Up until now it has been trivial to substitute values into \textit{formulas} and have the answer pop out quite simply. But now we shall need to do some work to find the answers - we will need to \textbf{rearrange} the \textbf{expression}.

\begin{boxdef}
    \large{\textbf{Trivial: }}
    \begin{solutionordottedlines}[1cm]
        Something that is easy, simple or self-evident.
    \end{solutionordottedlines}
\end{boxdef}


\begin{examplebox}
    \begin{questions}
        \Question[1] The manager of a bed-and-breakfast guest house finds that the weekly profit \(\$ P\) is given by the formula \[ P=40 G-600 \] where \(G\) is the number of guests who stay during the week. Make \(G\) the subject of the formula and use the result to find the number of guests needed to make a profit of \(\$ 800\).
        \begin{solutionordottedlines}[3cm]
            \[
            \begin{aligned}
            P & =40 G-600 \\
            P+600 & =40 G \\
            G & =\frac{P+600}{40} \\
            \text { When } P & =800, \\
            G & =\frac{800+600}{40} \\
            & =\frac{1400}{40}=35
            \end{aligned}
            \]

            Thirty-five guests are required to make a profit of \(\$ 800\).
        \end{solutionordottedlines}
        \Question[3] Given the formula \(v^{2}=u^{2}+2 a s\) :
        \begin{parts}
            \part rearrange the formula to make \(s\) the subject
            \begin{solutionordottedlines}[2cm]
                \(\quad v^{2}=u^{2}+2 a s\)

                \[
                \begin{array}{rlrl}
                v^{2}-u^{2} & =2 a s & & \text { (Subtract } u^{2} \text { from both sides of the formula.) } \\
                s & =\frac{v^{2}-u^{2}}{2 a} & \text { (Divide both sides of the equation by } 2 a .)
                \end{array}
                \]
            \end{solutionordottedlines}
            \part find the value of \(s\) when \(u=4, v=10\) and \(a=2\)
            \begin{solutionordottedlines}[2cm]
                When \(u=4, v=10\) and \(a=2\).

                \[
                \begin{aligned}
                s & =\frac{10^{2}-4^{2}}{2 \times 2} \\
                & =\frac{100-16}{4} \\
                & =\frac{84}{4} \\
                & =21
                \end{aligned}
                \]
            \end{solutionordottedlines}
            \part find the value of \(s\) when \(u=4, v=12\) and \(a=3\)
            \begin{solutionordottedlines}[2cm]
                When \(u=4, v=12\) and \(a=3\).

                \[
                \begin{aligned}
                s & =\frac{12^{2}-4^{2}}{2 \times 3} \\
                & =\frac{144-16}{6} \\
                & =\frac{128}{6} \\
                & =21 \frac{1}{3}
                \end{aligned}
                \]
            \end{solutionordottedlines}
        \end{parts}
        \Question[4] Rearrange each of these formulas to make the pronumeral in brackets the subject.
        \begin{parts}
            \part \(E=\frac{p^{2}}{2 m}\) \hfill\((m)\)
            \begin{solutionordottedlines}[2cm]
                \(\quad E=\frac{p^{2}}{2 m}\)

                \[
                \begin{aligned}
                m E & =\frac{p^{2}}{2} & & \text { (Multiply both sides of the equation by } m .) \\
                m & =\frac{p^{2}}{2 E} & & \text { (Divide both sides by } E .)
                \end{aligned}
                \]
            \end{solutionordottedlines}
            \part \(T=2 \pi \sqrt{\frac{\ell}{g}}\) \hfill\((\ell)\)
            \begin{solutionordottedlines}[2cm]
                \[
                \begin{array}{rlr}
                T & =2 \pi \sqrt{\frac{\ell}{g}} & \\
                \frac{T}{2 \pi} & \left.=\sqrt{\frac{\ell}{g}} \quad \text { (Divide both sides of the equation by } 2 \pi .\right) \\
                \frac{\ell}{g} & =\left(\frac{T}{2 \pi}\right)^{2} & \text { (Square both sides of the equation.) } \\
                & =\frac{T^{2}}{4 \pi^{2}} & \\
                \ell & \left.=\frac{T^{2}}{4 \pi^{2}} \times g \quad \text { (Multiply both sides of the equation by } g .\right)
                \end{array}
                \]

                That is, \(\ell=\frac{T^{2} g}{4 \pi^{2}}\)
            \end{solutionordottedlines}
            \part \(\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\) \hfill\((u)\)
            \begin{solutionordottedlines}[2cm]
                \(\quad \frac{1}{f}=\frac{1}{u}+\frac{1}{v}\)
                \(\frac{1}{f}-\frac{1}{v}=\frac{1}{u}\)
                (Subtract \(\frac{1}{v}\) from both sides.)
                \(\frac{v-f}{f v}=\frac{1}{u}\)
                (common denominator on LHS of equation)

                \[
                u=\frac{f v}{v-f}
                \]

                (Take reciprocals of both sides.)

                Note: \(\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\) does not imply \(f=u+v\).
            \end{solutionordottedlines}
            \part \(P=\sqrt{h+c}-a\)\hfill\((h)\)
            \begin{solutionordottedlines}[2cm]
                \[
                P=\sqrt{h+c}-a
                \]

                \[
                \begin{array}{rlrl}
                P+a & =\sqrt{h+c} \quad & \text { (Add } a \text { to both sides of the equation. }) \\
                (P+a)^{2} & =h+c \quad \quad \text { (Square both sides. }) \\
                h & =(P+a)^{2}-c
                \end{array}
                \]

                The previous example shows some of the techniques that can be used to rearrange a formula.
            \end{solutionordottedlines}
            \part \(\frac{3 p}{4}-\frac{5}{q}=\frac{p^{2}}{3 q}\)\hfill\((q)\)
            \begin{solutionordottedlines}[2cm]
                \[
                \begin{gathered}
                \frac{3 p}{4}-\frac{5}{q}=\frac{p^{2}}{3 q} \\
                12 q\left(\frac{3 p}{4}-\frac{5}{q}\right)=12 q \times \frac{p^{2}}{3 q}
                \end{gathered}
                \]

                \(3 p \times 3 q-5 \times 12=p^{2} \times 4 \quad\) (Multiply both sides by the lowest common denominator, \(12 q\).)

                \[
                \begin{aligned}
                9 p q-60 & =4 p^{2} \\
                9 p q & =4 p^{2}+60 \\
                q & =\frac{4 p^{2}+60}{9 p}
                \end{aligned}
                \]
            \end{solutionordottedlines}
        \end{parts}
    \end{questions}
\end{examplebox}


\begin{exercisebox}
      \begin{questions}
          \Question The profit \( P \) made each day by a store owner who sells CDs is given by the
  formula \( P=5n-150 \), where \( n \) is the number of CDs sold.
          \begin{parts}
              \Part[1] What profit is made if the store owner sells 60 CDs?
              \begin{solutionordottedlines}
                  \( P = 5n - 150 \) \\
                  \( P = 5(60) - 150 \) \\
                  \( P = 300 - 150 \) \\
                  \( P = \$150 \)
              \end{solutionordottedlines}
              \Part[1] Make \( n \) the subject of the formula.
              \begin{solutionordottedlines}
                  \( P = 5n - 150 \) \\
                  \( P + 150 = 5n \) \\
                  \( n = \frac{P + 150}{5} \)
              \end{solutionordottedlines}
              \Part[2] How many CDs were sold if the store made:
              \begin{subparts}
                  \subpart a profit of \( \$275 \) ?
                  \begin{solutionordottedlines}
                      \( n = \frac{P + 150}{5} \) \\
                      \( n = \frac{275 + 150}{5} \) \\
                      \( n = \frac{425}{5} \) \\
                      \( n = 85 \)
                  \end{solutionordottedlines}
                  \subpart a profit of \( \$400 \) ?
                  \begin{solutionordottedlines}
                      \( n = \frac{P + 150}{5} \) \\
                      \( n = \frac{400 + 150}{5} \) \\
                      \( n = \frac{550}{5} \) \\
                      \( n = 110 \)
                  \end{solutionordottedlines}
                  \subpart a loss of \( \$100 \) ?
                  \begin{solutionordottedlines}
                      \( n = \frac{P + 150}{5} \) \\
                      \( n = \frac{-100 + 150}{5} \) \\
                      \( n = \frac{50}{5} \) \\
                      \( n = 10 \)
                  \end{solutionordottedlines}
                  \subpart no profit?
                  \begin{solutionordottedlines}
                      \( n = \frac{P + 150}{5} \) \\
                      \( n = \frac{0 + 150}{5} \) \\
                      \( n = \frac{150}{5} \) \\
                      \( n = 30 \)
                  \end{solutionordottedlines}
              \end{subparts}
          \end{parts}
          \Question The cost \( C \) of hiring a reception room for a function is given by the formula \( C=
  n+250 \), where \( n \) is the number of people attending the function.
          \begin{parts}
              \Part[1] Rearrange the formula to make \( n \) the subject.
              \begin{solutionordottedlines}
                  \( C = 12n + 250 \) \\
                  \( C - 250 = 12n \) \\
                  \( n = \frac{C - 250}{12} \)
              \end{solutionordottedlines}
              \Part[2] How many people attended the function if the cost of hiring the reception room was:
              \begin{subparts}
                  \subpart \( \$730 \) ?
                  \begin{solutionordottedlines}
                      \( n = \frac{C - 250}{12} \) \\
                      \( n = \frac{730 - 250}{12} \) \\
                      \( n = \frac{480}{12} \) \\
                      \( n = 40 \)
                  \end{solutionordottedlines}
                  \subpart \( \$1090 \) ?
                  \begin{solutionordottedlines}
                      \( n = \frac{C - 250}{12} \) \\
                      \( n = \frac{1090 - 250}{12} \) \\
                      \( n = \frac{840}{12} \) \\
                      \( n = 70 \)
                  \end{solutionordottedlines}
                  \subpart \( \$1210 \) ?
                  \begin{solutionordottedlines}
                      \( n = \frac{C - 250}{12} \) \\
                      \( n = \frac{1210 - 250}{12} \) \\
                      \( n = \frac{960}{12} \) \\
                      \( n = 80 \)
                  \end{solutionordottedlines}
                  \subpart \( \$1690 \) ?
                  \begin{solutionordottedlines}
                      \( n = \frac{C - 250}{12} \) \\
                      \( n = \frac{1690 - 250}{12} \) \\
                      \( n = \frac{1440}{12} \) \\
                      \( n = 120 \)
                  \end{solutionordottedlines}
              \end{subparts}
          \end{parts}
         \Question Given the formula \( t=a+(n-1)d \) :
          \begin{parts}
              \Part[1] rearrange the formula to make \( a \) the subject
              \begin{solutionordottedlines}
                  \( t = a + (n - 1)d \) \\
                  \( a = t - (n - 1)d \)
              \end{solutionordottedlines}
              \Part[2] find the value of \( a \) when:
              \begin{subparts}
                  \subpart \( t=11, n=4 \) and \( d=3 \)
                  \begin{solutionordottedlines}
                      \( a = t - (n - 1)d \) \\
                      \( a = 11 - (4 - 1) \cdot 3 \) \\
                      \( a = 11 - 3 \cdot 3 \) \\
                      \( a = 11 - 9 \) \\
                      \( a = 2 \)
                  \end{solutionordottedlines}
                  \subpart \( t=8, n=5 \) and \( d=-3 \)
                  \begin{solutionordottedlines}
                      \( a = t - (n - 1)d \) \\
                      \( a = 8 - (5 - 1) \cdot (-3) \) \\
                      \( a = 8 - 4 \cdot (-3) \) \\
                      \( a = 8 + 12 \) \\
                      \( a = 20 \)
                  \end{solutionordottedlines}
              \end{subparts}
              \Part[1] rearrange the formula to make \( d \) the subject
              \begin{solutionordottedlines}
                  \( t = a + (n - 1)d \) \\
                  \( d = \frac{t - a}{n - 1} \)
              \end{solutionordottedlines}
              \Part[2] find the value of \( d \) when:
              \begin{subparts}
                  \subpart \( t=48, a=3 \) and \( n=16 \)
                  \begin{solutionordottedlines}
                      \( d = \frac{t - a}{n - 1} \) \\
                      \( d = \frac{48 - 3}{16 - 1} \) \\
                      \( d = \frac{45}{15} \) \\
                      \( d = 3 \)
                  \end{solutionordottedlines}
                  \subpart \( t=120, a=-30 \) and \( n=101 \)
                  \begin{solutionordottedlines}
                      \( d = \frac{t - a}{n - 1} \) \\
                      \( d = \frac{120 - (-30)}{101 - 1} \) \\
                      \( d = \frac{150}{100} \) \\
                      \( d = 1.5 \)
                  \end{solutionordottedlines}
              \end{subparts}
              \Part[1] rearrange the formula to make \( n \) the subject and find the value of \( n \) when
  \( t=150, a=5 \) and \( d=5 \)
              \begin{solutionordottedlines}
                  \( t = a + (n - 1)d \) \\
                  \( n = \frac{t - a}{d} + 1 \) \\
                  \( n = \frac{150 - 5}{5} + 1 \) \\
                  \( n = \frac{145}{5} + 1 \) \\
                  \( n = 29 + 1 \) \\
                  \( n = 30 \)
              \end{solutionordottedlines}
          \end{parts}
          \Question Rearrange each of these formulas to make the pronumeral in brackets the subject.
          \begin{parts}
              \part \( y=mx+c \) \hfill\(( c )\)
              \begin{solutionordottedlines}
                  \( c = y - mx \)
              \end{solutionordottedlines}
              \part \( A=\frac{1}{2}bh \)\hfill\(( x )\)
              \begin{solutionordottedlines}
                  % This part seems to have a typo, as there is no 'x' in the given formula.
                  % Assuming the subject to be made is 'b':
                  \( b = \frac{2A}{h} \)
              \end{solutionordottedlines}
              \part \( P=A+2\ell h \)\hfill\(( \ell )\)
              \begin{solutionordottedlines}
                  \( \ell = \frac{P - A}{2h} \)
              \end{solutionordottedlines}
              \part \( A=2\pi r^2 + 2\pi rh \)\hfill\(( h )\)
              \begin{solutionordottedlines}
                  \( h = \frac{A - 2\pi r^2}{2\pi r} \)
              \end{solutionordottedlines}
              \part \( s=\frac{n}{2}(a+\ell) \)\hfill\(( a )\)
              \begin{solutionordottedlines}
                  \( a = \frac{2s}{n} - \ell \)
              \end{solutionordottedlines}
              \part \( V=\pi r^{2}+\pi rs \)\hfill\(( s )\)
              \begin{solutionordottedlines}
                  \( s = \frac{V}{\pi r} - r \)
              \end{solutionordottedlines}
          \end{parts}
          \Question[3] The formula for the sum \( S \) of the interior angles in a convex \( n \)-sided polygon
  \( S=180(n-2) \). Rearrange the formula to make \( n \) the subject and use this to find the number of sides in
  the polygon if the sum of the interior angles is:
          \begin{parts}\begin{multicols}{3}
              \part \( 1080^\circ \)
              \begin{solutionordottedlines}
                  \( n = \frac{S}{180} + 2 \) \\
                  \( n = \frac{1080}{180} + 2 \) \\
                  \( n = 6 + 2 \) \\
                  \( n = 8 \)
              \end{solutionordottedlines}
              \part \( 1800^\circ \)
              \begin{solutionordottedlines}
                  \( n = \frac{S}{180} + 2 \) \\
                  \( n = \frac{1800}{180} + 2 \) \\
                  \( n = 10 + 2 \) \\
                  \( n = 12 \)
              \end{solutionordottedlines}
              \part \( 3240^\circ \)
              \begin{solutionordottedlines}
                  \( n = \frac{S}{180} + 2 \) \\
                  \( n = \frac{3240}{180} + 2 \) \\
                  \( n = 18 + 2 \) \\
                  \( n = 20 \)
              \end{solutionordottedlines}
          \end{multicols}\end{parts}
      \Question[2] 8 When an object is shot up into the air with a speed of \( u \) metres per second, its
  height above the ground \( h \) metres and time of flight \( t \) seconds are related (ignoring air resistance
  by \( h=ut-4.9t^{2} \).

          Find the speed at which an object was fired if it reached a height of 27.5 metres after 5 seconds.
          \begin{solutionordottedlines}
              \( h = ut - 4.9t^2 \) \\
              \( 27.5 = u(5) - 4.9(5)^2 \) \\
              \( 27.5 = 5u - 4.9(25) \) \\
              \( 27.5 = 5u - 122.5 \) \\
              \( 5u = 27.5 + 122.5 \) \\
              \( 5u = 150 \) \\
              \( u = \frac{150}{5} \) \\
              \( u = 30 \text{ metres per second} \)
          \end{solutionordottedlines}
          \Question[4] Rearrange each of these formulas to make the pronumeral in brackets the subject. (All
  pronumerals represent positive numbers.)
          \begin{parts}\begin{multicols}{2}
              \part \( c=a^{2}+b^{2} \)\hfill\(( a )\)
              \begin{solutionordottedlines}
                  \( a = \sqrt{c - b^2} \)
              \end{solutionordottedlines}
              \part \( x=\sqrt{ab} \)\hfill\(( b )\)
              \begin{solutionordottedlines}
                  \( b = \frac{x^2}{a} \)
              \end{solutionordottedlines}
              \part \( T=\frac{2\pi}{n} \)\hfill\(( n )\)
              \begin{solutionordottedlines}
                  \( n = \frac{2\pi}{T} \)
              \end{solutionordottedlines}
              \part \( E=\frac{m}{2r^2} \)\hfill\(( r )\)
              \begin{solutionordottedlines}
                  \( r = \sqrt{\frac{m}{2E}} \)
              \end{solutionordottedlines}
          \end{multicols}\end{parts}
      \end{questions}
  \end{exercisebox}