International Mathematics (0607) | Equations

Welcome to the Equations Chapter!

Hello mathematicians! This chapter, nestled right in the heart of Algebra, is perhaps the most practical skill you will learn: how to solve mathematical puzzles.

An equation is like a balanced scale: both sides must be exactly equal. Our job is to figure out the value (or values!) that make the scale balance. Mastering equations is crucial because they are the building blocks for modeling real-world situations, from finance to physics. Let's get solving!

Section 1: Constructing and Rearranging Formulas (C2.5.1, C2.5.5, E2.5.1)

1.1 Translating Words into Mathematical Language

Before we solve, we often need to set up the problem. This means converting a sentence into an algebraic expression, equation, or formula.

Key Definitions:

• Expression: A mathematical phrase that contains numbers, variables, and operations (e.g., \(2x + 5\)). It doesn't have an equals sign.
• Equation: A statement that two expressions are equal (e.g., \(2x + 5 = 11\)).
• Formula: A special type of equation that shows the relationship between different quantities, often represented by letters (e.g., Area \(A = \pi r^2\)).

Tip for Construction (Extended E2.5.1):
When setting up problems involving sequences or relationships, translate carefully:

• "A number that is 2 more than n" translates to \(n + 2\). (C2.5.1)
• "The product of two consecutive even numbers." If the first even number is \(x\), the next even number must be \(x + 2\). The product is \(x(x + 2)\).

1.2 Changing the Subject of a Formula (Rearranging)

Changing the subject means isolating a different variable on one side of the equals sign. Think of it as putting a specific variable under the spotlight.

Step-by-Step Guide: Simple Formulas (Core C2.5.5)

For Core students, the subject variable appears only once, and there are no powers or roots involved.

1. Identify the variable you want to isolate (the new subject).
2. Use inverse operations to move everything else away from it. Remember the "Golden Rule": Do the same thing to both sides!

Example: Change the subject of \(y = 3x - 5\) to make \(x\) the subject.

• \(y = 3x - 5\)
• Add 5 to both sides: \(y + 5 = 3x\)
• Divide by 3: \(\frac{y + 5}{3} = x\)

The new formula is \(x = \frac{y + 5}{3}\).

Section 2: Solving Linear Equations (C2.5.2)

2.1 Linear Equations in One Unknown

A linear equation means the highest power of the unknown variable (usually \(x\)) is 1. The solution is always a single number.

Step-by-Step Solving (The Balance Method)

The goal is to get all the terms with the unknown variable on one side, and all the constant numbers on the other side.

Example 1: Solve \(3x + 4 = 10\)

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

Example 2: Solve \(5 - 2x = 3(x + 7)\)

1. Expand the bracket first: \(5 - 2x = 3x + 21\)
2. Collect the \(x\) terms (it's often easier to move them to the side where they stay positive): Add \(2x\) to both sides.
   \(5 = 3x + 2x + 21 \rightarrow 5 = 5x + 21\)
3. Collect the constant terms: Subtract 21 from both sides.
   \(5 - 21 = 5x \rightarrow -16 = 5x\)
4. Divide by 5: \(x = -\frac{16}{5}\) or \(-3.2\)

Common Mistake: Forgetting to apply the inverse operation to every term on the other side, or messing up the signs when moving terms. Be careful with negative coefficients, like \(-2x\) in the example above!

2.2 Solving Fractional Equations (Extended E2.5.3)

If your equation includes fractions, the first priority is to eliminate the denominators.

Using Numerical Denominators

Example: Solve \(\frac{x}{2} + \frac{x}{3} = 5\)

1. Find the Lowest Common Multiple (LCM) of the denominators (2 and 3). The LCM is 6.
2. Multiply every single term in the equation by the LCM (6):
   \(6 \times \frac{x}{2} + 6 \times \frac{x}{3} = 6 \times 5\)
3. Simplify (cancel out the denominators):
   \(3x + 2x = 30\)
4. Solve the resulting linear equation: \(5x = 30 \rightarrow x = 6\)

Using Algebraic Denominators (Extended E2.5.3)

This is trickier! You must multiply by a term that cancels *all* denominators.

Example: Solve \(\frac{2}{x+2} + \frac{3}{2x-1} = 1\)

1. Multiply every term by both denominators, \((x+2)(2x-1)\):
   \((x+2)(2x-1) \times \frac{2}{x+2} + (x+2)(2x-1) \times \frac{3}{2x-1} = 1 \times (x+2)(2x-1)\)
2. Cancel out terms:
   • \(2(2x-1) + 3(x+2) = (x+2)(2x-1)\)
3. Expand all brackets:
   • \(4x - 2 + 3x + 6 = 2x^2 - x + 4x - 2\)
4. Simplify and gather terms (this usually results in a quadratic equation):
   • \(7x + 4 = 2x^2 + 3x - 2\)
   • \(0 = 2x^2 - 4x - 6\) (Now solve this quadratic, perhaps by factorising or using the formula!)

Section 3: Simultaneous Linear Equations (C2.5.3, E2.5.4)

When you have two or more unknown variables (e.g., \(x\) and \(y\)), you need the same number of equations to find a unique solution. We are looking for the point \((x, y)\) where the two lines intersect.

We have two main methods: Elimination and Substitution.

3.1 Method 1: Elimination

This method involves adding or subtracting the two equations so that one variable "disappears" (is eliminated).

Step-by-Step Elimination

Consider these equations:
(1) \(3x + 2y = 19\)
(2) \(x + 2y = 13\)

1. Match the Coefficients: Check if the coefficients (the numbers in front) of \(x\) or \(y\) are the same. (In this example, the coefficients of \(y\) are both 2).
2. Add or Subtract: Since the signs for the \(y\) terms are the same (\(+2y\)), we subtract Equation (2) from Equation (1) to eliminate \(y\).
   \((3x - x) + (2y - 2y) = (19 - 13)\)
   \(2x = 6\)
3. Solve for the First Variable: \(x = 3\)
4. Substitute Back: Put the value of \(x\) into either original equation to find \(y\). Using (2):
   \(3 + 2y = 13 \rightarrow 2y = 10 \rightarrow y = 5\)
5. Final Answer: \(x = 3\) and \(y = 5\). (Write the solution as a coordinate \((3, 5)\) if requested.)

Memory Aid: Same Signs Subtract (SSS), Different Signs Add (DSA). This helps you remember whether to add or subtract the equations in Step 2.

3.2 Method 2: Substitution

This method is generally best if one of the variables is already isolated or has a coefficient of 1.

Step-by-Step Substitution

Consider these equations:
(1) \(y = x + 1\)
(2) \(4x + y = 16\)

1. Isolate a Variable: (1) is already solved for \(y\).
2. Substitute: Replace that variable in the *other* equation. Substitute \((x + 1)\) for \(y\) in Equation (2):
   \(4x + (x + 1) = 16\)
3. Solve for the First Variable:
   \(5x + 1 = 16 \rightarrow 5x = 15 \rightarrow x = 3\)
4. Substitute Back: Put \(x = 3\) back into the easiest equation (1):
   \(y = 3 + 1 \rightarrow y = 4\)
5. Final Answer: \(x = 3\) and \(y = 4\).

Section 4: Solving Quadratic Equations (Extended E2.5.5)

A quadratic equation contains an \(x^2\) term and can be written in the standard form: \(ax^2 + bx + c = 0\), where \(a \neq 0\). These equations usually have two solutions.

4.1 Method 1: Solving by Factorisation

If you can factorise the quadratic expression, you can use the fact that if \(A \times B = 0\), then either \(A = 0\) or \(B = 0\) (or both).

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

1. Factorise the quadratic: \((x - 2)(x - 3) = 0\)
2. Set each factor to zero:
   • \(x - 2 = 0 \rightarrow x = 2\)
   • \(x - 3 = 0 \rightarrow x = 3\)
3. Solutions: \(x = 2\) or \(x = 3\).

4.2 Method 2: The Quadratic Formula

When a quadratic cannot be factorised easily (or at all), you must use the formula. This formula is provided in the exam paper list!

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

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

1. Identify \(a\), \(b\), and \(c\): Here, \(a=2\), \(b=5\), \(c=-3\).
2. Substitute the values into the formula: \[ x = \frac{-5 \pm \sqrt{5^2 - 4(2)(-3)}}{2(2)} \]
3. Simplify the discriminant (\(b^2 - 4ac\)): \(25 - (-24) = 49\). \[ x = \frac{-5 \pm \sqrt{49}}{4} \]
4. Calculate the two solutions using \(\pm\):
   \(x_1 = \frac{-5 + 7}{4} = \frac{2}{4} = 0.5\)
   \(x_2 = \frac{-5 - 7}{4} = \frac{-12}{4} = -3\)

Did you know? The term \(\sqrt{b^2 - 4ac}\) is called the discriminant. If it's negative, there are no real solutions!

Note on Surd Form (E2.5.5): Sometimes you will be asked for the "exact value" or to leave the answer in surd form. If \(\sqrt{b^2 - 4ac}\) is not a perfect square, simplify the surd but don't calculate the decimal.
e.g., if you get \(\frac{2 + \sqrt{20}}{2}\), simplify \(\sqrt{20} = 2\sqrt{5}\), giving the final answer as \(1 + \sqrt{5}\).

4.3 Method 3: Solving Quadratics using the GDC (C2.5.4, E2.5.5)

Your Graphic Display Calculator (GDC) is a powerful tool for solving equations quickly by graphing the function.

To solve \(ax^2 + bx + c = 0\), you graph the function \(y = ax^2 + bx + c\). The solutions are the points where the graph crosses the \(x\)-axis. These points are called the zeros (or roots).

GDC Steps:

1. Enter the equation into the calculator's graph function (e.g., Y1).
2. Graph the function.
3. Use the "Zero" or "Root" function (usually found in the CALC or G-Solve menu) to find the \(x\)-intercepts.

Section 5: Advanced Equations and Rearranging (Extended Content E2.5.6, E2.5.7)

5.1 Complex Changing the Subject (Extended E2.5.6)

For extended mathematics, you must be able to rearrange formulas where the subject appears more than once, or where powers and roots are involved.

Case 1: Powers and Roots

To undo a power, use the inverse root. To undo a root, use the inverse power.

Example: Make \(r\) the subject of the formula for the volume of a cone, \(V = \frac{1}{3}\pi r^2 h\)

1. Multiply by 3: \(3V = \pi r^2 h\)
2. Divide by \(\pi h\): \(\frac{3V}{\pi h} = r^2\)
3. Take the square root: \(r = \sqrt{\frac{3V}{\pi h}}\)

Case 2: Subject Appears Twice

You must first gather all terms containing the new subject onto one side, and then factorise it out.

Example: Make \(t\) the subject of \(A = \frac{t+k}{t}\)

1. Multiply by \(t\) to clear the fraction: \(At = t + k\)
2. Gather terms containing \(t\) on the left: \(At - t = k\)
3. Factorise \(t\) out (this is the crucial step!): \(t(A - 1) = k\)
4. Isolate \(t\) by dividing by the bracket \((A - 1)\): \(t = \frac{k}{A - 1}\)

5.2 Solving Unfamiliar Equations using the GDC (C2.5.4, E2.5.7)

The syllabus requires you to solve equations that might be non-linear or unfamiliar, like trigonometric or reciprocal equations, using your GDC.

Solving by Finding Intersections

When faced with an equation like \(2x - 1 = \frac{1}{x}\), solving algebraically can be difficult. The GDC makes this easy by splitting the equation into two functions:

• Function 1: \(y_1 = 2x - 1\)
• Function 2: \(y_2 = \frac{1}{x}\)

The solution to the original equation is the \(x\)-coordinate of the intersection point(s) of the two graphs.

GDC Steps:

1. Enter \(y_1\) and \(y_2\) into the GDC graph menu.
2. Adjust the window settings (zoom) until you clearly see where the graphs cross.
3. Use the "Intersection" function (usually in CALC or G-Solve) to find the intersection point(s). The calculator will give you the \(x\) and \(y\) values; you need the \(x\) values as the solution.