Video: Pack 1 β€’ Paper 3 β€’ Question 1

Pack 1 β€’ Paper 3 β€’ Question 1

02:34

Video Transcript

Make π‘š the subject of the formula 𝑝 equals two 𝐿 minus π‘š over four.

Well, what I’m gonna do is actually demonstrate a couple of methods to actually solving this problem. Okay, so we’ve got 𝑝 equals two 𝐿 minus π‘š over four. So first of all, what I’m gonna do is actually subtract two 𝐿 from each side. And when I actually do that, I get 𝑝 minus two 𝐿 is equal to negative π‘š over four.

The next stage is actually to multiply each side of the equation by four, because obviously we want π‘š to be on its own. So when we do that, we get four multiplied by 𝑝 minus two 𝐿 is equal to negative π‘š.

So now if we actually look back at the question, we can see that we want π‘š as the subject of the formula, not negative π‘š. So therefore, what we’re gonna do is actually multiply each side of the equation by negative one. So therefore, we’re gonna get negative four multiplied by 𝑝 minus two 𝐿 equals π‘š.

So now if we actually expand the bracket, first of all we’re gonna get negative four multiplied by 𝑝, which gives us negative four 𝑝. And then we have negative four multiplied by negative two 𝐿, which is gonna give us eight 𝐿. That’s because a negative multiplied by a negative is a positive. So therefore, we get negative four 𝑝 plus eight 𝐿 is equal to π‘š. And then if we just swapped the equation around just so we can have π‘š on the left-hand side, we see that π‘š is equal to negative four 𝑝 plus eight 𝐿. So we’ve made π‘š the subject of the formula.

Okay, great! I said we shall use a couple of methods. So I’m just using another method just to check. So again, we’re gonna start with 𝑝 is equal to two 𝐿 minus π‘š over four. So then the first stage is gonna be to add π‘š over four, which will give us 𝑝 plus π‘š over four equals two 𝐿. And the reason we’re doing this with this method is what we’re trying to do is actually avoid having a negative π‘š, cause sometimes some students make mistakes when it comes to having a negative π‘š.

Then next, what we’re gonna do is subtract 𝑝 from each side. So therefore, we’re gonna have π‘š over four is equal to two 𝐿 minus 𝑝. Then we multiply each side of the equation by four. And we get π‘š is equal to four multiplied by two 𝐿 minus 𝑝. And then if we expand the bracket, we get π‘š is equal to eight 𝐿, because we got four multiplied by two 𝐿, minus four 𝑝, and that’s because we have four multiplied by negative 𝑝. So therefore, we get that π‘š is equal to eight 𝐿 minus four 𝑝.

Let’s just rearrange this to have it in the same form as we had the first way working it out. So therefore, again, we get π‘š is equal to negative four 𝑝 plus eight 𝐿. So therefore, we’ve made π‘š the subject of the formula and shown you two methods to do so.

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