Rearrengement of a formula
Very often you will find that a formula is expressed with one variable written alone in one side of the ‘=’ symbol (usually on the left but not always). The variable that is written alone is known as the subject of the formula.
Consider each of the following formulas
$ \displaystyle s=ut+\frac{1}{2}a{{t}^{2}}$ ($ \displaystyle s$ is the subject)
$ \displaystyle F=ma$ ($ \displaystyle F$ is the subject)
$ \displaystyle x=\frac{{-b\pm \sqrt{{{{b}^{2}}-4ac}}}}{{2a}}$ ($ \displaystyle x$ is the subject)
Now that you can recognize the subject of a formula, you must look at how you change the subject of a formula.
If you take the formula $ \displaystyle v=u+at$ and note that $ \displaystyle v$ is currently the subject, you can change the subject by rearranging the formula.
To make $ \displaystyle a$ the subject of this formula:
$ \displaystyle v=u+at$ Write down the starting formula.
$ \displaystyle v-u=at$ Subtract $ \displaystyle u$ from both sides (to isolate the term containing $ \displaystyle a$)
$ \displaystyle \frac{{v-u}}{t}=a$ Divide both sides by $ \displaystyle t$ (Notice that everything on the left is divided by t). You now have $ \displaystyle a$ on its own and it is the new subject of the formula.
$ \displaystyle a=\frac{{v-u}}{t}$ This is usually rewritten so the subject is on the left.
Rules to obtain the subject of a formula
1. The same number can be added to each side of a formula.
2. The same number can be subtracted from each side of a formula.
3. Both sides of a formula can be divided by the same number.
4. Both sides of a formula can be multiplied by the same number.
Note! Remember that what you do to one side of the formula must be done to the other side. This ensures that the formula you produce still represents the same relationship between the variables.
This process is very similar to solving equations.
Example 1: Make the variable shown in brackets the subject of the formula in each case.
a) $ \displaystyle x+y=c$ $ \displaystyle (y)$
Write down the starting formula
$ \displaystyle x+y=c$
Subtract $ \displaystyle x$ from both sides
$ \displaystyle x+y-x=c-x$
The subject we wanted
$ \displaystyle y=c-x$
b) $ \displaystyle \sqrt{x}+y=z$ $\displaystyle (x)$
Write down the starting formula
$ \displaystyle \sqrt{x}+y=z$
Subtract $ \displaystyle y$ from both sides
$ \displaystyle \sqrt{x}+y-y=z-y$
$ \displaystyle \sqrt{x}=z-y$
Square both sides
$ \displaystyle {{\left( {\sqrt{x}} \right)}^{2}}={{\left( {z-y} \right)}^{2}}$
The subject we wanted
$ \displaystyle x={{\left( {z-y} \right)}^{2}}$
c) $ \displaystyle \frac{{a-b}}{c}=d$ $\displaystyle (b)$
Write down the starting formula
$ \displaystyle \frac{{a-b}}{c}=d$
Multiply both sides by $ \displaystyle c$ to clear the fraction
$ \displaystyle c\cdot \frac{{a-b}}{c}=c\cdot d$
$ \displaystyle a-b=cd$
Make $ \displaystyle b$ positive by adding $ \displaystyle b$ to both sides
$ \displaystyle a-b+b=cd+b$
$ \displaystyle a=cd+b$
Subtract $ \displaystyle cd$ from both sides
$ \displaystyle a-cd=cd-cd+b$
$ \displaystyle a-cd=b$
Rewrite so the subject is on the left
$ \displaystyle b=a-cd$
Example 2: Make the variable shown in brackets the subject of the formula in each case.
a) $ \displaystyle a(n-m)=t$$ \displaystyle (m)$
Write down the starting formula
$ \displaystyle a(n-m)=t$
Open the brackets
$ \displaystyle an-am=t$
Subtract $ \displaystyle an$ from both sides
$ \displaystyle an-am-an=t-an$
$ \displaystyle -am=t-an$
Divide both sides with $ \displaystyle -a$
$ \displaystyle \frac{{-am}}{{-a}}=\frac{{t-an}}{{-a}}$
The subject we wanted
$ \displaystyle m=\frac{{an-t}}{a}$
b) $ \displaystyle \frac{{xy}}{z}=t$ $\displaystyle (z)$
Write down the starting formula
$ \displaystyle \frac{{xy}}{z}=t$ $\displaystyle (z)$
Multiply both sides with $ \displaystyle z$
$ \displaystyle z\frac{{xy}}{z}=zt$
$ \displaystyle xy=zt$
Divide both sides with $ \displaystyle t$
$ \displaystyle \frac{{xy}}{z}=\frac{{zt}}{z}$
$ \displaystyle \frac{{xy}}{z}=t$
Rewrite so the subject is on the left
$ \displaystyle t=\frac{{xy}}{z}$
c) $ \displaystyle \sqrt{{b+c}}=c$ $\displaystyle (b)$
Write down the starting formula
$ \displaystyle \sqrt{{b+c}}=c$
Square both sides
$ \displaystyle {{\left( {\sqrt{{b+c}}} \right)}^{2}}={{c}^{2}}$
$ \displaystyle b+c={{c}^{2}}$
Subtract $ \displaystyle c$ both sides
$ \displaystyle b+c-c={{c}^{2}}-c$
The subject we wanted
$ \displaystyle b={{c}^{2}}-c$
d) $ \displaystyle \frac{x}{{\sqrt{y}}}=c$ $\displaystyle (b)$
Write down the starting formula
$ \displaystyle \frac{x}{{\sqrt{y}}}=c$
Multiply with $ \displaystyle \sqrt{y}$ both sides
$ \displaystyle \sqrt{y}\frac{x}{{\sqrt{y}}}=\sqrt{y}c$
$ \displaystyle x=\sqrt{y}c$
Square both sides
$ \displaystyle {{x}^{2}}={{\left( {\sqrt{y}c} \right)}^{2}}$
$ \displaystyle {{x}^{2}}=y{{c}^{2}}$
Divide with $ \displaystyle {{c}^{2}}$ both sides
$ \displaystyle \frac{{{{x}^{2}}}}{{{{c}^{2}}}}=y$
Rewrite so the subject is on the left
$ \displaystyle y=\frac{{{{x}^{2}}}}{{{{c}^{2}}}}$
Example 3: If the length of a pendulum is $ \displaystyle l$ meters, the acceleration due to gravity is $ \displaystyle g{}^{m}\!\!\diagup\!\!{}_{{{{s}^{2}}}}\;$ and $ \displaystyle T$ is the period of oscillation in seconds then:
$ \displaystyle T=2\pi \sqrt{{\frac{l}{g}}}$
Rearrange the formula to make $ \displaystyle l$ the subject.
Firstly we write down the starting formula
$ \displaystyle T=2\pi \sqrt{{\frac{l}{g}}}$
After observing we need to remove the root so we square both sides
$ \displaystyle {{\left( T \right)}^{2}}={{\left( {2\pi \sqrt{{\frac{l}{g}}}} \right)}^{2}}$
$ \displaystyle {{T}^{2}}=4{{\pi }^{2}}\frac{l}{g}$
Then multiply with $ \displaystyle g$ both sides
$ \displaystyle {{T}^{2}}\cdot g=4{{\pi }^{2}}\frac{l}{g}\cdot g$
$ \displaystyle {{T}^{2}}=4{{\pi }^{2}}\cdot l$
Finally divide with $ \displaystyle 4{{\pi }^{2}}$ both sides
$ \displaystyle \frac{{{{T}^{2}}}}{{4{{\pi }^{2}}}}=l$
Rewrite so the subject is on the left
$ \displaystyle l=\frac{{{{T}^{2}}}}{{4{{\pi }^{2}}}}$
