Video: GCSE Mathematics Foundation Tier Pack 5 β€’ Paper 3 β€’ Question 16

GCSE Mathematics Foundation Tier Pack 5 β€’ Paper 3 β€’ Question 16

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Video Transcript

Part a) Factorise 49 plus 14𝑧.

To factorise an algebraic expression like the one we have here means to write it as the product of terms or expressions which multiply together to give the original expression. We need to look for common factors between the terms and then take these common factors out.

First, let’s look at the number part of each term. We have 49 and 14. We need to find the highest common factor of these two numbers. 49 and 14 both appear in the seven times table. 49 is seven multiplied by seven, and 14 is seven multiplied by two. And this is the highest common factor of these two numbers. So we can take seven out the front of our bracket.

Usually, we would then look at the letter part of each term. But in fact, only one of these two terms has any letters at all. So this tells us that we’ve already taken out the highest common factor of these two terms.

Next, we need to work out what goes inside our bracket. And this needs to be the numbers or terms that we multiply seven by to give the original expression. First, we need to make 49, and seven multiplied by seven is 49 as we’ve already said. So we just need a seven for the first term in the bracket. There’s an addition sign between the two terms. And then we need to make 14𝑧. Well, seven multiplied by two gives 14, so seven multiplied by two 𝑧 will give 14𝑧, so this is the second term in the bracket.

Looking at the terms inside our bracket, so that’s seven and two 𝑧, these terms have no common factors other than one. So this tells us that we fully factorised our expression. And our answer is seven multiplied by seven plus two 𝑧. We can of course perform a quick check by expanding the bracket out again, and it gives 49 plus 14𝑧.

Part b) Simplify nine 𝑦 minus three π‘₯ minus eight 𝑦 minus seven π‘₯ plus 𝑦.

To simplify an expression of this form means to group like terms together. Like terms are those which have the same letters with the same powers. For example, three 𝑦 and seven 𝑦 are like terms, as they have the same letter 𝑦 and although it isn’t written they have the same power of 𝑦, which is one. However, three 𝑦 and seven 𝑦 squared are not like terms, as although they have the same letters, they have different powers.

We need to identify the like terms in this question. We can see that we have three terms that all have 𝑦, and this is 𝑦 to the power of one. So we have nine 𝑦 minus eight 𝑦 plus 𝑦. So nine 𝑦 minus eight 𝑦 just gives one 𝑦, which remember we always just write as 𝑦, and then 𝑦 plus 𝑦 gives two 𝑦. So all of the 𝑦 terms can be simplified to just two 𝑦.

Next, we look at the π‘₯ terms. And we have negative three π‘₯ minus seven π‘₯. Well, if we start at negative three and then we subtract a further seven, this makes our value more negative. So negative three π‘₯ minus seven π‘₯ gives negative 10π‘₯. We can’t simplify this expression any further as our 𝑦s and π‘₯s are not like terms. So our final answer is two 𝑦 minus 10π‘₯.

Another way to approach this question would be to reorder the terms in the original expression so that all of the like terms are together. So here we’ve reordered so that we have all the 𝑦s first and then all of the π‘₯s. If you’re going to do this, you must make sure that you bring the sign in front of each term with it. We could then look at the 𝑦s and the π‘₯s separately. And we would see again that we have two 𝑦 and negative 10π‘₯ giving us the same simplified answer of two 𝑦 minus 10π‘₯.

Part c) Solve 49 plus two 𝑀 equals 100𝑀.

In this part of the question, we’re being asked to solve an equation. So we need to find the value of 𝑀 that makes this equation work. We notice first of all that 𝑀 is currently on both sides of the equation. So we want to rearrange the equation slightly so that 𝑀 is only on one side.

We have a larger number of 𝑀s on the right of the equation. So we’ll collect the 𝑀 terms on this side as this means we’ll be working with a positive rather than a negative number of 𝑀s. In order to eliminate the positive two 𝑀 on the left of the equation, we need to subtract two 𝑀. But whatever we do to one side of the equation we must also do to the other in order to keep the equation balanced. So we’re subtracting two 𝑀 from both sides.

On the left of the equation, we now just have 49. And on the right, 100𝑀 minus two 𝑀 gives 98𝑀. To find the value of one 𝑀, we need to divide the right of the equation by 98, but we also need to do the same on the left. So this gives 49 over 98 is equal to 𝑀.

We can actually simplify this fraction quite a lot as both 49 and 98 can be divided by 49. 49 divided by 49 is one, and 98 divided by 49 is two. So this fraction just simplifies to one over two or one-half.

Be careful here. A common mistake that’s often made is to think that the solution to an equation needs to be a whole number. So instead of dividing 49 by 98, some students will divide the bigger number by the smaller number, so that’s 98 by 49, and get an answer of two. However, this is incorrect. So our solution to part c) is that 𝑀 is equal to a half.

It’s always a good idea to check our answers where possible. So let’s substitute this value that we found for 𝑀 back into each side of the equation. On the left, we have 49 plus two 𝑀, which is now equal to 49 plus two multiplied by a half. Two multiplied by a half is just one, so we have 49 plus one, which is 50. On the right of the equation, we have 100𝑀, which is now equal to 100 multiplied by a half. Remember, multiplying by a half is the same as dividing by two. So 100 multiplied by a half is the same as 100 divided by two, which is 50. Therefore, we found that both sides of this equation give the value 50 when we substitute our value of 𝑀. And this tells us that our solution of 𝑀 equals a half is correct.

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