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

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

04:08

Video Transcript

Part a) Solve three 𝑔 minus two 𝑔 plus four 𝑔 equals 30.

The first step in solving this equation is to collect like terms on the left of the equation. Like terms are those which have the same letters with the same powers. So in fact, all three terms on the left are like terms, as they’re all some number of 𝑔. And the power of 𝑔 here, although it’s not written, is one in every case.

First, we have three 𝑔 minus two 𝑔. And if you have three lots of something and then take away two lots of the same thing, you’re left with just one lot of that thing. So we have one 𝑔. But remember, we never write one 𝑔. We just write 𝑔. The equation therefore simplifies to 𝑔 plus four 𝑔 equals 30.

Next, we need to add 𝑔 and four 𝑔 together, so one lot of 𝑔 plus four lots of 𝑔 gives five lots of 𝑔. And now we have five 𝑔 is equal to 30. This equation can now be solved in one step. Remember, five 𝑔 means five multiplied by 𝑔. So if we want to know what just 𝑔 is equal to, we need to divide by five. But whatever we do to one side of the equation we must also do to the other. So we’re dividing both sides of the equation by five. Five 𝑔 divided by five is 𝑔, and 30 divided by five is six. So the solution to this equation is 𝑔 equals six.

Part b) Solve seven minus π‘₯ equals 23.

Now at the moment in this equation, we have a negative number of π‘₯s, and it’s always easier to work with a positive number of π‘₯s. So what we’re going to do first of all is add π‘₯ to both sides, as this will eliminate the negative π‘₯ on the left and leave us with positive π‘₯ terms on the right. So adding π‘₯ to both sides just leaves seven on the left. And on the right, we now have π‘₯ plus 23. To leave just π‘₯ on its own on the right of the equation, we need to subtract 23. But remember, whatever we do to one side we must also do to the other.

On the left, we have negative 16, and that’s because we have to subtract seven from seven to get to zero and then we need to subtract a further 16 so that we’ve subtracted 23 overall. We’re left with negative 16 equals π‘₯. So our solution to part b) is π‘₯ equals negative 16.

Part c) Simplify π‘Ž to the power of six over π‘Ž to the power of three.

To answer this question, we need to recall one of our laws of indices or powers. And it’s this: π‘₯ to the power of π‘š divided by π‘₯ to the power of 𝑛 is equal to π‘₯ to the power of π‘š minus 𝑛. This means that if two terms have the same base, which means the number or letter being raised to the power β€” so in our rule, that’s π‘₯, but in our question, it’s π‘Ž β€” then if we are dividing one by the other, we can subtract the powers. So π‘Ž to the power of six divided by π‘Ž to the power of three is equal to π‘Ž to the power of six minus three. Six minus three is three, so this simplifies to π‘Ž to the power of three, which we can also read as π‘Ž cubed.

We can also just look briefly at why this rule works. π‘Ž to the power of six means π‘Ž multiplied together six times. And π‘Ž cubed or π‘Ž to the power of three means π‘Ž multiplied together three times. We can then cancel factors of π‘Ž in the denominator of this fraction with factors of π‘Ž in the numerator. And as there are three π‘Žs in the denominator, these will cancel three of the π‘Žs that we’re multiplying in the numerator, leaving us with π‘Ž multiplied by π‘Ž multiplied by π‘Ž overall. This is why we can subtract the powers, because whatever the power of π‘Ž is in the denominator, we’ll now have that many less for the power of π‘Ž in the numerator.

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