Video: AQA GCSE Mathematics Foundation Tier Pack 4 β€’ Paper 3 β€’ Question 5

(a) simplify π‘₯ + π‘₯ + π‘₯ + π‘₯ Γ— π‘₯ Γ— 𝑦. (b) simplify 7π‘Ž + 6 βˆ’ 3(2π‘Ž + 1).

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Video Transcript

There are two parts to this question. Part a) simplify π‘₯ plus π‘₯ plus π‘₯ plus π‘₯ multiplied by π‘₯ multiplied by 𝑦. Part b) simplify 7π‘Ž plus six minus three multiplied by two π‘Ž plus one.

When dealing with part a, it is important that we look at the signs carefully. The last two signs are multiplication ones. Therefore, we will need to consider π‘₯ multiplied by π‘₯ multiplied by 𝑦 first. We do this because of our order of operations or BIDMAS. Multiplication comes before addition. π‘₯ multiplied by π‘₯ is equal to π‘₯ squared. This means that π‘₯ multiplied by π‘₯ multiplied by 𝑦 is equal to π‘₯ squared 𝑦. When multiplying two letters or terms, we can just put the two letters together. For example, π‘Ž multiplied by 𝑏 is equal to π‘Žπ‘.

Let’s now consider the first three terms in our expression, π‘₯ plus π‘₯ plus π‘₯. This is equal to three π‘₯. Therefore, our expression has been simplified to three π‘₯ plus π‘₯ squared 𝑦. As the two terms have different powers, one is π‘₯ and one is π‘₯ squared 𝑦, we cannot simplify this any further. You cannot add or subtract two terms with different indices or powers. The expression π‘₯ plus π‘₯ plus π‘₯ plus π‘₯ multiplied by π‘₯ multiplied by 𝑦 in its simplest form is three π‘₯ plus π‘₯ squared 𝑦.

When trying to simplify the expression in part b, we once again need to use BIDMAS to work out which operation to do first. As the B stands for brackets, we need to expand or multiply out the bracket first. In order to expand or multiply out a bracket, we need to multiply the term outside by each of the terms inside the bracket. In this case, we firstly need to multiply negative three by two π‘Ž. This is equal to negative six π‘Ž, as negative three multiplied by two is equal to negative six. Multiplying a negative number and a positive number gives us a negative answer. We then need to multiply negative three by one or plus one. This is equal to negative three, as once again multiplying a negative by a positive number gives us a negative answer. The bracket expands or simplifies to negative six π‘Ž minus three. Therefore, we are left with seven π‘Ž plus six minus six π‘Ž minus three.

Our next step is to group or collect like terms. When doing this, it is important that you group the sign that is in front of the term. In this case, we need to group seven π‘Ž and negative six π‘Ž. We also need to group the constants plus six and negative three. Seven π‘Ž minus six π‘Ž is equal to one π‘Ž, as seven minus six is equal to one. This is just written as π‘Ž. Positive six minus three is equal to three or positive three. This means that our expression simplifies to π‘Ž plus three. Seven π‘Ž plus six minus three multiplied by two π‘Ž plus one is equal to π‘Ž plus three.

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