Video: AQA GCSE Mathematics Foundation Tier Pack 4 β€’ Paper 1 β€’ Question 7

Simplify 8π‘š + 3𝑛 βˆ’ 2π‘š + 5𝑛.

04:29

Video Transcript

Simplify eight π‘š plus three 𝑛 minus two π‘š plus five 𝑛.

So now, if we’re trying to simplify an expression like this, what we want to do is we want to collect like terms. And a like term is a term that shares the same variable or letter. However, they must be of the same order or same power. For example, three 𝑛 squared and two 𝑛 are not like terms, because even though they contain the same variables, so they both got 𝑛, they have different powers because the first term is 𝑛 squared. And the second term is just 𝑛 or 𝑛 to the power of one. Okay, so now we know what like terms are. Let’s collect the like terms in our expression.

So when we’re dealing with like terms, what I say is to circle them. So I’ve circled eight π‘š. And then, I’ve circled negative two π‘š because these are like terms, cause they both contain π‘š. And it’s π‘š to the power of one. Also as you note, I have also included the sign in front of the two π‘š. And that’s because the signs in front of the ones are relevant and refer to our term. So we’ve got a negative two π‘š. And we’ve got eight π‘š.

Well, if we have eight π‘š minus two π‘š, what do we actually mean? Well, eight π‘š means eight multiplied by π‘š. So it means we’ve got eight π‘šs, which I’ve drawn here. So if we take away two π‘šs, we then see how many we’re left with, which is six. So therefore, we can say that eight π‘š minus two π‘š is six π‘š. And the way that we do this without having to show like I have is just to look at the coefficients. So the coefficients are the numbers before their variable. And we subtract or add these differences. Eight minus two is six. So we’ve got six π‘š.

And then next, we have positive three 𝑛 plus five 𝑛. So again, we look at the coefficients. So we have positive three or just three and then add five because we’ve got positive five. So three add five is eight. So therefore, we’ve got six π‘š plus eight 𝑛. So great! We’ve simplified our expression. But have we finished? Well no, we haven’t quite finished because if we look at six π‘š plus eight 𝑛, it can actually be simplified further. And we do that using factorisation. And what is factorising? Well, to factorise our expression, it means putting it into brackets.

Well, to enable us to put it into a bracket, what we want to do is find a factor of the coefficients, so six and eight. Well, in fact, we want to find the highest common factor of six and eight. Well, if we look at the factors of six and eight, we’ve got for the six, six and one, cause six multiplied by one is six, and two and three, cause two multiplied by three is six. And for eight, we’ve got eight and one, cause eight multiplied by one is eight, and four and two, cause four multiplied by two is eight. So therefore, the highest common factor of six and eight is two, because the only other common factor is one.

So then, what we do is we put two outside of our brackets. And then, inside the brackets, we need to decide what the two terms are going to be. Well, when we have a factor outside the brackets, it means that we multiply this by each of the terms inside the brackets to get our original expression. So therefore, the first term in our bracket is gonna be three π‘š. And that’s because if we divide six π‘š by two, we get three π‘š because if we multiply two by three π‘š, we get six π‘š. And six π‘š is the term that we want from the original expression. Okay, great.

So now let’s find the second term in our brackets. Well, we know it’s gonna be three π‘š plus our second term. And that’s because, in the original expression, we had a plus or positive sign between the two terms. Well, our second term is gonna be four 𝑛. And that’s because two multiplied by four 𝑛 gives us eight 𝑛. And that’s the original value that we’d have in our expression.

So therefore, we can say that eight π‘š plus three 𝑛 minus two π‘š plus five 𝑛, fully simplified, is two multiplied by three π‘š plus four 𝑛.

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