Rearranging Harder Formulae (Subject Appears Twice, or Under a Power/Root)

Picture a stripe of paint running the same width along two walls. One wall is four metres long, the other is two. You have exactly enough paint for six square metres, and you want to use all of it. So how wide can the stripe be? The area is the width times four, plus the width times two. So the width you are chasing is sitting in two places at once, and dividing by four only clears one of them. The other one is still standing there.

Straightforward rearranging undoes one operation at a time, and that is exactly what breaks here. This is Higher tier work, and it assumes you can already change the subject when your letter appears only once. So watch what happens when it appears twice.

Here is the formula. Capital A equals x y plus x z, and you have to make x the subject. Both x's are marked in the same colour, because where they sit is the entire problem. Try the usual move and divide both sides by y. Every term gets divided, so you end up with A over y equals x plus x z over y. There is still an x in two places, and the second one is messier than when you started. That is why undoing one step at a time cannot finish this. An inverse operation acts on a whole side at once, and y is only attached to one of the two x terms. Nothing you divide by reaches both of them.

So here are three possible first moves. A, divide both sides by y. B, subtract x z from both sides. C, factorise the right hand side into x times the bracket y plus z. Which one actually puts x within reach? Pick one. I'll wait. The answer is C. Factorising is the move, because it takes two x's and writes them as one. x y plus x z becomes x, times the bracket y plus z, and now there is exactly one x on the page.

Watch the bracket close around both terms. The formula now reads A equals x, times the bracket y plus z. That bracket is a single object multiplying x, so you divide both sides by the whole of it. x equals A over the bracket y plus z. And the bracket stays whole on the bottom. You cannot cancel the y on its own, and you cannot cancel the z on its own, because neither of them is a factor of A by itself.

Here is your handle for this video, and you have just earned the first half of it. Two x's, one bracket. When the letter you want turns up twice, the bracket is the thing that makes it one.

Back to the paint, then. The width is six divided by four plus two, so your stripe can be exactly one metre wide.

Your letter can also sit in one place and still be out of reach, because it is squared.

Make r the subject of A equals pi r squared, where r is positive. Think about what was done to r: it was squared, and then multiplied by pi. To get it back, you travel that route backwards. So divide both sides by pi first, giving A over pi equals r squared. Now the square is sitting on its own, and a square root is exactly what undoes it. r equals the square root of A over pi. And because the question tells you r is positive, you write the positive root only. Without that condition, a square root would give you a plus or minus pair. Leave the answer as a root. Tidying a root into a neater surd form is a different skill, and nothing in this question asks for it.

Flip that round now, so the root is wrapped around the letter you want. This is the shape with a trap in it, so we will do it twice, once badly and then properly.

Make g the subject of y equals three plus the square root of g minus four. The g is under the root, and there is a three sitting outside it, on the same side.

Two candidate first moves. A, square both sides right now. B, subtract the three from both sides first. Only one of them leads anywhere. Take your pick. I'll wait. The answer is B, subtract the three first, and here is exactly what goes wrong if you do not.

Square straight away and you get y squared equals nine plus g minus four, which tidies to g equals y squared minus five. It looks like working, and it is wrong, because squaring acts on a whole side at once. Three plus something, all squared, is not nine plus that something. Test it with numbers and it falls apart. If g is eight, the square root of eight minus four is two, so y is five. Feed five into that wrong answer and it hands you twenty, not eight. Now do it properly. Take the three across first, so y minus three equals the square root of g minus four. The root is alone on its side, and squaring has nothing else to trip over. Square both sides. The bracket y minus three, squared, equals g minus four. That bracket is not decoration. It is what says the whole of y minus three was squared, not just the y. Add four to both sides, and g equals the bracket y minus three, squared, plus four. The four stays outside the bracket. It was never under the root, so it never gets squared. Leave it exactly in that form. Expanding the bracket is perfectly allowed, but it takes longer, and every extra line you write is one more place where a slip can creep in.

That is the second half of your handle: root alone, then square. Put both halves together and you have the whole video. Two x's, one bracket. Root alone, then square.

One sentence from an examiner report names the first of those blockages precisely. It comes from a report on a Higher tier paper, and we are picking it up part way through. Full marks required an ability to rearrange equations. The latter was beyond many students, often because they did not grasp the principal of using factorisation to isolate the intended subject of the formula. That is the same wall you hit with A equals x y plus x z. The rearranging is what stopped them, and factorising is the step that gets past it. The root case has its own line, from a report on a different Higher paper, about a change of subject question where the letter was g. This was very typical of the changing the subject questions on this award but still many students seem to struggle to gain more than 1 mark for squaring both sides. When students gathered the terms in g on one side of the equation, they often included the 5 and then gained an incorrect result. The five in that report is a constant that had no business inside the squared bracket, and it went in anyway. Different letters, identical error. So write the isolating line down every time. The line where the bracket appears, or the line where the root sits alone, is the line that shows you knew which move to make.

Two x's, one bracket. Root alone, then square. That is the chant, and here is the run through that sits underneath it. When your letter appears twice, no single inverse operation can reach both places. Factorise it into one bracket, divide by the whole bracket, and leave that bracket whole in your answer. When your letter is squared, undo the multiplying first, then take the root, and check whether the question has told you the letter is positive. And when your letter is under a root, get that root alone before you square, because every constant that started outside the root stays outside the bracket. So if you can make x the subject of y equals a x plus b x, and make r the subject of A equals pi r squared for positive r, you have this.

Next in the chapter: Writing an Expression or Formula from a Context.

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