Question

How do you solve `| x - 3 | + 2 >7`?

Answer 1

`x in (-oo, -2) uu (8, + oo)`

Explanation:

The first thing to do here is isolate the modulus on one side of the inequality by subtracting `2` from both sides

`|x-3| + color(red)(cancel(color(black)(2))) - color(red)(cancel(color(black)(2))) > 7 - 2`

`|x-3| > 5`

Now, because you're dealing with an absolute value inequality, you must account for two possible cases

`color(white)(a)`

• `color(purple)(ul(color(black)(x - 3 >= 0 implies |x- 3| = x-3))`

When this is the case, the inequality becomes

`x - 3 > 5`

`x > 8`

`color(white)(a)`

• `color(purple)(ul(color(black)(x - 3 <0 implies | x -3| = -(x-3)))`

This time, you will have

`-(x-3) > 5`

`-x + 3 >5`

`-x > 2 implies x < -2`

So, you need to have `x < -2` and `x > 8`, which means that the solution interval for this inequality will be

`color(green)(|bar(ul(color(white)(a/a)color(black)(x in (-oo, -2) uu (8, + oo))color(white)(a/a)|)))`