Question

Which compound inequality is equivalent to |ax-b|>c for all real numbers a, b, and c, where c>0

Answer 1

Keywords

compound inequality, absolute value, equivalent

we have

$\left|ax-b\right| > c$

we know that

To find the compound inequality  calculate the two solutions of the absolute value

First solution (case positive)

$(ax-b) > c$ -------> inequality A

Second solution (case negative)

$-(ax-b) > c$ -------> inequality B

Multiply by $-1$ all expression

$(ax-b) < -c$

therefore

$\left|ax-b\right| > c$ is equivalent to

$(ax-b) > c$ and $(ax-b) < -c$

The answer is

The equivalent compound inequality is

$(ax-b) > c$

$(ax-b) < -c$

Answer 2

I think the answer is ax - b <  - c