# solve the inequality x/2x-3 < 5/x?

Relevance

Inequalities work just like equations.

Assuming this is x/(2x-3) < 5/x

Multiply LHS by x and RHS by (2x -3)

x^2 < 5(2x -3)

x^2 < 10x -15

Now transfer sides just as if < was an =

x^2 -10x -15 < 0

Now that doesn't factorize but using quadratic formula if that said x^2 -10x -15 = 0 then either

either x = 11.3246 or x = -1.3246.

This means that x/(2x-3) , 5/x in the range -1.3246 to 11.3246. If you say put -1 and 12 into the inequality you should see you've got the range in the right side.

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• First solve x/(2x-3) = 5/x for x:

x/(2x-3) = 5/x

x/(2x-3) - 5/x = 0

x(x) - 5(2x-3) = 0 (note that 2x-3 and x cannot equal zero, that is x can't be 3/2 or 0)

x^2 - 10x + 15 = 0

(x-5)^2 - 25 + 15 = 0

(x-5)^2 - 10 = 0

(x-5)^2 = 10

|x-5| = sqrt(10)

x = sqrt(10) + 5 or x = 5 - sqrt(10)

To find the range of x-values for which x/(2x-3) < 5/x hold we have to consider the following intervals:

(-infinity, 0), (0, 3/2), (3/2, 5 - sqrt(10)), (5 - sqrt(10), sqrt(10) + 5), and (sqrt(10) + 5, infinity).

First let's choose x=-1 to check whether (-infinity, 0) is part of the range:

-1/(2(-1)-3) <? 5/(-1)

-1/(-5) <? -5

1/5 is not less than 5, so (-infinity, 0) is not part of the range. Next choose x=1 to check whether (0, 3/2) is part of the range:

1/(2-3) <? 5/1

-1 < 5 so (0, 3/2 is part of the range). Next choose x=1.7 to see if (3/2, 5 - sqrt(10)) is part of the range:

1.7 / (3.4-3) <? 5/1.7

4.25 is not less than 5/1.7 so (3/2, 5 - sqrt(10)) is not part of the range. Next choose x=2 to see if (5 - sqrt(10), sqrt(10) + 5) is part of the range:

2/(4-3) <? 5/2

2 is less than 5/2 so (5 - sqrt(10), sqrt(10) + 5) is part of the range. Finally choose x=10 to check whether (5 + sqrt(10), infinity) is part of the range:

10/(20-3) <? 5/10

10/17 is not less than 5/10 so (5 + sqrt(10), infinity is not part of the range. So going back we see that the inequality holds for x-values in the following range:

(0, 3/2) U (5 - sqrt(10), sqrt(10) + 5)

NOTE: Don't forget that you have to pay attention to x-values that lead to division by zero. These lead to discontinuities that can cause changes in the inequality. For example 5/x becomes very a very large negative number as x approaches zero from the left and a very large positive number as x approaches zero from the right.

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• a million. y = 7 - 5x 2. -3y = -6x + a million y = 2x - a million/3 3. x(2y + a million) = 12 2y + a million = 12/x 2y = 12/x - a million y = 6/x - a million/2 4. 4x <= 20 x <= 5 5. -8 < 2x < 6 -4 < x < 3 6. 8x < a million OR x - 9 > -5 x < a million/8 OR x > 4 7. 3x - a million < -7 OR 3x - a million > 7 3x < -6 OR 3x > 8 x < -2 OR x > 8/3 8. x + 3 <= -4 OR x+3 >= 4 x <= -7 OR x >= a million 9. a million - 2x >= -3 AND a million - 2x <= 3 -2x >= -4 AND -2x <= 2 x <= 2 AND x >= -a million -a million <= x <= 2

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