Forming & Solving Equations (Edexcel IGCSE Maths): Revision notes

Section 1

How do you substitute into a formula?

Substitution means replacing letters in a formula with given numbers, then working out the value using the correct order of operations (BIDMAS).

Given the formula $v = u + at$ , find $v$ when $u = 5$ , $a = 2$ , $t = 3$ :

$v = 5 + (2 \times 3) = 5 + 6 = 11$

Always substitute the whole number (including its sign) into the formula, and use brackets around negative values to avoid sign errors, e.g. $a = -4$ becomes $(-4)$ .

Key termssubstitutionformulaBIDMAS

Common mistake

A common error is substituting $-4$ into $x^2$ and writing $-4^2 = -16$ . Always bracket first: $(-4)^2 = 16$ .

Exam tip

Write out the formula first, then substitute values underneath before calculating — this keeps your working clear for method marks.

Section 2

How do you form an equation from a worded problem?

To form an equation, translate the words into algebra step by step, using a letter for the unknown quantity.

Example: "I think of a number, double it, then add 7. The result is 25." Let the number be $x$ .

$2x + 7 = 25$

For shape problems, use given facts (e.g. angles in a triangle sum to $180°$ , or perimeter = sum of sides) to build the equation. Example: a triangle has angles $x$ , $2x$ and $x + 30$ . Since angles sum to $180°$ :

$x + 2x + (x + 30) = 180$

Key termsunknownexpressionequation

Think of it like this

Forming an equation is like translating a sentence into a different language — each phrase ("double it", "add 7") becomes a piece of algebra, kept in the same order as the words.

Example

"The perimeter of a rectangle with length $x + 4$ and width $x$ is 28 cm" becomes $2(x + 4) + 2x = 28$ .

Section 3

How do you solve a linear equation once it's formed?

Solve by doing the same operation to both sides until $x$ is isolated.

Solve $3x - 5 = 16$ :

$3x = 21$ (add 5 to both sides)

$x = 7$ (divide both sides by 3)

If the unknown appears on both sides, e.g. $5x + 3 = 2x + 18$ , collect the $x$ terms on one side first:

$3x + 3 = 18$ then $3x = 15$ then $x = 5$

If there are brackets, expand them before collecting terms.

Key termslinear equationisolatecollect like terms

Exam tip

Whatever you do to one side, do exactly the same to the other side — this keeps the equation balanced.

Common mistake

Forgetting to multiply every term inside a bracket, e.g. expanding $3(x - 2)$ as $3x - 2$ instead of $3x - 6$ .

Section 4

How do you change the subject of a formula?

Changing the subject means rearranging a formula so a different letter is on its own on one side.

Rearrange $v = u + at$ to make $t$ the subject:

$v - u = at$ (subtract $u$ from both sides)

$t = \dfrac{v - u}{a}$ (divide both sides by $a$ )

For formulae with squares or square roots, undo them last. Rearrange $A = \pi r^2$ to make $r$ the subject:

$r^2 = \dfrac{A}{\pi}$ then $r = \sqrt{\dfrac{A}{\pi}}$

If the new subject appears twice, factorise it out. Rearrange $ax + b = cx + d$ to make $x$ the subject:

$ax - cx = d - b$ then $x(a - c) = d - b$ then $x = \dfrac{d - b}{a - c}$

Key termssubject of the formularearrangefactorise

Exam tip

Perform inverse operations in reverse order (undo addition/subtraction before multiplication/division, and undo powers/roots last).

Common mistake

When the new subject appears on both sides, forgetting to factorise it out before dividing — you cannot isolate $x$ until it's written as a single term.

Must Know

Substitute values as whole numbers with correct signs, bracketing negatives before applying powers.
Translate worded problems into algebra phrase by phrase, keeping the same order as the sentence.
Solve equations by applying identical operations to both sides until the unknown is isolated.
Collect unknown terms on one side first when the unknown appears twice, then solve.
To change the subject, undo operations in reverse order (BIDMAS backwards), doing squares/roots last.
If the new subject appears more than once, factorise it out before dividing.

That's the notes covered.

Carry on to the next subtopic.