【PY100】Mathematics – (2) Change of Subject of Formulas

Last time we covered the laws of indices, and promised that this time we would move formulas around. Physics gives you $v = u + at$. The area of a circle is $A = \pi r^2$. A formula is usually written as "one letter = an expression". But a question often gives you $A$ and asks for $r$. Then you must rewrite the formula so that $r$ stands alone on one side. This is called changing the subject. You will use it for solving equations, for areas and volumes, and in physics, all the time.

Subject: the letter that stands alone on one side of a formula. For example, the subject of $A = \pi r^2$ is $A$. 1. Changing the subject means rewriting the formula so that a different letter becomes the subject. 2. When you finish, the new subject appears on one side only, and only once. The other side does not contain it.

The rule: do the same thing to both sides

A formula is like a balance. The two sides of the equals sign weigh the same. Add 3 to the left and you must add 3 to the right. Multiply the left by 5 and you must multiply the right by 5. Only then is the formula still true. Teachers often say "move it across and change the sign". That is a shortcut for the same idea: subtract 32 from both sides, the $+32$ on the left disappears, and a $-32$ appears on the right. It looks as if the number jumped across. Really, both sides did the same thing.

Which operation should you use? Look at what has been done to the letter you want. In $F = \frac{9C}{5} + 32$, $C$ is multiplied by 9, then divided by 5, then 32 is added. Each step wraps another layer around it. To get it out, use the opposite operations in the reverse order: the last layer on is the first layer off. It is like getting dressed. You put on a shirt, then a jacket. To undress, the jacket comes off first.

Inverse operation: an operation that undoes another one. 1. Addition and subtraction are inverses. 2. Multiplication and division are inverses. 3. Squaring and taking the square root are inverses (for a number that is not negative). To change the subject, start from the outermost layer and remove the layers one by one with inverse operations.

Example 1: Celsius and Fahrenheit

A temperature $F$ in degrees Fahrenheit and $C$ in degrees Celsius are related by $F = \frac{9C}{5} + 32$. Make $C$ the subject. $C$ was multiplied by 9, divided by 5, then had 32 added, so we undo those steps in reverse:

1. The last step was adding 32, so first subtract 32 from both sides: $F - 32 = \frac{9C}{5}$
2. Multiply both sides by 5. This is the step most people get wrong: the 5 multiplies the whole left side, so $F - 32$ must go in brackets: $5(F - 32) = 9C$
3. Divide both sides by 9, then swap the sides so that $C$ is on the left: $C = \frac{5(F - 32)}{9}$

The answer is $C = \frac{5(F - 32)}{9}$. If you drop the brackets in step 2 and write $5F - 32$, only $F$ has been multiplied by 5. The 32 has not, and everything after that is wrong. Check: water boils at 212 degrees Fahrenheit. Substitute $F = 212$ to get $C = \frac{5 \times 180}{9} = 100$, which is exactly 100 degrees Celsius.

Example 2: a formula with a square root

The period $T$ of a simple pendulum and its length $L$ are related by $T = 2\pi\sqrt{\frac{L}{g}}$, where $g$ is the acceleration due to gravity. Make $L$ the subject. $L$ was divided by $g$, square-rooted, then multiplied by $2\pi$. So we start with the $2\pi$:

1. Divide both sides by $2\pi$: $\frac{T}{2\pi} = \sqrt{\frac{L}{g}}$
2. The inverse of a square root is squaring, so square both sides. The whole left side is squared, top and bottom, so $2\pi$ squared is $4\pi^2$, not $2\pi^2$: $\frac{T^2}{4\pi^2} = \frac{L}{g}$
3. Multiply both sides by $g$, and $L$ is on its own: $L = \frac{gT^2}{4\pi^2}$

The answer is $L = \frac{gT^2}{4\pi^2}$. To check, you do not need real values. Pick easy ones: let $g = \pi^2$ and $L = 4$. The original formula gives $T = 2\pi\sqrt{\frac{4}{\pi^2}} = 2\pi \times \frac{2}{\pi} = 4$. Put $T = 4$ into the answer: $L = \frac{\pi^2 \times 16}{4\pi^2} = 4$. We get back the number we started with.

Example 3: the new subject appears twice

DSE Paper 1 likes this type: make $x$ the subject of $y = \frac{2x + 1}{x - 3}$. The difficulty is that $x$ is in both the numerator and the denominator, so removing layers one by one will not work. Instead, collect every term with $x$ on one side, then take $x$ out as a common factor.

1. Multiply both sides by $(x - 3)$ to clear the fraction: $y(x - 3) = 2x + 1$
2. Expand the brackets. $y$ multiplies both terms inside: $xy - 3y = 2x + 1$
3. Move the terms with $x$ to the left and the terms without $x$ to the right, changing signs as they cross: $xy - 2x = 3y + 1$
4. This is the key step. Both terms on the left contain $x$, so take $x$ out as a common factor. Now $x$ appears only once: $x(y - 2) = 3y + 1$
5. Divide both sides by $(y - 2)$: $x = \frac{3y + 1}{y - 2}$

The answer is $x = \frac{3y + 1}{y - 2}$, with $y \neq 2$, because a denominator cannot be 0. Check: put $x = 4$ into the original formula to get $y = \frac{9}{1} = 9$. Put $y = 9$ into the answer: $x = \frac{28}{7} = 4$. We are back where we started.

The most common mistake is stopping halfway. From $xy - 3y = 2x + 1$, a student writes $x = \frac{2x + 3y + 1}{y}$. It looks finished, but there is still an $x$ on the right. $x$ is not the subject yet, and the answer scores nothing. So every time you finish, look at the other side: is the new subject still there? If it is, you are not done.

To sum up: do the same thing to both sides. Undo the operations in reverse order with inverse operations. If the new subject appears more than once, collect it on one side first, then take it out as a common factor. Example 3 used expanding brackets and taking out a common factor without much explanation. Next time: polynomial expansion and operations. We start with how to multiply out brackets, and go through those steps properly.

Learn more: DSE Mathematics tutoring

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