Subtract mixed numbers (exchange with 1 whole number to make improper fraction) - vTomb
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Now look at this question: at first, it looks easier than those we have completed so far, because we have the same denominators in the question. But what problem will we have if we try to take 2 and 2 fifths away from 5 and 1 fifth? Well, we can do 5 take away 2, but if we only have 1 fifth, we can’t take 2 fifths away. Now, here’s what we can do instead: rather than taking away 2 fifths from 1 fifth, []we can take 2 fifths away from 1 and 1 fifth. []As an improper fraction, 1 and 1 fifth is the same as 6 fifths. []So now we can write 4 and 6 fifths []minus []2 and 2 fifths. Why is it only 4 and 6 fifths? Well, it’s a bit like exchanging: we exchanged one of our whole numbers for 5 more fifths. So now we’ve got 1 less unit – we’ve got 4 instead of 5 – but that gave us 5 more fifths, so as we had 1 already, we now have 6 fifths. []So now we can answer our question: []our answer is 2 and 4 fifths, because we can do 4 take away 2 to make 2, and we can do 6 fifths take away 2 fifths to make 4 fifths.[]
Now we have 4 and 1 fifth take away 3 and 1 third. When subtracting fractions, we need our denominators to be the same, [][]so let’s change our fractions to fifteenths. [][][][]1 fifth is 3 fifteenths, and [][][][]1 third is 5 fifteenths. So we can re-write our question: []4 and 3 fifteenths []minus []3 and 5 fifteenths. []But we can’t answer this, because we can’t take 5 fifteenths away from 3 fifteenths. So we need to exchange with our whole numbers. []We need to take 5 fifteenths away from 1 and 3 fifteenths, []which is the same as []18 fifteenths. So we need to re-write our question again: []3 and 18 fifteenths []minus []3 and 5 fifteenths. So, basically, to get from 4 and 3 fifteenths to 3 and 18 fifteenths, we have lost a whole number, but that means we can add the number in our denominator to our numerator. []So now we can answer the question: []we have 13 fifteenths, because 3 minus 3 is 0, and 18 minus 5 is 13.[]
So now, pause the video and try 6 and 3 quarters minus 3 and 4 fifths. We need the denominators the same, [][]so let’s change both fractions into twentieths [][][][][][][][] so now we can re-write our question, []6 and 15 twentieths []minus []3 and 16 twentieths. []We can’t answer yet, because we can’t take 16 twentieths away from 15 twentieths. So rather than 15 twentieths, we need to take the fraction away from []1 and 15 twentieths, [][]which is 35 twentieths as an improper fraction. []So now we have 5 and 35 twentieths []minus []3 and 16 twentieths, []and that gives us []2 and 19 twentieths.[]
Now pause the video and work out 4 and 2 sevenths minus 2 and 2 fifths. We need the denominators the same, [][]so let’s change both fractions into thirty-fifths – because that’s the lowest common multiple of 7 and 5. [][][][]2 sevenths is 10 thirty-fifths. [][][][]2 fifths is 14 thirty-fifths. So let’s re-write our question as []4 and 10 thirty-fifths []minus []2 and 14 thirty-fifths. []We can’t answer yet, because we can’t take 14 away from 10. So we need to exchange with our whole number: rather than 10 thirty-fifths, we want to take our fraction away from []1 and 10 thirty-fifths. [][]This is the same as 45 thirty-fifths. So we can re-write our question again, as []3 and 45 thirty-fifths []minus []2 and 14 thirty-fifths. []So our answer is []1 and 31 thirty-fifths.