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# Solve for y if |2y + 3| + 1 5?

Solve for y if

|2y + 3| + 1 5

A) (-∞, -1/2] U [7/2, ∞)

B) (-∞, -7/2] U [1/2, ∞)

C) (-∞, -2/3] U [2, ∞)

D) (-∞, -2] U [2/3, ∞)

E) (-∞, -4/3] U [8/3, ∞)

F) (-∞, -8/3] U [4/3, ∞)

G) (-∞, -3/2] U [9/2, ∞)

H) (-∞, -9/2] U [3/2, ∞)

I) (-∞, ∞)

J) None of these.

### 1 Answer

Relevance

- 1 decade agoFavorite Answer
|2y + 3| + 1 ≥ 5

<=> |2y + 3| ≥ 4

<=> ( 2y + 3 )² ≥ 4²

<=> 4y² + 12y + 9 ≥ 16

<=> 4y² + 12y - 7 ≥ 0

If f(x) = 4y² + 12y - 7 = 0

=> y = -7/2 and y = 1/2

=> B) (-∞, -7/2] U [1/2, ∞)

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