Make $b$ the subject of $\frac{9}{a}=\frac{9}{b}-\frac{1}{c}$
To make \(b\) the subject of the formula \[\frac{9}{a} = \frac{9}{b} - \frac{1}{c},\] we need to isolate \(b\) on one side of the equation. We assume that \(a, b, c eq 0\) for the expressions to be well-defined. We start by gathering the terms that involve \(b\) on one side.
We add \(\frac{1}{c}\) to both sides of the equation to isolate the term containing \(b\):
\[\frac{9}{b} = \frac{9}{a} + \frac{1}{c}.\]
Next, we simplify the right-hand side by finding a common denominator, which is \(ac\):
\[\frac{9}{b} = \frac{9c}{ac} + \frac{a}{ac} = \frac{9c + a}{ac}.\] Now we have a simple proportional relationship: \(\frac{9}{b} = \frac{9c + a}{ac}\).
To solve for \(b\), we can take the reciprocal of both sides:
\[\frac{b}{9} = \frac{ac}{9c + a}.\] Multiplying both sides by 9 gives the final expression for \(b\):
\[b = \frac{9ac}{9c + a}.\] Thus, \(b\) is isolated as the subject of the formula.
In physics, formulas often describe relationships between variables like resistance, voltage, or distance. For instance, the thin lens equation \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\) helps determine the distance of an image (\(d_i\)) based on the focal length (\(f\)) and the object distance (\(d_o\)). By rearranging this formula to isolate \(d_i = \frac{f \cdot d_o}{d_o - f}\), you can easily calculate where an image will form. Similarly, rearranging the equation provided here allows for quick calculation of a specific variable when the others are known, a fundamental skill in scientific and engineering problem-solving.