# examplebox


\section*{Exercise 6A}

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t_letter
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\begin{sol}

6 Given  \(\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\), find the value of \(f\) when:
a \(u=2\) and \(v=4\)
b \(u=6\) and \(v=9\)
c \(u=12\) and \(v=11\)
d \(u=2\) and \(v=15\)





\section*{\(6 B\) 
 Changing the subject of a formula}
Sometimes the pronumeral whose value is to be determined is not the subject of the formula. In this situation you have a choice. You can either:

\begin{itemize}
  \item rearrange the formula to make the unknown pronumeral the subject and then substitute values for the known pronumerals (method 1), or
  \item substitute the values for the known pronumerals and then solve the resulting equation for the unknown pronumeral (method 2). This was the approach taken in Section 6A.
\end{itemize}

In either case, an equation has to be solved. In method 2, the equation involves numbers. In method 1, the equation involves pronumerals.

It is preferable to use method 1 when you are asked to find several values of a pronumeral that is not the subject of the formula.

\section*{Example 7}
The cost \(\$ C\) of hiring Scott's car is given by the formula \(C=\frac{1}{4} x+40\), where \(x\) is the number of kilometres driven. Find the number of kilometres driven by a person who is charged \(\$ 130\) for hiring the car.

\section*{Solution}
\section*{Method 1}
Rearrange the formula to make \(x\) the subject and then substitute the given value of \(C\), as follows.

\[
\begin{array}{rlrl}
C & =\frac{1}{4} x+40 & \\
C-40 & =\frac{1}{4} x & & \text { (Subtract } 40 \text { from both sides of the formula.) } \\
x & =4 C-160 & & \text { (Multiply both sides of the formula by 4.) }
\end{array}
\]

When \(C=130, x=4 \times 130-160\)

\[
=360
\]

Thus the person drove \(360 \mathrm{~km}\).

\section*{Method 2}
Substitute the numbers and then solve the resulting equation, giving

\[
C=\frac{1}{4} x+40
\]

When \(C=130\),

\[
\begin{array}{rlrl}
130 & =\frac{1}{4} x+40 & \\
90 & =\frac{1}{4} x & & \text { (Subtract } 40 \text { from both sides of the equation.) } \\
x & =360 & & \text { (Multiply both sides of the equation by } 4 .)
\end{array}
\]

Thus the person drove \(360 \mathrm{~km}\).

\section*{Example 8}

\section*{Solution}

\section*{Example 9}

\section*{Solution}

\section*{Example 10}

\section*{Solution}

b



d


\section*{Example 11}

\section*{Solution}

\section*{Changing the subject of a formula}
When rearranging a formula, the basic strategy is to move all terms involving the new subject to one side, and all the other terms to the other side. To do this:

\begin{itemize}
  \item fractions can be eliminated by multiplying both sides of the formula by a common denominator
  \item all like terms should be collected
  \item the same operation(s) must be performed on both sides of the formula.
\end{itemize}

\section*{Exercise 6B}



5 Rearrange each of these formulas to make the pronumeral in brackets the subject.


c 57.6 joules, \(2.4 \mathrm{~m} / \mathrm{s}\)






\section*{Constructing formulas}
In this section we shall learn how to create formulas from given information.

\section*{Example 12}

\section*{Example 13}

\section*{Exercise 6C}