Video: AQA GCSE Mathematics Higher Tier Pack 2 β€’ Paper 3 β€’ Question 28

𝑝, π‘ž, and π‘Ÿ are integers. 𝑝 is three more than double π‘Ÿ. π‘ž is three less than double π‘Ÿ. Prove that (π‘π‘ž + 9)/4 = π‘ŸΒ².

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Video Transcript

𝑝, π‘ž, and π‘Ÿ are integers. 𝑝 is three more than double π‘Ÿ. π‘ž is three less than double π‘Ÿ. Prove that π‘π‘ž plus nine over four equals π‘Ÿ squared.

To answer this question, we need to use algebra. And the first thing that we’ll do is find expressions for 𝑝 and π‘ž in terms of π‘Ÿ. We’re told that 𝑝 is three more than double π‘Ÿ. So we can express 𝑝 as two π‘Ÿ plus three. Two π‘Ÿ is double π‘Ÿ. And as 𝑝 is three more than this, we need to add three. π‘ž however is three less than double π‘Ÿ. So we can express this as π‘ž equals two π‘Ÿ minus three. We now have expressions for both 𝑝 and π‘ž in terms of π‘Ÿ.

Next, we’ll take this expression π‘π‘ž plus nine over four and substitute both our expressions for 𝑝 and π‘ž in terms of π‘Ÿ into this expression. So we have two π‘Ÿ plus three multiplied by two π‘Ÿ minus three, that’s for π‘π‘ž, and then plus nine over four as before. The next step is to expand the brackets in the numerator so that we can then simplify the result. We can use the memory aid FOIL to help us with this expansion. Each letter in the word FOIL stands for something, and it tells us which terms we need to multiply together.

The F stands for firsts, so we need to multiply the first term in each bracket together. That gives two π‘Ÿ multiplied by two π‘Ÿ, which is four π‘Ÿ squared because two multiplied by two is four and π‘Ÿ multiplied by π‘Ÿ is π‘Ÿ squared. The O stands for outers, so we multiply together the terms on the outside of the expression. That’s the two π‘Ÿ in the first bracket and the negative three in the second. Two π‘Ÿ multiplied by negative three gives negative six π‘Ÿ. Next, the I stands for inners, so we multiply together the terms on the inside of the expression. That’s the positive three in the first bracket and the two π‘Ÿ in the second. Multiplying these gives six π‘Ÿ. Finally, L stands for last, so we need to multiply the last term in each bracket together. That’s the positive three in the first bracket and the negative three in the second, which gives negative nine.

Our expansion of the bracket then is four π‘Ÿ squared minus six π‘Ÿ plus six π‘Ÿ minus nine. The next step is to simplify this by collecting like terms, and you’ll notice that we have negative six π‘Ÿ plus six π‘Ÿ in the center of the expansion. Well these terms just cancel each other out to zero, which means the expansion simplifies to four π‘Ÿ squared minus nine. Let’s now substitute this expansion back into our expression for π‘π‘ž plus nine over four. So we’ve found that two π‘Ÿ plus three multiplied by two π‘Ÿ minus three expands to four π‘Ÿ squared minus nine, which means the expression becomes four π‘Ÿ squared minus nine plus nine all over four. In the numerator, the negative nine and the plus nine will just cancel each other out when added to give zero. So the numerator simplifies to four π‘Ÿ squared, and we now have four π‘Ÿ squared over four. The final step is to cancel the factor of four in the numerator and denominator, which just leaves us with one π‘Ÿ squared over one, which is just equal to π‘Ÿ squared. Notice that this is what we we’re trying to show, that π‘π‘ž plus nine over four, simplified too. So we’re done. All of the stages of algebraic working are an important part of our solution to this question.

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