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A little guidance will be very much appreciated :)

- Thread starter gikiian
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- #1

- 98

- 0

A little guidance will be very much appreciated :)

- #2

Stephen Tashi

Science Advisor

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If you solve the inequality in a deductive manner, you don't only look at the case 2x -3 > 0. You look at all the possible cases for the sign of the numerator and denominator of the fraction (x+1)/(2x+3). You say that the cases when one is positive and the other is negative can't povide any solutions since the left hand side would be negative, so less than 2. In another case, both the numerator and denominator are positive. In that case x+1 > 0 and 2x+3 > 0, so x> -1 and 2x > -3, x > -3/2.Solving (x+1)/(2x-3) > 2; why do we take 2x-3>0

You have to look at all the cases to work the problem in a systematic manner. Perhaps the materials you are looking at know some clever shortcut.

- #3

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The above is not one of the correct approaches.If you solve the inequality in a deductive manner, you don't only look at the case 2x -3 > 0.

You look at all the possible cases for the sign of the numerator and denominator of

the fraction (x+1)/(2x+3).

You say that the cases when one is positive and the other is negative can't povide

any solutions since the

left hand side would be negative, so less than 2. In another case, both the

numerator and denominator are positive.

In that case x+1 > 0 and 2x+3 > 0, so x> -1 and 2x > -3, x > -3/2.

You have to look at all the cases to work the problem in a systematic manner.

Perhaps the materials you are looking at know some clever shortcut.

Mine:

Subract 2 from each side, and that gives a new fraction that is greater

than 0. Then find the critical numbers to establish the candidate

intervals from which to choose for the solution:

After subtracting 2 and simplifying the fraction:

[itex]\dfrac{-3x + 7}{2x - 3} > 0[/itex]

Critical numbers:

-3x + 7 = 0

3x = 7

x = 7/3

-------------------

2x - 3 = 0

2x = 3

x = 3/2

-------------------

Then the candidate intervals to choose from for the solution are:

(-oo, 3/2), (3/2, 7/3), and (7/3, oo).

Upon checking test values, it is shown that the solution is:

(3/2, 7/3),

or equivalently, as

3/2 < x < 7/3

- #4

Deveno

Science Advisor

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let's look at the case where both the numerator and the denominator are < 0:

x+1 < 0 => x < -1.

2x - 3 < 0 => x < 3/2.

so for both these to be true, we take the more restrictive requirement: x < -1.

since with that assumption, we have 2x - 3 < 0 as well, we have:

x + 1 < 4x - 6 (when we multiply by a negative number, we reverse the inequality)

1 < 3x - 6

7 < 3x

7/3 < x

but this is a contradiction, x can't be BOTH < -1 AND > 7/3.

so the case where both numerator and denominator are negative, does not lead to any valid solutions.

so we may assume, without loss of generality (in THIS particular case) that both numerator and denominator are > 0.

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