1. Rearrange [latex]j=3w-5[/latex] in terms of [latex]w[/latex].

Strategy:

We first get [latex]3w[/latex] as the subject and then divide by [latex]3[/latex].

Solution:
\begin{alignat*}{1}
j&=3w-5\\
\implies j+5 & =3w-5+5\mathrm{\quad(add}\,5\,\mathrm{to\,both\,sides)}\\
\implies j+5 & =3w\quad \text{(giving } 3w \text{ as the subject})\\
\implies \frac{j+5}{3} & =\frac{3w}{3}\mathrm{\quad(dividing\ both\ sides\ by\ } 3)\\
\implies\frac{j+5}{3} & =w \quad \text{canceling the }3 \text{'s}\\
\implies w & =\frac{j+5}{3}. \quad\text{(rewriting to get new subject on left)}
\end{alignat*}

2. Make [latex]c[/latex] the subject of [latex]E=mc^{2}[/latex].

Strategy:

First, get [latex]c^2[/latex] as the subject by dividing both sides by [latex]m[/latex], then take the square root of both sides.

Solution:
\begin{align*}
E & =mc^{2}\\
\implies \frac{E}{m} & =\frac{mc^{2}}{m}\quad \text{(divide both sides by }m\text{)}\\
\implies \frac{E}{m} & =c^{2} \quad \text{(canceling }m)\\
\implies \sqrt{\frac{E}{m}} & =\sqrt{c^{2}}\quad\text{(take square root of both sides)}\\
\implies \sqrt{\frac{E}{m}} & =c\mathrm{\quad(remember}\,\sqrt{n^{2}}=n)\\
\implies c & =\sqrt{\frac{E}{m}}.\quad \text{(rewriting to get new subject on left)}
\end{align*}

3. Transpose [latex]S=ut+\frac{1}{2}at^2[/latex] to make [latex]u[/latex] the subject.

Strategy:

First, get the [latex]ut[/latex] term on its own and then divide by [latex]t[/latex].

Solution:
\begin{alignat*}{1}
S & =ut+\frac{1}{2}at^2\\
\implies S- \frac{1}{2}at^2 & =ut+\frac{1}{2}at^2-\frac{1}{2}at^2\quad \text{(subtract }ut+\frac{1}{2}at^2 \text{ from both sides)}\\
\implies S- \frac{1}{2}at^2 & =ut\\
\implies \frac{S- \frac{1}{2}at^2}{t} & =\frac{ut}{t} \quad \text{(dividing both sides by }t \text{)}\\
\implies \frac{S- \frac{1}{2}at^2}{t} & =u \quad \text{(canceling } t \text{ on the right side)}\\
\implies u & =\frac{S- \frac{1}{2}at^2}{t}. \quad \text{(rewriting to get new subject on left)}
\end{alignat*}
The appearance of this formula may be improved by removing the fraction in the numerator on the right-hand side. This is done by multiplying the top and bottom of the right-hand side by [latex]2[/latex] as follows:
\begin{alignat*}{1}
u & =\frac{S- \frac{1}{2}at^2}{t} \times \frac{2}{2}\\
&=\frac{2S- at^2}{2t} .
\end{alignat*}

1. Transform the formula [latex]P=2\left(L-W\right)[/latex] to make [latex]L[/latex] the subject.

Strategy:

First Divide both sides by [latex]2[/latex] to remove the bracket and then use the usual techniques to determine [latex]L[/latex].

Solution:
\begin{align*}
P & =2\left(L-W\right)\\
\implies \frac{P}{2} & =L-W\\
\implies \frac{P}{2}+W & =L\\
\implies L & =\frac{P}{2}+W.
\end{align*}

2. Transform the formula [latex]P=2(L-W)[/latex] to make [latex]W[/latex] the subject.

Solution:

\begin{align*}
P & =2\left(L-W\right)\\
\implies \frac{P}{2} & =L-W\quad\left(\textrm{cancel the 2 and remove the brackets}\right)\\
\implies \frac{P}{2}-L & =-W\\
\implies -W & =\frac{P}{2}-L\\
\implies W & =L-\frac{P}{2}
\end{align*}

3. Rearrange the formula [latex]L=\dfrac{\left(Mt-g\right)}{b}[/latex] to make [latex]M[/latex] the subject.

Solution:

\begin{align*}
L & =\frac{Mt-g}{b}\\
\implies Lb & =Mt-g\\
\implies Lb+g & =Mt\\
\implies \frac{Lb+g}{t} & =M\\
\implies M & =\frac{Lb+g}{t}
\end{align*}

4. Make [latex]v[/latex] the subject of [latex]E=mgh+\dfrac{1}{2}mv^{2}.[/latex]

Solution:

\begin{align*}
E & =mgh+\frac{1}{2}mv^{2}\\
\implies E-mgh & =\frac{1}{2}mv^{2}\\
\implies 2\left(E-mgh\right) & =mv^{2}\\
\implies \frac{2\left(E-mgh\right)}{m} & =v^{2}\\
\implies v^{2} & =\frac{2\left(E-mgh\right)}{m}\\
\implies v & =\pm\sqrt{\frac{2\left(E-mgh\right)}{m}}\\
& =\sqrt{\frac{2\left(E-mgh\right)}{m}}
\end{align*}
where we assume the velocity [latex]v[/latex] is positive.

5. If [latex]\dfrac{2}{k}=\dfrac{j+1}{3}[/latex], find [latex]k[/latex].

Strategy: Multiply both sides by [latex]k[/latex] to remove the [latex]k[/latex] on the bottom of the left-hand side. Then use the usual methods to get [latex]k[/latex] on its own.

Solution:

Note that when we multiply both sides by [latex]k[/latex] the [latex]j+1[/latex] term is put into a bracket
\begin{align*}
\frac{2}{k} & =\frac{j+1}{3}\\
\implies 2 & =\frac{k\left(j+1\right)}{3}\\
\implies 6 & =k\left(j+1\right)\\
\implies \frac{6}{j+1} & =k\\
\implies k & =\frac{6}{j+1}.
\end{align*}
It is possible to do the first two steps in one action as follows:
\begin{align*}
6 & =k\left(j+1\right).
\end{align*}
This is called cross-multiplication.

1. Transpose [latex]E  =mgh+\frac{1}{2}mv^2[/latex] to make [latex]m[/latex] the subject.

Solution:
In this case, the terms involving [latex]m[/latex] are already on the right-hand side of the formula so we can factorise immediately:
\begin{align*}
E & =mgh+\frac{1}{2}mv^2\\
& =m\left( gh+\frac{1}{2}v^2 \right).\\
\end{align*}
Now we can divide both sides by [latex]\left( gh-\frac{1}{2}v^2 \right)[/latex] to get [latex]m[/latex]:\begin{align*}
E& =m\left( gh+\frac{1}{2}v^2 \right)\\
\frac{E}{\left( gh+\frac{1}{2}v^2 \right)} &=m\\
m&=\frac{E}{\left( gh+\frac{1}{2}v^2 \right)} .
\end{align*}This can be made to look a bit better by multiplying the top and bottom of the right-hand side to remove the fraction in the denominator:\begin{align*}
m&=\frac{E}{\left( gh+\frac{1}{2}v^2 \right)} \\
&=\frac{E}{\left( gh+\frac{1}{2}v^2 \right)} \times \frac{2}{2}\\
&=\frac{2E}{\left( 2gh+v^2 \right)}.
\end{align*}

2. Make [latex]I[/latex] the subject of the formula [latex]Ir=E-IR[/latex].

Solution:
First get all terms involving [latex]I[/latex] to one side by adding [latex]IR[/latex] to
both sides:
\begin{align*}
Ir+IR & =E.\\
\end{align*}
Take [latex]I[/latex] out as a common factor:
\begin{align*}
I\left(r+R\right) & =E.\\
\end{align*}
Now divide both sides by [latex]r+R[/latex]:
\begin{align*}
Ir+IR & =E.\\
\end{align*}

1. If
\begin{align*}
T & =1+\frac{1}{n+P}
\end{align*}
make [latex]P[/latex] the subject.

Solution:
The new subject [latex]P[/latex] is in the denominator of the second term on the RHS. To get this on the top, we multiply both sides by [latex]n+P\,[/latex]:
\begin{align*}
T \left( n+P \right) & =1 \times \left( n+P \right) +\frac{1}{n+P} \times \left( n+P \right)\\
\implies T n+TP & = n+P+1.\\
\end{align*}
Now get all the terms involving P on one side (we will put them on the left):
\begin{align*}
T n+TP-P &= n+1\\
\implies TP-P &= n+1 -Tn.\\
\end{align*}
Factorise [latex]TP-P \;[/latex]:
\begin{align*}
P\left( T-1 \right) &= n+1 -Tn.\\
\end{align*}

Divide both sides by [latex]\left( T-1 \right)[/latex]:
\begin{align*}
P &= \frac{n+1 -Tn}{\left( T-1 \right)}.\\
\end{align*}

It is possible to factorise the right-hand side to get:
\begin{align*}
P &= \frac{n\left(1-T \right)+1 }{\left( T-1 \right)}.\\
\end{align*}

2. If
\begin{align*}
\frac{1}{R}&=\frac{1}{R_1}+\frac{1}{R_2}
\end{align*}
make [latex]R_1[/latex] the subject.

Solution:
Multiplying both sides by [latex]R_{1}[/latex]:
\begin{align*}
\frac{1}{R} \times R_{1}&=\frac{1}{R_1} \times R_{1} +\frac{1}{R_2}\times R_{1}\\
\frac{R_1}{R}&=1+\frac{R_1}{R_2}.
\end{align*}

Bring all terms involving [latex]R_1[/latex] to the left:

\begin{align*}
\frac{R_1}{R}-\frac{R_1}{R_2}&=1.
\end{align*}

Factorise and obtain a common denominator:
\begin{align*}
R_1 \left( \frac{1}{R}-\frac{1}{R_{2}} \right)&=1\\
\implies R_1 \left( \frac{R_{2}-R}{RR_{2}} \right)&=1.
\end{align*}

Divide both sides by [latex]\dfrac{R_{2}-R}{RR_{2}}[/latex]
\begin{align*}
R_1 &=1 \div \left( \frac{R_{2}-R}{RR_{2}} \right)\\
&=1 \times \left( \frac{RR_{2}}{R_{2}-R} \right)\\
&= \frac{RR_{2}}{R_{2}-R}.
\end{align*}

3. Transpose
\begin{align*}
m=\sqrt{\frac{d-s}{s\left(e-f\right)}}\\
\end{align*}
to make [latex]s[/latex] the subject.

Solution:
The square root sign acts as a bracket and we want to remove it so we can get at the subject term [latex]s[/latex].
We remove it by squaring both sides:
\begin{align*}
m^2 & =\left(\sqrt{\frac{d-s}{s\left(e-f\right)}}\right)^{2}\\
& =\frac{d-s}{s\left(e-f\right)}.
\end{align*}
Now multiply both sides by [latex]s\left(e-f\right)[/latex]:
\begin{align*}
s\left(e-f\right)m^{2}& =  d-s.
\end{align*}
Now gather all the [latex]s[/latex] terms to one side.
\begin{align*}
s\left(e-f\right)m^{2}+s&=  d.
\end{align*}
Take a common factor of [latex]s[/latex] out of the right-hand side:
\begin{align*}
s\left(\left(e-f\right)m^{2}+1\right) & =d.
\end{align*}
Now divide by [latex]\left(\left(e-f\right)m^{2}+1\right)[/latex] to get the
result
\begin{align*}
s & =\frac{d}{\left(\left(e-f\right)m^{2}+1\right)}.
\end{align*}

1. The formula for complex interest is

\begin{align*}
A=P\left( 1 + \dfrac{r}{n} \right)^{nt}
\end{align*}

where [latex]A[/latex] is the final amount, [latex]P[/latex] is the initial amount, [latex]r[/latex] is the rate of interest per time period, [latex]n[/latex] is the number of times the interest is calculated per time period and [latex]t[/latex] is the time of the investment.

a) Transpose the formula to make t the subject
b) Hence calculate the time, in years, it takes for an initial investment of $5000 to become $10000 at an interest rate of 10% per year when the interest is calculated annually.

Solution a):

Taking logs (in base 10) of both sides we have,

\begin{align*}
\log \left(A \right)&=\log \left( P\left( 1 + \frac{r}{n} \right)^{nt} \right)\\
&=\log \left( P\right) +\log \left(  1 + \frac{r}{n} \right)^{nt}  \text{ using the log multiplication law}\\
&=\log \left( P\right) +nt\log\left(1+\frac{r}{n}  \right) \text{ using the log power law}\\
\end{align*}

Rearrange to get the terms containing [latex]t[/latex] on the left hand side:

\begin{align*}
nt\log\left(1+\frac{r}{n}  \right)&=\log \left( A\right)-\log \left(P \right). \\
\end{align*}

Now get [latex]t[/latex] on its own:

\begin{align*}
t&=\frac{\log \left( A\right)-\log \left(P \right)}{n\log\left(1+\frac{r}{n}  \right)}\\
&=\frac{\log \left( \frac{A}{P} \right)}{n\log\left(1+\frac{r}{n}  \right)}\\\end{align*}

where, in the final step, we used the log division law on the numerator.

Solution b):

We have [latex]A=$10000[/latex], [latex]P=$5000[/latex], [latex]r=0.1[/latex] and [latex]n=1[/latex] as the interest is paid annually. Substituting into the answer from part a) gives:

\begin{align*}
t&=\frac{\log \left( \frac{10000}{5000} \right)}{\log\left(1+\frac{0.1}{1}  \right)}\\
&=\frac{\log \left( 2 \right)}{\log\left(1.1 \right)}\\
&\approx 7.3\,.
\end{align*}

Hence, it will take 7.3 years to get $10000 from an initial investment of $5000 at an interest rate of 10% per year.

2. Given the formula

\begin{align*}
T&=\frac{1}{c}\log_{10}\left(m-A \right),
\end{align*}

make [latex]m[/latex] the subject.

Solution:

We have

\begin{align*}
T&=\frac{1}{c} \log_{10} \left( m-A \right) \\
\implies cT&=\log_{10} \left( m-A \right). \\
\end{align*}

Hence

\begin{align*}
10^{cT}&=10^{\log_{10}\left(m-A \right)} \\
&=m-A\\
\implies m&=A+10^{cT}.
\end{align*}