Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 decade ago

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

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  • 1 decade ago
    Favorite 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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