Manipulation of Formulae

In many cases, algebraic terms in a formula may need to be moved around and rearranged meaning we have to manipulate the entire formula.

Consider the following formula that derives the area of a trapezoid :

A=\frac{(a+b)h}{2}

Where $a$ is the length of the top of the trapezoid, $h$ is the vertical height and $b$ is the base.

We say $A$ is the subject of the formula. In a way, the formula is made in mind of finding the area. Another way of phrasing it is the $A$ is in terms of $a,b$ and $h$ . That is, $A$ is on one side, and everything else is on the other.

But what if want to find $h$ given $A,a,b$ . This would mean that we would have to put $h$ in terms of $A,a,b$ , which make $h$ the subject of the formula. We can manipulate.

Example

infoNote

Express $h$ in terms of $A,a,b$ in the following equation :

A=\frac{(a+b)h}{2}

Manipulating the formula simply means rearranging the variable terms such that we isolate $h$ . All basic algebraic techniques we have seen so far apply.

\begin{align*} A&=\frac{(a+b)h}{2} & \\\\ 2A&=(a+b)h & \text{\footnotesize\textcolor{gray}{(\(\cdot2\))}} \\\\ \frac{2A}{(a+b)}&=h & \text{\footnotesize\textcolor{gray}{(\( \div (a+b) \))}} \\ \end{align*}

$h$ is now the subject of the formula.

Example

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Express $k$ in terms of $h$ and $j$ in the following expression :

\frac{1}{h}=\frac{k}{j+k}

\begin{align*} \frac{1}{h}&=\frac{k}{j+k} & \\\\ 1(j+k)&=h(k) & \text{\footnotesize\textcolor{gray}{(cross multiply)}} \\\\ j+k&=hk & \\\\ j&=hk-k & \text{\footnotesize\textcolor{gray}{(\( -k\))}} \\\\ j&=k(h-1) & \text{\footnotesize\textcolor{gray}{(factor out \(k \))}} \\\\ \frac{j}{h-1}&=k & \text{\footnotesize\textcolor{gray}{(\( \div (h-1) \))}} \end{align*}

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Manipulation of Formulae

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