Quick questions on Inequations with absolute values and with the unknown in the denominator

4short Q&A pairs drawn directly from our worked dot-point answer. For full context and worked exam questions, read the parent dot-point page.

What is method 2, stage by stage (the distance method)?

Show answer

The distance method is worth drilling because it is the fastest and the least error-prone. Take

|2x - 3| < 5

. The plan is: make the coefficient of

x

equal to

1

, read the result as a distance, then mark the interval.

What is method 1, graph it?

Show answer

Solve the equation

|ax + b| = k

first, draw

y = |ax + b|

(a V shape) and the horizontal line

y = k

on one set of axes, and read the solution off the picture. Where the V is below the line solves

<

; where it is above solves

>

What is method 2, distance on the number line?

Show answer

Rewrite the inequation so the coefficient of

x

1

, then read it as a distance. The two preparatory steps are: force

a

to be positive (for example write

|{-2x} + 3|

|2x - 3|

, since

|-u| = |u|

), then divide through by the now-positive

a

What is method 3, the algebraic rewrite?

Show answer

Strip the bars by splitting into the two cases the absolute value stands for:

Have a question we have not covered?

This dot-point answer is short enough that we have not extracted many short questions yet. Read the full dot-point answer or ask Mo, our study assistant, in the chat for follow ups.

All Maths Extension 1Q&A pages