E2.5 – Equations

Table of Contents

An equation is a mathematical sentence that says two things are equal, with an equals sign between them. “Solving” an equation means finding the value of the unknown letter that makes the sentence true. In this topic you will learn how to build equations from words, and how to solve many different types — linear equations, fractional equations, simultaneous equations, and quadratic equations — as well as how to rearrange a formula.

1. Constructing Expressions, Equations and Formulas #

Before solving, you often have to build the mathematics from words. It helps to know the difference between three things:

Name What it is Example
Expression A group of terms with no equals sign. It cannot be solved, only simplified. $2n + 3$
Equation Two expressions joined by an equals sign. It can be solved to find the unknown. $2n + 3 = 7$
Formula A rule that connects two or more quantities (letters). $A = l \times w$

Writing an Expression from Words #

Choose a letter for the unknown, then turn the words into algebra one step at a time.

Worked Example — The product of two consecutive even numbers
  1. Any even number can be written as $2n$ (2 times a whole number is always even).
  2. The next even number is 2 more, so it is $2n + 2$.
  3. “Product” means multiply, so the expression is: $$2n(2n + 2)$$
  4. If needed, expand the brackets: $\quad 2n(2n + 2) = 4n^2 + 4n$

Constructing Simultaneous Equations #

Sometimes a problem has two unknowns. Then you write two equations — one for each piece of information. (You will solve this pair later in Section 4.)

Worked Example — Build two equations

A pen costs $x$ and a notebook costs $y$.

  1. “Two pens and three notebooks cost \$12”: $$2x + 3y = 12$$
  2. “One pen costs \$1 more than one notebook”: $$x = y + 1 \quad\Longrightarrow\quad x – y = 1$$

We now have a pair of simultaneous equations ready to solve.

2. Solving Linear Equations in One Unknown #

A linear equation has the unknown to the power 1 only (no $x^2$). To solve it, keep the equation balanced: whatever you do to one side, you must do to the other. Use inverse operations (opposite operations) to get the letter on its own.

Worked Example 1 — Solve $3x + 4 = 10$
  1. Subtract 4 from both sides: $\quad 3x = 6$
  2. Divide both sides by 3: $\quad x = 2$
Worked Example 2 — Solve $5 – 2x = 3(x + 7)$
  1. Expand the bracket: $\quad 5 – 2x = 3x + 21$
  2. Add $2x$ to both sides (collect the $x$ terms together): $\quad 5 = 5x + 21$
  3. Subtract 21 from both sides: $\quad -16 = 5x$
  4. Divide both sides by 5: $\quad x = -\dfrac{16}{5} = -3.2$
Tip: When there are brackets, expand them first. When the unknown appears on both sides, move all the unknown terms to one side and all the numbers to the other side.

3. Solving Fractional Equations #

These are equations where the unknown appears in the denominator (the bottom of a fraction). The denominators may be numbers or simple linear expressions such as $2x + 1$ or $x + 2$.

Method Multiply every term by the denominators to clear the fractions. This turns the fractional equation into an ordinary equation you already know how to solve.
Worked Example 1 — Solve $\dfrac{x}{2x + 1} = 4$
  1. Multiply both sides by $(2x + 1)$: $\quad x = 4(2x + 1)$
  2. Expand the bracket: $\quad x = 8x + 4$
  3. Subtract $8x$ from both sides: $\quad -7x = 4$
  4. Divide by $-7$: $\quad x = -\dfrac{4}{7}$
Worked Example 2 — Solve $\dfrac{2}{x + 2} + \dfrac{3}{2x – 1} = 1$
  1. Multiply every term by both denominators $(x + 2)(2x – 1)$: $$2(2x – 1) + 3(x + 2) = (x + 2)(2x – 1)$$
  2. Expand each part: $$4x – 2 + 3x + 6 = 2x^2 + 3x – 2$$
  3. Simplify the left side: $\quad 7x + 4 = 2x^2 + 3x – 2$
  4. Move everything to one side: $\quad 0 = 2x^2 – 4x – 6$
  5. Divide by 2: $\quad x^2 – 2x – 3 = 0$
  6. Factorise and solve (see Section 6): $\quad (x – 3)(x + 1) = 0$ $$x = 3 \quad \text{or} \quad x = -1$$
Worked Example 3 — Solve $\dfrac{x}{x + 2} = \dfrac{3}{x – 6}$
  1. When one fraction equals another, cross-multiply: $$x(x – 6) = 3(x + 2)$$
  2. Expand both sides: $\quad x^2 – 6x = 3x + 6$
  3. Move everything to one side: $\quad x^2 – 9x – 6 = 0$
  4. This does not factorise, so use the quadratic formula (see Section 6). The exact answers are left in surd form: $$x = \frac{9 \pm \sqrt{105}}{2}$$
Watch out: Clearing the fractions often produces a quadratic equation. Solve it using the methods in Section 6.

4. Simultaneous Linear Equations in Two Unknowns #

“Simultaneous” means the two equations are true at the same time. The solution is the pair of values ($x$ and $y$) that fits both equations. A common method is elimination: make the number in front of one letter match, then add or subtract to remove that letter.

Worked Example — Solve the pair from Section 1

$2x + 3y = 12 \quad \text{…(1)}$
$x – y = 1 \qquad\;\; \text{…(2)}$

  1. Make the $y$ terms match. Multiply equation (2) by 3: $$3x – 3y = 3 \quad \text{…(3)}$$
  2. Add (1) and (3) so the $y$ terms cancel: $$2x + 3y + 3x – 3y = 12 + 3 \quad\Longrightarrow\quad 5x = 15$$
  3. Divide by 5: $\quad x = 3$
  4. Substitute $x = 3$ into equation (2): $\quad 3 – y = 1 \Longrightarrow y = 2$
  5. Solution: $\quad x = 3, \; y = 2$  (each pen costs \$3 and each notebook costs \$2)

Check in equation (1): $\;2(3) + 3(2) = 6 + 6 = 12$ ✓

Other method: You can also use substitution — rearrange one equation to make one letter the subject, then put that into the other equation. Always find both $x$ and $y$.

5. Simultaneous Equations: One Linear and One Non-Linear #

Here one equation is linear (a straight line) and the other is non-linear, containing a squared term. The highest power is two — there are no cubes or higher powers. The best method is substitution.

IMAGE NEEDED: Graph of a straight line crossing a parabola at two points, showing the two solution pairs

Google Images Search: “IGCSE maths simultaneous equations line and parabola two intersection points graph”

Worked Example — Solve $y = x + 2$ and $y = x^2$
  1. Both equations equal $y$, so set them equal to each other: $$x^2 = x + 2$$
  2. Move everything to one side: $\quad x^2 – x – 2 = 0$
  3. Factorise and solve: $\quad (x – 2)(x + 1) = 0 \Longrightarrow x = 2 \;\text{or}\; x = -1$
  4. Find $y$ for each $x$ using $y = x + 2$:
    • When $x = 2$: $\;y = 2 + 2 = 4$
    • When $x = -1$: $\;y = -1 + 2 = 1$
  5. The two solution pairs are: $\quad (2,\; 4) \quad \text{and} \quad (-1,\; 1)$
Remember: A linear and non-linear pair usually has two solution pairs. Each $x$-value has its own $y$-value, so always give your answers in matching pairs.

6. Solving Quadratic Equations #

A quadratic equation contains an $x^2$ term and can be written in the standard form:

Standard form $$ax^2 + bx + c = 0$$

There are three methods. Use whichever fits the equation best.

Method 1 — Factorisation #

Best when the quadratic factorises easily. Write it as two brackets, then each bracket can equal zero.

Worked Example — Solve $x^2 + 5x + 6 = 0$
  1. Find two numbers that multiply to $+6$ and add to $+5$: these are $+2$ and $+3$.
  2. Write as two brackets: $\quad (x + 2)(x + 3) = 0$
  3. Set each bracket to zero: $\quad x = -2 \quad \text{or} \quad x = -3$

Method 2 — Completing the Square #

This method rewrites the quadratic in completed square form $(x + p)^2 + q$, which is also useful on its own.

Worked Example — Solve $x^2 + 6x + 5 = 0$
  1. Halve the number in front of $x$ (half of 6 is 3) and write $(x + 3)^2$.
  2. $(x + 3)^2$ expands to $x^2 + 6x + 9$, which is 4 too big, so subtract 4. Completed square form: $$x^2 + 6x + 5 = (x + 3)^2 – 4$$
  3. Set equal to zero: $\quad (x + 3)^2 – 4 = 0$
  4. Rearrange: $\quad (x + 3)^2 = 4$
  5. Square root both sides (remember $\pm$): $\quad x + 3 = \pm 2$
  6. Solve: $\quad x = -1 \quad \text{or} \quad x = -5$

Some textbooks set this out slightly differently: they move the constant to the other side first, then complete the square. The steps and the answer are exactly the same — only the order on the page changes. Use whichever layout you prefer.

Same equation, alternative layout — Solve $x^2 + 6x + 5 = 0$
  1. Move the constant to the right side: $\quad x^2 + 6x = -5$
  2. Complete the square on the left side: $\quad (x + 3)^2 – 9 = -5$
  3. Add 9 to both sides: $\quad (x + 3)^2 = 4$
  4. Square root both sides (remember $\pm$): $\quad x + 3 = \pm 2$
  5. Solve: $\quad x = -1 \quad \text{or} \quad x = -5$

Method 3 — The Quadratic Formula #

This always works, even when the quadratic does not factorise. The formula is given to you in the exam List of formulas, so you do not need to memorise it:

Quadratic formula $$x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$$
Worked Example — Solve $x^2 – 4x + 1 = 0$
  1. Identify the values: $\quad a = 1, \; b = -4, \; c = 1$
  2. Substitute into the formula: $$x = \frac{-(-4) \pm \sqrt{(-4)^2 – 4(1)(1)}}{2(1)} = \frac{4 \pm \sqrt{12}}{2}$$
  3. Simplify the surd: $\quad \sqrt{12} = 2\sqrt{3}$, so $$x = \frac{4 \pm 2\sqrt{3}}{2} = 2 \pm \sqrt{3}$$
  4. The exact answers in surd form: $\quad x = 2 + \sqrt{3} \quad \text{or} \quad x = 2 – \sqrt{3}$
Choosing a method: Try factorisation first. If it will not factorise, use the quadratic formula or completing the square. When a question asks for an exact answer, leave it in surd form (do not round).

7. Changing the Subject of a Formula #

The subject is the letter on its own, usually before the equals sign. “Changing the subject” means rearranging the formula so a different letter is on its own. Use inverse operations, just like solving an equation.

When the Subject Appears Twice #

Collect all the terms with the wanted letter on one side, then factorise to write the letter once, and divide.

Worked Example — Make $x$ the subject of $y = \dfrac{x + 1}{x – 1}$
  1. Multiply both sides by $(x – 1)$: $\quad y(x – 1) = x + 1$
  2. Expand the bracket: $\quad yx – y = x + 1$
  3. Collect the $x$ terms on one side, the rest on the other: $\quad yx – x = 1 + y$
  4. Factorise out $x$ (it now appears once): $\quad x(y – 1) = 1 + y$
  5. Divide by $(y – 1)$: $\quad x = \dfrac{1 + y}{y – 1}$

When There is a Power or Root of the Subject #

Use the opposite operation: a square is undone by a square root, and a square root is undone by squaring.

Worked Example A — Make $r$ the subject of $A = \pi r^2$ (a power)
  1. Divide both sides by $\pi$: $\quad r^2 = \dfrac{A}{\pi}$
  2. Square root both sides: $\quad r = \sqrt{\dfrac{A}{\pi}}$
Worked Example B — Make $x$ the subject of $y = \sqrt{x + 3}$ (a root)
  1. Square both sides to remove the square root: $\quad y^2 = x + 3$
  2. Subtract 3 from both sides: $\quad x = y^2 – 3$
Tip: Deal with a power or root last, once the subject is on its own. If the subject appears twice, factorising is the key step that lets you write it once.
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