into the second equation: $(5 - y) - y = 1 \Rightarrow 5 - 2y = 1 \Rightarrow -2y = -4 \Rightarrow y = 2$

Step 3: Substitute the value of $y$ back into either equation to find $x$: $x + 2 = 5 \Rightarrow x = 3$

Therefore, the solution is $x = 3$ and $y = 2$.

3. Quadratic Equations

A quadratic equation is an equation of the form $ax^2 + bx + c = 0$, where $a \neq 0$.

Solving Quadratic Equations

There are three main methods for solving quadratic equations:

• Factorisation: Express the quadratic expression as a product of two linear factors.
• Quadratic Formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
• Completing the Square: Transform the quadratic equation into a form where it can be solved by taking the square root.

Example (Quadratic Formula): Solve for $x$: $2x^2 + 5x - 3 = 0$

Here, $a = 2$, $b = 5$, and $c = -3$.

Using the quadratic formula: $x = \frac{-5 \pm \sqrt{5^2 - 4(2)(-3)}}{2(2)} = \frac{-5 \pm \sqrt{25 + 24}}{4} = \frac{-5 \pm \sqrt{49}}{4} = \frac{-5 \pm 7}{4}$

So, $x_1 = \frac{-5 + 7}{4} = \frac{2}{4} = \frac{1}{2}$ and $x_2 = \frac{-5 - 7}{4} = \frac{-12}{4} = -3$.

Therefore, the solutions are $x = \frac{1}{2}$ and $x = -3$.

4. Linear Inequalities

Linear inequalities are inequalities where the highest power of the variable is 1. They are used to represent relationships between values, such as $ax + b > 0$ or $y < mx + c$.

Solving Linear Inequalities

The process for solving linear inequalities is similar to solving linear equations, but with an additional rule: when multiplying or dividing by a negative number, the inequality sign must be reversed.

Example: Solve for $x$: $2x - 3 < 5$

Step 1: Add 3 to both sides: $2x - 3 + 3 < 5 + 3 \Rightarrow 2x < 8$

Step 2: Divide both sides by 2: $\frac{2x}{2} < \frac{8}{2} \Rightarrow x < 4$

The solution is $x < 4$. This can be represented on a number line with an open circle at 4.

5. Quadratic Inequalities

Quadratic inequalities are inequalities involving quadratic expressions. To solve them, we first find the roots of the corresponding quadratic equation ($ax^2 + bx + c = 0$). These roots divide the number line into intervals. We then test a value from each interval to determine where the quadratic expression is positive or negative.

Example: Solve $x^2 - 4 > 0$

Step 1: Find the roots of the equation $x^2 - 4 = 0$. This gives $x = 2$ and $x = -2$

Step 2: Consider the intervals $(-\infty, -2)$, $(-2, 2)$, and $(2, \infty)$

Test $x = -3$: $(-3)^2 - 4 = 9 - 4 = 5 > 0$. So, $(-\infty, -2)$ is a solution.

Test $x = 0$: $(0)^2 - 4 = -4 < 0$. So, $(-2, 2)$ is not a solution.

Test $x = 3$: $(3)^2 - 4 = 9 - 4 = 5 > 0$. So, $(2, \infty)$ is a solution.

Therefore, the solution is $x < -2$ or $x > 2$.

Topic	Key Concepts	Methods
Linear Equations	$ax + b = 0$, $y = mx + c$	Substitution, Elimination
Simultaneous Equations	Set of two or more equations with the same variables	Substitution, Elimination
Quadratic Equations	$ax^2 + bx + c = 0$	Factorisation, Quadratic Formula, Completing the Square
Linear Inequalities	$ax + b > 0$, $y < mx + c$	Similar to solving linear equations, reverse sign when multiplying/dividing by a negative number
Quadratic Inequalities	$ax^2 + bx + c > 0$ or $ax^2 + bx + c < 0$	Find roots, test intervals