Rearranging formulae

How do you make one particular variable the subject of a more complicated relationship or fraction? This technique is used especially when working with formulae that contain several variables. Combined with other formulae, it can achieve remarkable things – we can derive everything from nothing.

Simple linear expressions

Take this formula: ax + b = c. How do we isolate x in this imaginary formula? We always start by putting all the terms containing the variable we are isolating on the left-hand side and everything else on the right. In theory it doesn’t matter which side is which, but the usual convention is to have the unknown we are solving for on the left. So first we move b to the right-hand side, that is, we add −b to both sides of the equation:

$$\begin{array}{rcl}ax+b& =& c/-b\\ ax& =& c-b\end{array}$$

The terms with the variable are now on one side, but the a is still in the way. How do we get rid of it neatly? What do we have to do to the expression ax to get just x? Yes, we divide the expression – and therefore the whole equation – by a, which is the same as multiplying by $\frac{1}{a}$. Note that a must not be zero (we can’t divide by zero). If we do that, we get:

$$\begin{array}{rcl}ax& =& c-b\cdot \frac{1}{a}(a\ne 0)\\ x& =& \frac{c-b}{a}\end{array}$$

Done. Another example:

$$2ax-3bx=10a$$

Again we want to isolate x. The terms with the unknown are already on the left-hand side, so we have a little less work to do. The complication is that there are two terms with the unknown. We solve this by taking out a common factor. We take x out of the left-hand side of the equation and get:

$$\begin{array}{rcl}2ax-3bx& =& 10a\\ x(2a-3b)& =& 10a\end{array}$$

In the previous example we had the expression ax and wanted just x, so we multiplied the whole equation by $\frac{1}{a}$. Now we have the expression x(2a − 3b) in the equation and again we want to know x. We do the same thing and divide the whole equation by the expression (2a − 3b). Don’t be afraid – we really can divide an equation by a whole bracket, as long as 2a − 3b ≠ 0.

$$\begin{array}{rcl}x(2a-3b)& =& 10a/\cdot \frac{1}{2a-3b}\\ x& =& \frac{10a}{2a-3b}\end{array}$$

Fractions

If an expression contains fractions, we use the same rearrangements of equations, but we must be careful not to divide or multiply by zero. Example:

$$\frac{a}{x}=\frac{b}{c}$$

We want to isolate x. We multiply the whole equation by x – we just have to add the assumption that x ≠ 0. That holds anyway, because the equation contains the fraction $\frac{a}{x}$ with x in the denominator, and we can’t divide by zero. We get the equation:

$$\begin{array}{rcl}\frac{a}{x}& =& \frac{b}{c}/\cdot x(x\ne 0)\\ a& =& \frac{bx}{c}/\cdot c(c\ne 0)\\ ac& =& bx\cdot \frac{1}{b}(b\ne 0)\\ \frac{ac}{b}& =& x\\ x& =& \frac{ac}{b}\end{array}$$

When isolating an unknown, simply use the usual equivalent rearrangements of equations.