SOMEONE PLEASE HELP WITH THIS AND SHOW STEPS PLS

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  a) (x < -3) ∪ (-1 < x)

  b) all real numbers

  c) x ≤ 1

  d) no solution

Step-by-step explanation:

An absolute value inequality of the form |f(x)| ? a resolves to two inequalities:

  ±f(x) ? a

where ? is some comparison symbol.

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a) |x +2| > 1

resolves to the inequalities ...

  x + 2 > 1   ⇒   x > -1

and the other inequality this resolves to is ...

  -(x +2) > 1

  -x -2 > 1

  -x > 3

  x < -3

Then the solution is the union of the disjoint solution spaces:

  (x < -3) ∪ (-1 < x)

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b) |2(x -1)| ≥ 0

  2(x -1) ≥ 0   ⇒   x ≥ 1

and the other inequality this resolves to is ...

  -2(x -1) ≥ 0

  x -1 ≤ 0 . . . . . divide by -2

  x ≤ 1

The union of these solutions is ...

  x ∈ all real numbers

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c) 9x -4 ≤ 6 -x

  10x ≤ 10 . . . . . . add x+4 to both sides

  x ≤ 1 . . . . . . . . . divide by 10

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d) No solution. (An absolute value cannot be negative.)

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The attachment shows the solutions plotted against a number line.

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Additional comment

When the inequality is of the form ...

  |f(x)| < a

it can be written as the compound inequality ...

  -a < f(x) < a

If the comparison is > instead of <, using the same "expansion" results in an inconsistency:

  -a > f(x) > a . . . . . . never true for a > 0

If you ignore that technicality (which your teacher may not understand), you can do the solution in the same way you would for the other inequality symbol(s). Just remember the result should be the union of the two disjoint solution spaces.

Using (a) as an example:

  -1 > x+2 > 1 . . . . . to solve this, we subtract 2 everywhere

 -3 > x > -1   ⇒   the solution is (-3>x) ∪ (x>-1)