Taking it further

Two graphs of modulus functions where the first touches the x-axis at a negative value and the second touches the x-axis at a positive value

Solve the inequality \(|x+1|>2|x-1|\).

Is it possible to find the solutions without solving any equations?

There are several approaches that could be taken to solve \(|x+1|>2|x-1|\). If we don’t want to solve any equations we might think about the geometry of the lines. We could start by labelling the graph with any information we know.

The same graphs as before however the values that the functions touch the x-axis at are given as -1 and 1

At the first intersection, we can create two similar triangles where the corresponding sides are length \(1\) and length \(2\).

At the first intersection of the graphs two similar triangles are drawn and corresponding lengths are labelled with side lengths 1 and 2

How does the ratio of the sides help us to find the \(x\)-coordinate of the intersection point?

For the second intersection we can think about the following triangles.

We know the gradients of the lines, and that the triangles are the same height. What can we deduce about their base lengths?

If we did solve \(|x+1|>2|x-1|\) algebraically we could use the graph to look at the two intersection points and think about which branches of the modulus functions are meeting. This would give us the equations \(x+1 = 2(1-x)\) and \(x+1 = 2(x-1)\) to solve.

Alternatively, we could square both sides to get \((x+1)^2 = 4(x-1)^2\). What does this look like graphically? How do the graphs of \(y = (x+1)^2\) and \(y = 4(x-1)^2\) relate to the graphs of \(y = |x+1|\) and \(y = 2|x-1|\)?