Equations

1. Overview

Equations are the foundation of algebra, representing a balance between two expressions. This topic covers everything from forming equations from word problems to solving complex systems, including linear, quadratic, and simultaneous equations. You'll learn to manipulate formulas and find unknown values, skills essential for success in IGCSE Mathematics and beyond. Mastering equations allows you to model and solve real-world problems.

Key Definitions

• Expression: A collection of symbols (numbers, variables, operators) without an equals sign (e.g., $3x + 5$).
• Equation: A statement that two expressions are equal (e.g., $3x + 5 = 11$).
• Formula: A rule or relationship represented by symbols (e.g., $A = lw$).
• Subject: The single variable isolated on one side of a formula (usually the left).
• Unknown: The variable you are trying to solve for.
• Simultaneous Equations: A set of equations with multiple variables that are solved together to find a common solution.

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Core Content

3.1 Constructing Expressions, Equations, and Formulas

To translate word problems into algebraic expressions, equations, or formulas, carefully identify the unknown quantities and represent them with variables (often $x$ or $y$). Look for keywords that indicate mathematical operations:

• Sum/More than/Increased by: Addition ($+$)
• Difference/Less than/Decreased by: Subtraction ($-$)
• Product/Times/Multiplied by: Multiplication ($\times$)
• Quotient/Over/Divided by: Division ($\div$)

Worked example 1 — Forming an equation

A rectangle has a width of $w$ cm. Its length is 3 cm more than twice its width. The perimeter of the rectangle is 42 cm. Form an equation for the perimeter in terms of $w$.

• Step 1 (Express the length): Length $= 2w + 3$
• Step 2 (Write the perimeter formula): Perimeter $= 2(\text{length} + \text{width})$
• Step 3 (Substitute): $42 = 2(2w + 3 + w)$
• Step 4 (Simplify): $42 = 2(3w + 3)$
• Step 5 (Final Equation): $42 = 6w + 6$

3.2 Solving Linear Equations

The objective is to isolate the unknown variable on one side of the equation. This is achieved by performing the same operation on both sides of the equation to maintain equality. Remember to use inverse operations.

Method:

1. Expand any brackets.
2. Collect all terms containing the unknown on one side of the equation.
3. Collect all constant terms on the other side of the equation.
4. Divide both sides by the coefficient of the unknown variable.

Worked example 2 — Solving a linear equation

Solve the equation $5(x + 2) - 3 = 2x + 11$.

• Step 1 (Expand brackets): $5x + 10 - 3 = 2x + 11$
• Step 2 (Simplify): $5x + 7 = 2x + 11$
• Step 3 (Subtract $2x$ from both sides): $3x + 7 = 11$
• Step 4 (Subtract $7$ from both sides): $3x = 4$
• Step 5 (Divide both sides by $3$): $x = \frac{4}{3}$
• Final Answer: $x = \frac{4}{3}$

3.3 Simultaneous Linear Equations

Simultaneous equations involve two or more equations with two or more unknowns. The goal is to find values for the unknowns that satisfy all equations simultaneously. Two common methods are Elimination and Substitution.

Worked example 3 — Elimination method

Solve the following simultaneous equations:

1. $2x + y = 7$
2. $x - y = -1$

• Step 1 (Add equations 1 and 2 to eliminate $y$): $(2x + x) + (y - y) = 7 + (-1)$
• Step 2 (Simplify): $3x = 6$
• Step 3 (Solve for $x$): $x = \frac{6}{3} = 2$
• Step 4 (Substitute $x = 2$ into equation 1): $2(2) + y = 7$
• Step 5 (Solve for $y$): $4 + y = 7 \Rightarrow y = 3$
• Final Answer: $x = 2, y = 3$

Worked example 4 — Substitution method

Solve the following simultaneous equations:

1. $y = 2x + 1$
2. $3x + 2y = 16$

• Step 1 (Substitute equation 1 into equation 2): $3x + 2(2x + 1) = 16$
• Step 2 (Expand): $3x + 4x + 2 = 16$
• Step 3 (Simplify): $7x + 2 = 16$
• Step 4 (Subtract 2 from both sides): $7x = 14$
• Step 5 (Divide both sides by 7): $x = 2$
• Step 6 (Substitute $x = 2$ into equation 1): $y = 2(2) + 1$
• Step 7 (Solve for $y$): $y = 4 + 1 = 5$
• Final Answer: $x = 2, y = 5$

3.4 Changing the Subject (Simple)

Changing the subject of a formula involves rearranging the formula to isolate a specific variable on one side of the equation.

Worked example 5 — Changing the subject

Make $r$ the subject of the formula $A = \pi r^2$.

• Step 1 (Divide both sides by $\pi$): $\frac{A}{\pi} = r^2$
• Step 2 (Take the square root of both sides): $\sqrt{\frac{A}{\pi}} = r$
• Final Answer: $r = \sqrt{\frac{A}{\pi}}$

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Extended Content (Extended Curriculum Only)

4.1 Fractional Equations

Fractional equations contain variables in the denominator of one or more terms. To solve these, the general strategy is to eliminate the fractions by multiplying every term in the equation by the lowest common denominator (LCD).

Worked example 6 — Solving a fractional equation

Solve the equation $\frac{2}{x} + \frac{1}{3} = \frac{5}{2x}$.

• Step 1 (Identify the LCD): The LCD of $x$, $3$, and $2x$ is $6x$.
• Step 2 (Multiply every term by the LCD): $6x \cdot \frac{2}{x} + 6x \cdot \frac{1}{3} = 6x \cdot \frac{5}{2x}$
• Step 3 (Simplify): $12 + 2x = 15$
• Step 4 (Subtract 12 from both sides): $2x = 3$
• Step 5 (Divide both sides by 2): $x = \frac{3}{2}$
• Final Answer: $x = \frac{3}{2}$

4.2 Simultaneous Equations (Linear and Non-Linear)

These involve one linear equation and one non-linear equation (typically quadratic). The method of Substitution is always used. Solve the linear equation for one variable, then substitute that expression into the non-linear equation. This will result in a quadratic equation that can be solved using factorisation, the quadratic formula, or completing the square.

Worked example 7 — Linear and non-linear simultaneous equations

Solve the simultaneous equations:

1. $y = x - 1$
2. $x^2 + y^2 = 5$

• Step 1 (Substitute $y$ from equation 1 into equation 2): $x^2 + (x - 1)^2 = 5$
• Step 2 (Expand and simplify): $x^2 + x^2 - 2x + 1 = 5 \Rightarrow 2x^2 - 2x - 4 = 0$
• Step 3 (Divide by 2): $x^2 - x - 2 = 0$
• Step 4 (Factorise): $(x - 2)(x + 1) = 0 \Rightarrow x = 2$ or $x = -1$
• Step 5 (Find $y$ for each value of $x$):
  • When $x = 2$, $y = 2 - 1 = 1$
  • When $x = -1$, $y = -1 - 1 = -2$
• Final Answer: $(2, 1)$ and $(-1, -2)$

4.3 Quadratic Equations

Quadratic equations have the general form $ax^2 + bx + c = 0$, where $a$, $b$, and $c$ are constants and $a \ne 0$. There are three primary methods for solving quadratic equations:

Method A: Factorisation If the quadratic expression can be factored, set each factor equal to zero and solve for $x$.

Method B: Quadratic Formula (Must be memorised) $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Method C: Completing the Square Rewrite the quadratic in the form $(x + p)^2 + q = 0$, then solve for $x$.

Worked example 8 — Solving a quadratic equation

Solve the equation $2x^2 - 7x + 3 = 0$ using the quadratic formula.

• Step 1 (Identify $a$, $b$, and $c$): $a = 2$, $b = -7$, $c = 3$
• Step 2 (Substitute into the quadratic formula): $x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(2)(3)}}{2(2)}$
• Step 3 (Simplify): $x = \frac{7 \pm \sqrt{49 - 24}}{4} = \frac{7 \pm \sqrt{25}}{4} = \frac{7 \pm 5}{4}$
• Step 4 (Find the two solutions):
  • $x = \frac{7 + 5}{4} = \frac{12}{4} = 3$
  • $x = \frac{7 - 5}{4} = \frac{2}{4} = \frac{1}{2}$
• Final Answer: $x = 3$ or $x = \frac{1}{2}$

4.4 Advanced Changing the Subject

When the desired subject appears multiple times in the formula, factorisation is required to isolate it.

Worked example 9 — Advanced subject changing

Make $x$ the subject of the formula $ax + b = cx + d$.

• Step 1 (Group $x$ terms on one side): $ax - cx = d - b$
• Step 2 (Factorise $x$): $x(a - c) = d - b$
• Step 3 (Divide by $(a - c)$): $x = \frac{d - b}{a - c}$
• Final Answer: $x = \frac{d - b}{a - c}$

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Key Equations

Equation	Meaning	Notes
$\mathbf{x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}}$	Quadratic Formula	$a, b, c$ are coefficients from $ax^2 + bx + c = 0$. Must be memorised.
$\mathbf{b^2 - 4ac}$	The Discriminant	If $> 0$, two real solutions; if $= 0$, one real solution; if $< 0$, no real solutions.
$y = mx + c$	Gradient-intercept form	Used when equations involve straight lines.

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Common Mistakes to Avoid

• ❌ Wrong: Forgetting to multiply every term by the common denominator when solving fractional equations, only multiplying some terms.
• ✓ Right: Ensure every single term, including constants, is multiplied by the LCD to eliminate fractions correctly.
• ❌ Wrong: Writing the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ but skipping the substitution step and going straight to the answer.
• ✓ Right: Always show the substitution of values into the formula: $x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(6)}}{2(1)}$. This secures method marks even with calculation errors.
• ❌ Wrong: Incorrectly applying the distributive property when expanding brackets, such as $2(x+3) = 2x + 3$.
• ✓ Right: Remember to multiply each term inside the brackets by the term outside: $2(x+3) = 2x + 6$.
• ❌ Wrong: Forgetting to check solutions in simultaneous equations, especially when dealing with non-linear equations.
• ✓ Right: Substitute the values of $x$ and $y$ back into the original equations to verify they satisfy both.

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Exam Tips

1. Read Carefully: Always read the question twice, paying close attention to the last line. It often contains crucial information, such as the required degree of accuracy (e.g., "Give your answer to 2 decimal places").
2. Show All Working: Even if you use a calculator to solve a quadratic equation, write down the formula and the substituted values. This ensures you receive method marks even if you make a calculation error.
3. Verify Solutions: When solving simultaneous equations, substitute your solutions back into the original equations to check their validity. This helps identify and correct errors.
4. Standard Form: Before attempting to solve a quadratic equation, always rearrange it into the standard form $ax^2 + bx + c = 0$. This ensures correct identification of $a$, $b$, and $c$ for factorisation or the quadratic formula.
5. Command Words: Understand the meaning of command words:
   • "Solve": Find the value(s) of the variable(s) that satisfy the equation(s).
   • "Show that": Provide a clear and logical sequence of algebraic steps to arrive at a given equation or result. Do not simply substitute values.
   • "Rearrange/Express in terms of": Change the subject of the formula to the specified variable.