Video: AQA GCSE Mathematics Foundation Tier Pack 3 β€’ Paper 3 β€’ Question 5

Consider the equation 2π‘š βˆ’ 3𝑛 = 7. (a) Work out the value of 4(2π‘š βˆ’ 3𝑛). (b) Work out the value of 4π‘š βˆ’ 6𝑛. (c) Work out the value of 3𝑛 βˆ’ 2π‘š.

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Video Transcript

Consider the equation two π‘š minus three 𝑛 is equal to seven. Part a) Work out the value of four multiplied by two π‘š minus three 𝑛. Part b) Work out the value of four π‘š minus six 𝑛. Part c) Work out the value of three 𝑛 minus two π‘š.

We’re told in the question that the value of two π‘š minus three 𝑛 is seven. We need to use this fact to work out the value of parts a, b, and c. Part a wants us to work out the value of four multiplied by two π‘š minus three 𝑛. We know that the value of the bracket, two π‘š minus three 𝑛, is seven. Therefore, we need to calculate four multiplied by seven. Four multiplied by seven is equal to 28. Therefore, the value of four multiplied by two π‘š minus three 𝑛 is 28.

The second part of the question asks us to work out the value of four π‘š minus six 𝑛. Four and six have a common factor of two. This is because four and six are both in the two times table. This means that we can factorise out a two from the expression. Two multiplied by two π‘š is four π‘š. Therefore, the first term in the bracket is two π‘š. Two multiplied by three 𝑛 is equal to six 𝑛. Therefore, the second term in the bracket is three 𝑛. The sign in the middle, in this case the subtraction, must stay the same. Four π‘š minus six 𝑛 is equal to two multiplied by two π‘š minus three 𝑛.

Once again, as with part a, the bracket is two π‘š minus three 𝑛. This is equal to seven. We therefore need to calculate two multiplied by seven. This is equal to 14. Therefore, the value of four π‘š minus six 𝑛 is 14. An alternative way of solving part b would be to notice the link between the initial equation, two π‘š minus three 𝑛 equals seven, and the expression in part b, four π‘š minus six 𝑛.

The first term has been multiplied by two. The second term has also been multiplied by two. Negative three 𝑛 multiplied by two is equal to negative six 𝑛. Whatever we do to one side of an equation, we must do to the other. Therefore, we need to multiply seven by two to work out the value of four π‘š minus six 𝑛. Seven multiplied by two is equal to 14. So once again, the value of four π‘š minus six 𝑛 is 14.

The third part of the question asks us to work out the value of three 𝑛 minus two π‘š. As this has been written the other way round from the initial equation with the 𝑛 first instead of the π‘š, we can rewrite it as negative two π‘š plus three 𝑛. Notice that this is very similar to the initial equation. Two π‘š has become negative two π‘š. And negative three 𝑛 has become positive three 𝑛. We do this by multiplying both of the terms by negative one. As we need to do the same to the other side of the equation, we need to multiply seven by negative one. Seven multiplied by negative one is equal to negative seven. This means that the value of three 𝑛 minus two π‘š is negative seven.

An alternative method from our expression negative two π‘š plus three 𝑛 would be to factorise out a negative one. This would give us negative one multiplied by two π‘š minus three 𝑛. As two π‘š minus three 𝑛 is equal to seven, we’re left with negative one multiplied by seven, which we know equals negative seven.

As two π‘š minus three 𝑛 is equal to seven, four multiplied by two π‘š minus three 𝑛 is 28. Four π‘š minus six 𝑛 is 14. And three 𝑛 minus two π‘š is equal to negative seven.

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