Solving an equation with a second line

A straight-line graph of a law can solve a related equation. Rewrite the equation in the same plotted variables \(X\) and \(Y\), draw that straight line on the same axes, and read where the two lines cross.

Why: a point on the first line satisfies the law; a point on the second line satisfies the new equation. The crossing point satisfies both, so its \(X\)-value gives the solution once it is changed back to \(x\).

Method

1. Rewrite the equation to be solved so that one side is exactly the plotted \(Y\) (here \(\dfrac{y}{x}\), or \(\lg y\) for a logarithmic graph) and the other side is a straight line in the plotted \(X\): divide, multiply or take logarithms exactly as you did for the law.
2. State the second line in terms of the plotted variables, for example \(Y=8-X\).
3. Draw it with a ruler, using two points well apart.
4. Read \(X\) at the intersection and change it back to \(x\). Check the answer in the original equation.
5. If the crossing lies outside the range of the readings, extend both lines with a ruler and say that the answer is an extrapolation, so it is less reliable.

Example

Readings of \(y\) against \(x\) give a straight line \(Y=2X-1\) when \(Y=\dfrac{y}{x}\) and \(X=x\), so \(y=2x^2-x\). Find the value of \(x\) for which \(y=8x-x^2\).

Divide the new equation by \(x\) (allowed because \(x\ne0\); \(x=0\) is excluded): \(\dfrac{y}{x}=8-x\), which is the line \(Y=8-X\). Draw it on the same axes.

The lines meet at \((3,\,5)\), so \(x=3\). Check: \(2(3)^2-3=15\) and \(8(3)-3^2=15\).