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For example, in the formula for the area of a rectangle \(A = bh\) (area = base x height), the subject of the formula is \(A\).

In the formula \(A = bh\), the area (\(A\)) is the subject of the formula which means it is the area that is being worked out.

If the area and height of a rectangle were known and the base of the rectangle was required instead, the formula \(A = bh\) would need rearranging to make \(b\) the subject of the formula.

\(A = bh\) means \(A = b \times h\).

To make \(b\) the subject of the formula, \(b\) needs to be isolated. In the formula above, the letter \(b\) is multiplied by \(h\). The inverse of multiplying by \(h\) is dividing by \(h\), so divide both sides by \(h\) to isolate \(b\).

\(\begin{array}{ccc} A & = & bh \\ \div h && \div h \end{array}\)

\(\frac{A}{h} = b\)

The letter \(b\) is now isolated which means \(b\) is now the subject of the formula.

To work out the base of a rectangle the formula is now \(b = \frac{A}{h}\) so divide the area (\(A\)) by the height of the rectangle (\(h\)).

Rearrange the formula \(v = u + at\) to make \(u\) the subject of the formula.

To make \(u\) the subject of the formula means to rearrange the formula so it begins with \(u =\).

Answer this question by finding the letter \(u\) in the formula.

\(v = u + at\)

Isolate this letter by inversing any other items on this side of the equation. Next to the \(u\), there is also a \(+ at\). The inverse of adding \(at\) is subtracting \(at\), so subtract \(at\) from both sides.

\(\begin{array}{ccc} v & = & u + at \\ - at && -at \end{array}\)

\(v - at = u\)

Rearrange the formula \(v = u + at\) to make \(t\) the subject of the formula.

\(\begin{array}{ccc} v & = & u + at \\ -u && -u \end{array}\)

\(\begin{array}{ccc} v - u & = & at \\ \div a && \div a \end{array}\)

\(\frac{v-u}{a} = t\)

The letter \(t\) is now isolated, so \(t\) is now the subject of the formula.