Solve | 3 x - 2 | ≥ | x + 4 |

Given: | 3 x - 2 | ≥ | x + 4 | For | 3 x - 2 | ≥ | x + 4 | , look for the values of x for which the graph of y = | 3 x - 2 | is above of the graph of y = | x + 4 | For A: For B: x + 4 = - 3 x - 2 ⇒ x + 4 = - 3 x + 2 ⇒ 4 x = - 2 ⇒ x = - 2 4 ⇒ x = - 1 2 x + 4 = 3 x - 2 ⇒ x + 4 = 3 x - 2 ⇒ 2 x = 6 ⇒ x = 3 Therefore, the value of x ≤ - 1 2 or x ≥ 3

Solve | 3 x - 2 | ≥ | x + 4 |

Given: | 3 x - 2 | ≥ | x + 4 | For | 3 x - 2 | ≥ | x + 4 | , look for the values of x for which the graph of y = | 3 x - 2 | is above of the graph of y = | x + 4 | For A: For B: x + 4 = - 3 x - 2 ⇒ x + 4 = - 3 x + 2 ⇒ 4 x = - 2 ⇒ x = - 2 4 ⇒ x = - 1 2 x + 4 = 3 x - 2 ⇒ x + 4 = 3 x - 2 ⇒ 2 x = 6 ⇒ x = 3 Therefore, the value of x ≤ - 1 2 or x ≥ 3

Prove that | x + y | ≤ | x | + | y | .

An absolute value inequality is an expression with absolute functions as well as inequality signs. So, here we need to prove that | x + y | ≤ | x | + | y | . Consider an example with x = 3 and y = - 2 . We know that | x | = 3 and | y | = 2 Here, x + y = 3 + - 2 = 3 - 2 = 1 and | x + y | = 1 The value of | x | + | y | = 3 + 2 = 5 According to the condition 1 < 5 Thus, we can conclude that | x + y | ≤ | x | + | y |