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Answer :

βˆ’5(|xβˆ’3|)βˆ’5<15

Step 1: Add 5 to both sides.

βˆ’5(|xβˆ’3|)βˆ’5+5<15+5

βˆ’5(|xβˆ’3|)<20

Step 2: Divide both sides by -5.

βˆ’5(|xβˆ’3|)

βˆ’5

<

20

βˆ’5

|xβˆ’3|>βˆ’4

Step 3: Solve Absolute Value.

|xβˆ’3|>βˆ’4

We know eitherxβˆ’3>βˆ’4orxβˆ’3<βˆ’(βˆ’4)

xβˆ’3>βˆ’4(Possibility 1)

xβˆ’3+3>βˆ’4+3(Add 3 to both sides)

x>βˆ’1

xβˆ’3<βˆ’(βˆ’4)(Possibility 2)

xβˆ’3<4(Simplify both sides of the inequality)

xβˆ’3+3<4+3(Add 3 to both sides)

x<7

Answer:

x>βˆ’1 or x<7