Linear Inequations-16

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Solve: [|x + 3| + x]/(x + 2) > 1

Solution: We have

[|x + 3| + x]/(x + 2) > 1

Clearly, LHS of this inequation is meaningful for x ≠ -2

Now, [|x + 3| + x]/(x + 2) > 1

=>[|x + 3| + x]/(x + 2) - 1> 0

=> [|x + 3| + x – x - 2]/(x + 2) > 0

=> [|x + 3| - 2]/(x + 2)>0

Now, two cases arise:

CASE I

When x + 3 ≥ 0 i.e. x ≥ -3

In this case, |x + 3| = x + 3

Therefore, [|x + 3| - 2]/(x + 2)>0

=>[x + 3 - 2]/(x + 2) > 0

=> (x + 1)/(x + 2) > 0

=> x ∈ (-∞, -2) U (-1, ∞)

But, x ≥ -3. Therefore, the solution set of the given inequation in this case is

[-3, -2) U (-1, ∞)----------------------------------(i)

CASE II

When x + 3 < 0, i.e. x < -3;

In this case, |x + 3| = -(x + 3)

Therefore, [|x + 3| - 2]/(x + 2)>0

=>[-(x + 3) – 2]/(x + 2) > 0

=> -(x + 5)/(x + 2) > 0

=> (x + 5)/(x + 2) < 0

x ∈ (-5, -2)

But, x < -3. Therefore, the solution set of the given inequation in this case is the interval (-5, -3)---(ii)

From (i) and (ii), we obtain that the solution set of the given inequation is

[-3, -2) U (-1, ∞) U (-5, -3) = (-5, -2) U (-1, ∞)

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