Absolute Value Inequalities Part 2

Set 1.2
Questions
52 & 60
David Cohen

Sample: Solve | x | > 2

Apply this rule:

If | a | > b, where b > 0, then
a < - b or a > b.

x < - 2 or x > 2

52. Solve | x | > 0.

See attachment. Explain attachment.

60. Solve | x + 5 | ≥ 2

Apply this rule:

If | a | ≥ b, where b > 0, then
a ≤ - b or a ≥ b.

x + 5 ≤ - 2 or x + 5 ≥ 2

x ≤ - 2 - 5 or x ≥ 2 - 5

x ≤ - 7 or x ≥ -3

You say?

 

52. | x | > 0

=> x > 0 or -x > 0
make x positive: -x*-1 > 0*-1 (sign will change to <), so x < 0

solution:
=> x > 0 or x < 0
so, the value of x=0 is excluded because it would give you |0|>0 or 0>0 which is false

60. correct

 

Problem 52 looks so innocently simple but it's tricky in its own way. Tell me, why did you multiply both sides of -x > 0 by -1? I know the reason is to make x positive but why is this step needed in the solution process?

 

when you deal with variable, you always have positive sign in front of it

so, -x > 0 variable has negative sign and you have to say negative x is greater than zero
or much easier is have
x < 0 and say x is less than zero

 

Thanks. Can you check Rewrite Without Absolute Value Notation Part 1?