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Year 7 Maths Fractions

Subtracting Mixed Numerals

20 practice questions 2 video lessons Theory + worked examples
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Theory

To subtract mixed numerals, subtract the whole numbers, then subtract the fractions. The fractions must have the same denominator first, and if the first fraction is smaller you borrow a whole. This Year 7 maths page shows each step, another method using improper fractions, and how to check by adding back.

Subtract the whole numbers first and then the fractions: \(5\dfrac{1}{2} - 4\dfrac{1}{4} = (5 - 4) + \dfrac{2}{4} - \dfrac{1}{4} = 1\dfrac{1}{4}\).

Rename with the lowest common denominator (LCD) if the denominators are different.

Borrow if the first fraction is smaller than the second: take \(1\) whole and rename it as a fraction: \(5\dfrac{2}{7} = 4 + \dfrac{7}{7} + \dfrac{2}{7} = 4\dfrac{9}{7}\).

Whole number minus a fraction: write one whole as a fraction: \(5 - \dfrac{2}{3} = 4 + \dfrac{3}{3} - \dfrac{2}{3} = 4\dfrac{1}{3}\).

3 and 1 quarter take away 1 and 3 quarters Four bars cut into quarters: three whole bars and 1 quarter of a fourth bar are shaded. The last 7 shaded quarters are crossed out in red, leaving 1 whole bar and 2 quarters.
Taking \(1\dfrac{3}{4}\) (\(7\) quarters, crossed out) from \(3\dfrac{1}{4}\) leaves \(1\dfrac{2}{4} = 1\dfrac{1}{2}\). The \(3\) quarters had to come out of a whole bar: that is borrowing.

Borrowing and the improper fraction method:

\[ 4\dfrac{1}{2} - 3\dfrac{2}{3} = 3\dfrac{9}{6} - 3\dfrac{4}{6} = \dfrac{5}{6} \]
41/2−32/3=39/6−34/6=5/6
\[ 3\dfrac{1}{4} - 1\dfrac{2}{3} = \dfrac{39}{12} - \dfrac{20}{12} = \dfrac{19}{12} = 1\dfrac{7}{12} \]
31/4−12/3=39/12−20/12=19/12=17/12
Check by adding back. Answer \(+\) the number you took away \(=\) the number you started with.
When you borrow, the whole number goes down by 1. \(5\dfrac{2}{7} = 4\dfrac{9}{7}\), not \(5\dfrac{9}{7}\).

How to subtract mixed numerals

  1. Rename the fractions with the LCD.
  2. Borrow a whole if the first fraction is smaller than the second.
  3. Subtract the whole numbers, then the fractions.
  4. Simplify and check by adding back.
Example 1 — Rename first
Work out \(5\dfrac{2}{3} - 2\dfrac{1}{4}\).
Solution
\(5\dfrac{2}{3} - 2\dfrac{1}{4}\)\(=\)\(3 + \dfrac{8}{12} - \dfrac{3}{12}\)
\(\)\(=\)\(3\dfrac{5}{12}\)
52/3−21/4=3+8/12−3/12=35/12

\(3\dfrac{5}{12}\)

Example 2 — Borrow a whole
Work out \(4\dfrac{1}{2} - 3\dfrac{2}{3}\).
Solution

\(\dfrac{1}{2} = \dfrac{3}{6}\) is less than \(\dfrac{2}{3} = \dfrac{4}{6}\), so borrow: \(4\dfrac{3}{6} = 3\dfrac{9}{6}\).

\(3\dfrac{9}{6} - 3\dfrac{4}{6}\)\(=\)\(\dfrac{5}{6}\)
39/6−34/6=5/6

\(\dfrac{5}{6}\). Check: \(\dfrac{5}{6} + 3\dfrac{2}{3} = 4\dfrac{1}{2}\).

Example 3 — Same denominator and a whole
Work out
(i) \(6\dfrac{7}{10} - 2\dfrac{3}{10}\)
(ii) \(4 - \dfrac{3}{5}\)
Solution
\(6\dfrac{7}{10} - 2\dfrac{3}{10}\)\(=\)\(4\dfrac{4}{10} = 4\dfrac{2}{5}\)
\(4 - \dfrac{3}{5}\)\(=\)\(3 + \dfrac{5}{5} - \dfrac{3}{5} = 3\dfrac{2}{5}\)
67/10−23/10=42/5,4−3/5=32/5

(i) \(4\dfrac{2}{5}\)
(ii) \(3\dfrac{2}{5}\)

Example 4 — Different denominators
Work out \(5\dfrac{5}{6} - 2\dfrac{1}{4}\).
Solution
\(5\dfrac{10}{12} - 2\dfrac{3}{12}\)\(=\)\(3\dfrac{7}{12}\)
510/12−23/12=37/12

Check: \(3\dfrac{7}{12} + 2\dfrac{3}{12} = 5\dfrac{10}{12} = 5\dfrac{5}{6}\).

Example 5 — LCD and borrowing
Work out \(4\dfrac{1}{6} - 1\dfrac{3}{4}\).
Solution
\(4\dfrac{2}{12} - 1\dfrac{9}{12}\)\(=\)\(3\dfrac{14}{12} - 1\dfrac{9}{12}\)
\(\)\(=\)\(2\dfrac{5}{12}\)
42/12−19/12=314/12−19/12=25/12

\(\dfrac{2}{12} < \dfrac{9}{12}\), so borrow \(\dfrac{12}{12}\). The answer is \(2\dfrac{5}{12}\).

Example 6 — Time left at training
Training lasts \(2\) hours. The warm-up takes \(\dfrac{1}{3}\) hour and drills take \(1\dfrac{1}{4}\) hours. A game fills the rest. How long is the game, in minutes?
Solution
\(\dfrac{1}{3} + 1\dfrac{1}{4}\)\(=\)\(\dfrac{4}{12} + 1\dfrac{3}{12} = 1\dfrac{7}{12}\)
\(2 - 1\dfrac{7}{12}\)\(=\)\(1\dfrac{12}{12} - 1\dfrac{7}{12} = \dfrac{5}{12}\)
1/3+11/4=17/12,2−17/12=5/12

\(\dfrac{1}{12}\) of an hour is \(5\) minutes, so the game lasts \(25\) minutes.

Common pitfalls

Flipping the fractions. In \(4\dfrac{1}{3} - 1\dfrac{2}{3}\) you cannot do \(\dfrac{2}{3} - \dfrac{1}{3}\). Borrow: \(3\dfrac{4}{3} - 1\dfrac{2}{3} = 2\dfrac{2}{3}\).
Not lowering the whole number. \(5\dfrac{2}{7} = 4\dfrac{9}{7}\), not \(5\dfrac{9}{7}\).
Subtracting the denominators. \(\dfrac{4}{5} - \dfrac{1}{2}\) is not \(\dfrac{3}{3}\). It is \(\dfrac{8}{10} - \dfrac{5}{10} = \dfrac{3}{10}\).
Forgetting to rename the whole. \(7 - \dfrac{3}{4} = 6\dfrac{1}{4}\), not \(7\dfrac{1}{4}\).

Frequently asked questions

How do you subtract mixed numerals?

Rename the fractions with a common denominator, borrow if needed, then subtract the wholes and the fractions.

When do you need to borrow?

When the first fraction is smaller than the one you take away, as in \(4\dfrac{1}{6} - 1\dfrac{3}{4}\).

How do you borrow a whole?

Take \(1\) from the whole number and add it to the fraction: \(4\dfrac{2}{12} = 3\dfrac{14}{12}\).

Is there another method?

Yes. Change both to improper fractions, subtract, then change back.

How do you check a subtraction?

Add the answer to the number you took away. You should get the number you started with.