# Video: GCSE Mathematics Foundation Tier Pack 1 • Paper 2 • Question 17

GCSE Mathematics Foundation Tier Pack 1 • Paper 2 • Question 17

05:26

### Video Transcript

Part a) Simplify 𝑡 squared multiplied by 𝑡 cubed. Part b) Solve five minus 𝑝 equals 11. And Part c) Solve a half multiplied by 𝑚 minus three equals six.

So in part a, what we actually have is 𝑡 squared in brackets next to 𝑡 cubed in brackets. And what this actually means is 𝑡 squared multiplied by 𝑡 cubed. And therefore, to actually solve this problem, what we’re gonna have to use is one of our index laws. And the index law we’re gonna use is one that tells us that if we have 𝑥 to the power of 𝑎 multiplied by 𝑥 to the power of 𝑏, then it’s equal to 𝑥 to the power of 𝑎 plus 𝑏.

So what we do is we actually add the powers. The key to actually allow this to work — so this index rule to work — is to have the same base numbers. So we have so. We have 𝑥 and 𝑥 in our index law. And if we look to our question, our bases are actually 𝑡 and 𝑡. So we have the same base. So we can apply this law.

And when we do that, we get 𝑡 to the power of two plus three. And that’s because two was like our 𝑎 and three was like our 𝑏 and we add them together. So therefore, we’re going to get an answer of 𝑡 to the power of five. So we can say that 𝑡 squared multiplied by 𝑡 cubed fully simplified is equal to 𝑡 to the power of five.

Okay, great, but let’s have a look at why is this. So why does this work? Why does this index law solve this kind of problem? Well, we’re gonna start off by having a look at 𝑡 squared. What 𝑡 squared actually means is 𝑡 multiplied by 𝑡. And in the question, it said let’s multiply this by 𝑡 cubed. So I’ve done this and actually what I’ve got now is 𝑡 multiplied by 𝑡 — so our 𝑡 squared — then multiply it by 𝑡 multiplied by 𝑡 multiplied by 𝑡, our 𝑡 cubed. So as we can see, we actually have now five 𝑡s. And they’re actually all multiplied together, which gives us 𝑡 to the power of five.

So now, we can see why this index law works. Okay, great, let’s move on to part b. So in part b, we’re trying to solve an equation. So we have five minus 𝑝 is equal to 11.

Now, the first stage in order to do that is to actually add on 𝑝. And the reason we’re doing that is because first of all we want to make the 𝑝 positive because it’s easier to deal with. And it also means that actually there’s less chance of making an error with negative numbers. Then, this gives us five is equal to 11 plus 𝑝. And that’s because if we have negative 𝑝 plus 𝑝, it gives us just a zero. So that’s why the left-hand side is just five and then we just add 𝑝 to the right-hand side.

So now, to actually find the value of 𝑝, what we need to do is actually leave the 𝑝 on its own. And to do that, what we’re gonna do is actually subtract 11 from each side of the equation. Therefore, if we do that, we get negative six is equal to 𝑝. And that’s because five minus 11 gives us negative six. And if we have 11 plus 𝑝 and then we subtract 11 from that, we just get left with our 𝑝.

It’s also worth mentioning at this point when we take 11 away from five, I said we got to negative six. And a quick way of doing that if you’re not quite sure with the negative numbers is if you think, well, we take five away to get to zero, then there’s still six left from 11. So you take that six away. We get to negative six. Okay, great, so we can now say that the solution to five minus 𝑝 equals 11 is 𝑝 is equal to negative six.

What we can now do though is actually carry out a check to make sure that we’ve actually got the correct answer. And to do that, we actually substitute 𝑝 is equal to negative six back into the original equation. And when we do that, we get five minus negative six because negative six without 𝑝 is equal to 11.

And then, we need to remember that if we actually subtract a negative, that’s gonna give us a positive. So five minus negative six is gonna be the same as five add six. So then, that leaves us with five plus six is equal to 11, which is great cause it’s the same as we had in the original equation. So therefore, we can say you fully checked, we know that 𝑝 is equal to negative six.

Okay, now, let’s move on to part c. So now, in part c, what we want to do is solve a half multiplied by 𝑚 minus three equals six. And the first step is going to be multiplying both sides of the equation by two. And that’s because we’ve got a half 𝑚 minus three. So therefore, if we have a half multiplied by two, we’ll just get 𝑚 minus three. So it’s easier to deal with.

But of course, whatever we do to this side of the equation, we must do to the other side of the equation. So now, when we’ve actually done that and multiplied both sides of the equation by two, we get 𝑚 minus three is equal to 12. So now, it’s actually the value of 𝑚 that we’re looking to find.

So now, as we want to do that, what we’re gonna do is actually add three to each side of the equation. The reason we do that is because if we add three to negative three we get zero, so then it will leave 𝑚 on its own. And as we said before, whatever you do to one side, we gotta do to the other side of the equation. So when we do that, we get 𝑚 is equal to 15. And that’s because 12 add three is equal to 15.

So now, we found the solution to a half 𝑚 minus three equals six and that is 𝑚 is equal to 15. Well, what we can do now — like we’ve done before — is actually check to make sure this is correct. And the way we do that is by substituting in 𝑚 is equal to 15 back into the original equation. And when we do that, we get a half multiplied by 15 minus three is equal to six. So then, what we get is a half multiplied by 12 is equal to six. Well, this means half of 12 is six. So yes, this is correct. And it’s the same value that we got with the original equation.

So we can say that we’ve now solved part a, b, and c. And the answers are 𝑡 to the power of five, 𝑝 equals negative six, and 𝑚 equals 15, respectively.