Home / Teaching / 3r ESO / Rational numbers / 03 · Fractions: simplest form, equivalence, addition and subtraction
3r ESO · Rational numbers Notes 7 Oct 2026

Fractions: simplest form, equivalence, addition and subtraction

Fractions in their simplest form, equivalent fractions, and how to add and subtract fractions.

Simplest form

A fraction $\dfrac{a}{b}$ is in its simplest form (it is irreducible) if it cannot be simplified any further, that is, if $a$ and $b$ have no common factor:

$$\text{MCD}(a, b) = 1$$

To find it, divide the numerator and the denominator by $\text{HCF}(a, b)$.

Example. Let's find the simplest form of $\dfrac{48}{36}$. First break the numerator and the denominator into prime factors:

482
242
122
62
33
1
362
182
93
33
1
$$48 = 2^4 \cdot 3 \qquad 36 = 2^2 \cdot 3^2$$

The HCF is the common factors with the smallest power:

$$\text{MCD}(48, 36) = 2^2 \cdot 3 = 12$$

Divide the numerator and the denominator by $12$:

$$\dfrac{48}{36} = \dfrac{48 : 12}{36 : 12} = \dfrac{4}{3}$$

The LCM, which we will use for adding, is the common and non-common factors with the largest power: $\text{MCM}(48, 36) = 2^4 \cdot 3^2 = 16 \cdot 9 = 144$.

Equivalent fractions

Two fractions $\dfrac{a}{b}$ and $\dfrac{c}{d}$ are equivalent if they have the same simplest form.

To check it without simplifying, cross-multiply:

$$\dfrac{a}{b} = \dfrac{c}{d} \iff a \cdot d = b \cdot c$$

Example. Both simplify to $\dfrac{3}{4}$, and the cross products are equal:

$$\dfrac{15}{20} = \dfrac{6}{8} \qquad 15 \cdot 8 = 120 \qquad 20 \cdot 6 = 120$$

Example. Find $x$ so that the fractions are equivalent:

$$\dfrac{5}{x} = \dfrac{3}{12}$$

By simplifying. $\dfrac{3}{12} = \dfrac{1}{4}$. To go from $1$ to $5$ we multiply by $5$, so the denominator too:

$$\dfrac{5}{x} = \dfrac{1}{4} = \dfrac{1 \cdot 5}{4 \cdot 5} \;\Longrightarrow\; x = 4 \cdot 5 = 20$$

By cross-multiplying.

$$5 \cdot 12 = 3 \cdot x \;\Longrightarrow\; 60 = 3x \;\Longrightarrow\; x = \dfrac{60}{3} = 20$$

Addition and subtraction

  1. Find the LCM of all the denominators.
  2. Scale up the fractions so that the LCM is their denominator.
  3. Add or subtract the numerators.
  4. Simplify the result.

Example. $\dfrac{3}{8} - \dfrac{5}{6} + \dfrac{7}{12}$. Break the denominators into prime factors to find the LCM:

82
42
22
1
62
33
1
122
62
33
1
$$8 = 2^3 \quad 6 = 2 \cdot 3 \quad 12 = 2^2 \cdot 3$$
$$\text{LCM}(8, 6, 12) = 2^3 \cdot 3 = 24$$
$$\dfrac{3}{8} - \dfrac{5}{6} + \dfrac{7}{12} = \dfrac{9}{24} - \dfrac{20}{24} + \dfrac{14}{24} = \dfrac{3}{24} = \dfrac{1}{8}$$