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This page sets out the method for harder rearranging formulae in the order you would use it, then walks through the full lesson explanation. For three complete worked examples with the slips that cost marks, see the Harder rearranging formulae topic page.
When the letter you want appears twice or is trapped in a fraction, follow these five steps in order.
When the subject is under a square root or is squared, deal with the root or square before anything inside it.
From 5x+2 = yx+8, gather the x terms to get x(5-y) = 6.
x = 6/(5-y). The y stays in the denominator.
If y = √(x+1), then y² = x+1 and x = y²-1.
y(x-2) = 3 means x-2 = 3/y, so x = 2 + 3/y.
If y = x²+1, then x = √(y-1) for the positive root.
y = 2x+6 is the easy case x = (y-6)/2. The harder items have x twice, or a root, or a power.
If the subject appears once, peel the formula from the outside. y=(x+3)/5 becomes 5y=x+3, then x=5y-3. Check with y=4: x=17, and (17+3)/5=4. A square is undone by a square root, and you keep ± if both signs are possible. A square root is undone by squaring both sides. y=√((x+1)/3) gives y^2=(x+1)/3, then 3y^2=x+1, then x=3y^2-1. With y=2, x=11, and (11+1)/3=4=2^2.
Do not square root a sum term by term. √(a+b) is not √(a)+√(b). Square both sides of an equation as a whole, or square root both sides as a whole. And square before you try to sort out the inside of a root. The root is the outside operation. Also multiply every term when you clear a denominator. y(x-3) expands to xy-3y, not xy-3. The y multiplies both parts.
Method: 1. List the operations around the subject, from the outside in. 2. Undo the outermost operation first. 3. A square root is undone by squaring both sides. 4. Check by putting your result back into the original formula.
v^2=u^2+2as is four letters and a square. To make a the subject, subtract u^2 first: v^2-u^2=2as. Then divide by the whole of 2s, so a=(v^2-u^2)/2s. With u=3, v=5 and s=4, a=(25-9)/8=2. Put those numbers back into the original: 9+2(2)(4)=9+16=25=5^2. Dividing only by 2 and leaving s on the top, or dividing by s and leaving the 2, are both incomplete.
The same shape appears with pressure, density and speed. Name the subject, list what is in the way, and undo it. If the subject is squared, isolate that square before you take a root. If the question wants a positive length, say you are taking the positive root and why. Writing ± and then choosing is better than hiding the choice.
Method: 1. Isolate the term that contains the new subject. 2. Divide by everything that multiplies it, as one product. 3. If the subject is squared, take the square root only after it is alone. 4. Use the numbers in the question as a check, not as a replacement for the algebra.
You cannot "divide by x" when x is on the top and the bottom in different terms until those terms are together. For y=(2x+1)/(x-3), multiply by the denominator: y(x-3)=2x+1, so xy-3y=2x+1. Collect terms in x on one side and the rest on the other: xy-2x=3y+1. Factorise: x(y-2)=3y+1. Divide by the bracket: x=(3y+1)/(y-2). The subject now appears once.
Check with y=4. Then x=(12+1)/(4-2)=13/2. The original gives dfrac2×13/2+113/2-3=dfrac147/2=4. If the check fails, the usual cause is expanding y(x-3) as xy-3, or moving 2x without changing its side properly. Stopping at x=(2x+1)/y is not a rearrangement: the subject is still on the right.
Method: 1. Clear the denominator first if the subject is on the top and the bottom. 2. Expand, then collect every term with the subject on one side. 3. Factorise the subject out of those terms. 4. Divide by the bracket that is left.
The last step is almost always "divide by the bracket". For y=(3x-1)/(x+2), the same method gives x=(2y+1)/(3-y). Notice the sign in the denominator. It comes from factorising 3x-xy=x(3-y), not x(y-3). The two forms differ by a factor of -1 on the top and the bottom, so they can be written either way if you are consistent. Pick one and check it.
With y=1, x=(2+1)/(3-1)=3/2. The original is dfrac3×3/2-13/2+2=dfrac7/27/2=1. A rearrangement question is finished when the new subject is on the left, once, and a numerical check has been done or is obvious. Do not leave a root sign over only part of a sum, and do not leave the subject in a denominator.
Method: 1. The last step is usually a division by a bracket. 2. Keep the signs that the factorising produced. 3. The subject must appear only once on the left. 4. Test with a number that does not make a denominator zero.
A formula such as y=5/(x-2) has the subject at the bottom. The first move is to multiply both sides by the whole denominator, so the subject is no longer trapped: y(x-2)=5. Then divide by y to get x-2=5/y, and add 2 to both sides: x=5/y+2. Check with x=7: y=5/5=1, and the answer gives 5/1+2=7.
Do not just flip both sides without thinking. Writing x-2=y/5 is a frequent wrong turn, because the fractions were turned over but the 5 and y were not kept in the right place. A safe route is always the same: multiply by the denominator, expand if needed, then undo what is left around the subject step by step. Take care that the number added or subtracted at the end gets the correct sign.
Method: 1. Multiply both sides by the whole denominator, using brackets. 2. Divide both sides by whatever multiplies the subject bracket. 3. Add or subtract the remaining number to leave the subject alone. 4. Substitute a simple value to check the final formula.
Tip: Use brackets round any denominator that has more than one term before you multiply by it.
When the subject is in two places, such as y=(4x-3)/(2x+1), get rid of the fraction first: y(2x+1)=4x-3. Expand: 2xy+y=4x-3. Now gather every term containing x on one side and everything else on the other: 2xy-4x=-3-y. The subject is still in two terms, so factorise it out: x(2y-4)=-3-y. Finally divide: x=(-3-y)/(2y-4). This is equal to (y+3)/(4-2y), found by multiplying top and bottom by -1.
This four-stage pattern is worth learning by heart: clear the fraction, expand, collect the subject terms on one side, factorise. The factorising step is where the marks are, and it is the step that is often missed. A numerical check is quick. With x=1 the formula gives y=1/3, and substituting y=1/3 into the answer should return x=1. If it does not, look at the signs when moving terms across.
Method: 1. Multiply to remove any fraction, and expand all brackets. 2. Move every term with the subject to one side and the rest to the other. 3. Factorise the subject out as a common factor. 4. Divide by the bracket that is left, and tidy the signs.
To undo a square root, square both sides. In y=4+√(3x-2) first subtract 4 to isolate the root: y-4=√(3x-2). Then square both sides, including the whole of the left-hand side: (y-4)^2=3x-2. Finish with x=((y-4)^2+2)/3. Squaring only y instead of y-4 is a common slip. Isolate the root before you square, and square the whole side.
Formulae with constants such as π work the same way. For the period of a pendulum, T=2π√(l/g), divide by 2π to get T/(2π)=√(l/g), square both sides to get (T^2)/(4π^2)=l/g, and multiply by g to get l=(gT^2)/(4π^2). Treat π like any other number in the working. Leave answers in terms of π unless the question gives values and asks for a decimal.
Method: 1. Isolate the square root on its own side first. 2. Square both sides, using brackets on a side with more than one term. 3. Continue with the usual inverse steps to leave the subject alone. 4. Leave π in the answer unless told otherwise.
Some questions show a student's first line and ask what is wrong. For A=y/3-4, a student writes 3A=y-4. The error is that the whole right-hand side must be multiplied by 3, so the -4 should become -12. The correct line is 3A=y-12, so y=3A+12. A good answer names the term that was missed and says what it should have been.
A mark-scheme-style statement is short and specific: "the 4 should have been multiplied by 3 as well". Avoid vague words such as "they got it wrong". If you are then asked to make the subject, start from the original formula, not from the student's wrong line. A quick way to find the mistake yourself is to substitute a number into the formula and into the student's first line, and see which side stops matching.
Method: 1. Compare the student's line with a correct one, term by term. 2. Name the term that was not multiplied or divided properly. 3. Say what it should have been. 4. Start your own rearrangement from the original formula.
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