Mathematics (0580) | Equations

Welcome to the Equations Chapter!

Equations are the heart of Mathematics. They allow us to take real-world problems—like calculating a budget, figuring out how fast something is travelling, or even designing a bridge—and solve them precisely.
In this chapter, we will learn how to master solving different types of equations, from simple linear ones to more complex quadratics and fractional equations. Don't worry if this seems tricky at first; we will break down every method into simple, manageable steps!

Did you know?

The equals sign (\(=\)) was invented in 1557 by Robert Recorde, who thought: "no two things could be more equal than two parallel lines."

---

Part 1: Linear Equations in One Unknown (C2.5.2)

A linear equation in one unknown involves a variable (like \(x\)) raised only to the power of one. Our goal is always to isolate the unknown.

The Golden Rule of Equations

Whatever you do to one side of the equation, you must do to the other side to keep the equation balanced. Think of it like a seesaw!

Step-by-Step Solving Process

We use reverse operations (inverse operations) to undo the steps applied to the variable, working in reverse order of operations (often reverse PEMDAS/BODMAS).

1. Expand any brackets.
2. Collect all the variable terms (\(x\)) on one side.
3. Collect all the constant terms (numbers) on the other side.
4. Divide or multiply to find the value of the unknown.

Example 1: Simple Linear Equation
Solve \(3x + 4 = 10\).

• Subtract 4 from both sides: \(3x = 10 - 4\)
• \(3x = 6\)
• Divide both sides by 3: \(x = \frac{6}{3}\)
• \(x = 2\)

Example 2: Equations with Brackets (C2.5.2)
Solve \(5 - 2x = 3(x + 7)\).

• Step 1: Expand the bracket: \(5 - 2x = 3x + 21\)
• Step 2: Collect \(x\) terms (Add \(2x\) to both sides): \(5 = 3x + 2x + 21\)
  \(5 = 5x + 21\)
• Step 3: Collect constant terms (Subtract 21 from both sides): \(5 - 21 = 5x\)
  \(-16 = 5x\)
• Step 4: Divide by 5: \(x = -\frac{16}{5}\) or \(x = -3.2\)

Part 2: Simultaneous Linear Equations (C2.5.3, E2.5.4)

When you have two unknown variables (like \(x\) and \(y\)), you need at least two equations to solve for both values. This is called solving simultaneous equations.

We will focus on two main methods: Elimination and Substitution.

Method 1: Elimination

The goal is to eliminate one variable by adding or subtracting the two equations. This works best when the equations are neatly lined up.

Example:
(1) \(2x + y = 7\)
(2) \(3x - y = 8\)

1. Notice that the \(y\) terms have opposite coefficients (\(+y\) and \(-y\)).
2. Add Equation (1) and Equation (2):
   \((2x + 3x) + (y - y) = 7 + 8\)
   \(5x + 0 = 15\)
3. Solve for \(x\): \(x = 3\)
4. Substitute \(x = 3\) back into either original equation (use Equation 1 as it looks easier):
   \(2(3) + y = 7\)
   \(6 + y = 7\)
5. Solve for \(y\): \(y = 7 - 6\), so \(y = 1\).
6. Solution: \(x=3, y=1\)

Tip: If the coefficients are not the same (e.g., \(2x\) and \(3x\)), multiply one or both equations by a constant until one variable has matching coefficients.

Method 2: Substitution

The goal is to rearrange one equation to make one variable the subject, then substitute that expression into the other equation. This is often better if one variable already has a coefficient of 1.

Example:
(1) \(y = x + 3\)
(2) \(4x + 2y = 12\)

1. Equation (1) already has \(y\) as the subject.
2. Substitute the expression for \(y\) from (1) into (2):
   \(4x + 2(x + 3) = 12\)
3. Solve the resulting linear equation:
   \(4x + 2x + 6 = 12\)
   \(6x + 6 = 12\)
   \(6x = 6\)
   \(x = 1\)
4. Substitute \(x = 1\) back into Equation (1):
   \(y = (1) + 3\)
   \(y = 4\)
5. Solution: \(x=1, y=4\)

Part 3: Changing the Subject of a Formula (C2.5.5, E2.5.6)

A formula is an equation that connects two or more variables, defining a rule (e.g., \(A = \pi r^2\)). Changing the subject means rearranging the formula so a different variable stands alone on one side of the equals sign.

Case 1: Simple Formulas (Core Content - C2.5.5)

The subject only appears once, and there are no powers or roots.

Example: Make \(u\) the subject of \(v = u + at\).

• Goal: Get \(u\) alone.
• The term \(at\) is added to \(u\). Subtract \(at\) from both sides.
• \(v - at = u\)
• New formula: \(u = v - at\)

Example: Make \(R\) the subject of \(V = \frac{1}{3} \pi R^2 h\).

• Multiply by 3: \(3V = \pi R^2 h\)
• Divide by \(\pi h\): \(\frac{3V}{\pi h} = R^2\)
• Take the square root (since \(R\) is usually positive in geometry, we usually don't need \(\pm\)): \(R = \sqrt{\frac{3V}{\pi h}}\)

Case 2: Complex Formulas (Extended Content - E2.5.6)

The subject appears more than once, or involves a power/root.

1. Subject appears twice: Factorisation is Key!

Example: Make \(x\) the subject of \(y = \frac{x+a}{x}\).

1. Multiply by \(x\) to clear the denominator: \(yx = x + a\)
2. Collect all terms containing the desired subject (\(x\)) on one side: \(yx - x = a\)
3. Factorise out the desired subject (\(x\)): \(x(y - 1) = a\)
4. Divide by the bracket \((y-1)\): \(x = \frac{a}{y - 1}\)

2. Subject involves power or root: Use inverse powers!

Example: Make \(r\) the subject of \(t = 5 \sqrt{r^3 - k}\).

1. Divide by 5: \(\frac{t}{5} = \sqrt{r^3 - k}\)
2. Square both sides to remove the root: \((\frac{t}{5})^2 = r^3 - k\)
3. Add \(k\) to both sides: \((\frac{t}{5})^2 + k = r^3\)
4. Take the cube root: \(r = \sqrt[3]{(\frac{t}{5})^2 + k}\)

Part 4: Quadratic Equations (Extended Content - E2.5.5)

A quadratic equation is an equation where the highest power of the unknown variable (\(x\)) is 2. They have the general form: \(ax^2 + bx + c = 0\), where \(a \neq 0\).

Quadratic equations can have up to two solutions (or roots).

Method 1: Solving by Factorisation

If you can factorise the quadratic expression into two brackets, solving becomes very quick using the Zero Product Property: If \(A \times B = 0\), then either \(A=0\) or \(B=0\).

Example: Solve \(x^2 + 5x + 6 = 0\).

1. Factorise the expression: \((x + 2)(x + 3) = 0\)
2. Set each bracket equal to zero:
   \(x + 2 = 0\) OR \(x + 3 = 0\)
3. Solve: \(x = -2\) OR \(x = -3\)

Common Mistake to Avoid: Not factorising fully! Always check for common factors first.

Method 2: The Quadratic Formula

If a quadratic cannot be factorised easily (or at all), you must use the formula. This formula is provided in the exam booklet for Extended Candidates:

For \(ax^2 + bx + c = 0\), the solutions are:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

The term under the square root, \(b^2 - 4ac\), is called the discriminant.

Example: Solve \(2x^2 - 5x - 3 = 0\).

1. Identify \(a\), \(b\), and \(c\): \(a=2\), \(b=-5\), \(c=-3\).
2. Substitute into the formula:
   \[x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(2)(-3)}}{2(2)}\]
3. Simplify:
   \[x = \frac{5 \pm \sqrt{25 + 24}}{4}\]
   \[x = \frac{5 \pm \sqrt{49}}{4}\]
4. Calculate the two solutions:
   \[x = \frac{5 + 7}{4} = \frac{12}{4} = 3\]
   \[x = \frac{5 - 7}{4} = \frac{-2}{4} = -0.5\]

Remember Surd Form: If the question asks for an exact value, or if the number under the root is not a perfect square (and cannot be simplified to a simple decimal), you must leave the answer in surd form (like \(\sqrt{7}\) or \(3 + 2\sqrt{5}\)).

Part 5: Fractional and Graphical Solutions (Extended Content)

1. Solving Fractional Equations (E2.5.3)

These equations contain algebraic fractions. The main strategy is to multiply every term in the equation by the Lowest Common Multiple (LCM) of all denominators to eliminate the fractions.

Example: Solve \(\frac{x}{x+2} = \frac{3}{x-6}\).

1. The LCM is \((x+2)(x-6)\). Multiply both sides by the LCM (this is essentially cross-multiplication here):
   \(x(x-6) = 3(x+2)\)
2. Expand both sides:
   \(x^2 - 6x = 3x + 6\)
3. Rearrange to form a standard quadratic equation (\(ax^2 + bx + c = 0\)):
   \(x^2 - 9x - 6 = 0\)
4. Solve the resulting equation using the quadratic formula (since this does not factorise easily).

2. Using the Graphic Display Calculator (GCD) (C2.5.4, E2.5.7)

Your Graphic Display Calculator (GCD) is a powerful tool for solving equations, especially those that are non-linear or unfamiliar (like those involving functions you haven't studied yet).

Solving by Finding Zeros/Roots

To solve an equation like \(2x = x^2\) or \(x^3 - 4x + 1 = 0\):

1. Rearrange the equation so one side is zero. For \(2x = x^2\), write \(x^2 - 2x = 0\).
2. Define the left side as a function: Let \(Y_1 = x^2 - 2x\).
3. Sketch the graph.
4. Use the calculator's 'Find Zero' or 'Root' function to find the \(x\)-intercepts (where the graph crosses the \(x\)-axis, meaning \(Y_1 = 0\)).

Solving by Finding the Intersection

To solve an equation like \(2x - 1 = \frac{1}{x}\) (which is hard algebraically):

1. Define the left side as \(Y_1\) and the right side as \(Y_2\).
   \(Y_1 = 2x - 1\)
   \(Y_2 = \frac{1}{x}\)
2. Sketch both graphs.
3. Use the calculator's 'Find Intersection' function. The \(x\)-coordinates of the intersection points are the solutions to the equation.