Faded examples · Adding fractions with different denominators

Non-calculator

Each worked example shows a little less than the one before – you complete the faded (blank) steps yourself, following the same four steps every time. In each example also answer the check question – it tests why the step works.

①Example 1fully worked: read it through

Work out $\dfrac{1}{2}+\dfrac{1}{3}$.

1Find a common denominator

common denominator $=6$

2Rewrite each fraction over it

$\dfrac{1}{2}=\dfrac{3}{6},\;\; \dfrac{1}{3}=\dfrac{2}{6}$

3Add the numerators

$\dfrac{3}{6}+\dfrac{2}{6}=\dfrac{5}{6}$

4Simplify if you can

$\dfrac{5}{6}$ is already in its lowest terms

Check · Why can we not just add the numerators of $\tfrac{1}{2}+\tfrac{1}{3}$ to get $\tfrac{2}{5}$?

A the pieces are different sizes until the denominators matchB because $2$ and $3$ are not equal numbersC you must always add the denominators tooD adding fractions is not allowed

②Example 2you finish the last 1 step

Work out $\dfrac{2}{3}+\dfrac{1}{4}$.

1Find a common denominator

common denominator $=12$

2Rewrite each fraction over it

$\dfrac{2}{3}=\dfrac{8}{12},\;\; \dfrac{1}{4}=\dfrac{3}{12}$

3Add the numerators

$\dfrac{8}{12}+\dfrac{3}{12}=\dfrac{11}{12}$

4Simplify if you can

$\dfrac{11}{12}$ is already in its lowest terms

Check · A student used $7$ as the common denominator of $\tfrac{2}{3}$ and $\tfrac{1}{4}$. What did they do wrong?

A they should multiply the two fractions insteadB nothing – $7$ is correctC you use a common multiple of $3$ and $4$ (which is $12$), not $3+4$D they should have used $3$

③Example 3you finish the last 2 steps

Work out $\dfrac{3}{4}+\dfrac{1}{6}$.

1Find a common denominator

common denominator $=12$

2Rewrite each fraction over it

$\dfrac{3}{4}=\dfrac{9}{12},\;\; \dfrac{1}{6}=\dfrac{2}{12}$

3Add the numerators

$\dfrac{9}{12}+\dfrac{2}{12}=\dfrac{11}{12}$

4Simplify if you can

$\dfrac{11}{12}$ is already in its lowest terms

Check · What is the lowest common denominator of $4$ and $6$?

A $24$B $12$C $10$D $2$

④Example 4your turn: every step

Work out $\dfrac{1}{6}+\dfrac{3}{8}$.

1Find a common denominator

common denominator $=24$

2Rewrite each fraction over it

$\dfrac{1}{6}=\dfrac{4}{24},\;\; \dfrac{3}{8}=\dfrac{9}{24}$

3Add the numerators

$\dfrac{4}{24}+\dfrac{9}{24}=\dfrac{13}{24}$

4Simplify if you can

$\dfrac{13}{24}$ is already in its lowest terms

Check · How can you tell that $\tfrac{13}{24}$ cannot be simplified?

A $24$ is even, so it never simplifiesB $13$ is prime and is not a factor of $24$, so they share no common factorC a fraction with $24$ on the bottom can never simplifyD it can – it equals $\tfrac{1}{2}$

Answers · Fractions: Four Operations

Faded examples · Adding fractions with different denominators

① Example 1 common denominator $=6$→$\dfrac{1}{2}=\dfrac{3}{6},\;\; \dfrac{1}{3}=\dfrac{2}{6}$→$\dfrac{3}{6}+\dfrac{2}{6}=\dfrac{5}{6}$→$\dfrac{5}{6}$ is already in its lowest terms $\dfrac{5}{6}$

Check: A: the pieces are different sizes until the denominators match

② Example 2 common denominator $=12$→$\dfrac{2}{3}=\dfrac{8}{12},\;\; \dfrac{1}{4}=\dfrac{3}{12}$→$\dfrac{8}{12}+\dfrac{3}{12}=\dfrac{11}{12}$→$\dfrac{11}{12}$ is already in its lowest terms $\dfrac{11}{12}$

Check: C: you use a common multiple of $3$ and $4$ (which is $12$), not $3+4$

③ Example 3 common denominator $=12$→$\dfrac{3}{4}=\dfrac{9}{12},\;\; \dfrac{1}{6}=\dfrac{2}{12}$→$\dfrac{9}{12}+\dfrac{2}{12}=\dfrac{11}{12}$→$\dfrac{11}{12}$ is already in its lowest terms $\dfrac{11}{12}$

Check: B: $12$

④ Example 4 common denominator $=24$→$\dfrac{1}{6}=\dfrac{4}{24},\;\; \dfrac{3}{8}=\dfrac{9}{24}$→$\dfrac{4}{24}+\dfrac{9}{24}=\dfrac{13}{24}$→$\dfrac{13}{24}$ is already in its lowest terms $\dfrac{13}{24}$

Check: B: $13$ is prime and is not a factor of $24$, so they share no common factor