Inequality Help!
- Thread starter forsure15
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i got all the way to -4 < 1/2x - 3x/2.
there is prolly some special trick to take it from there.
the basic problem i see is that even if you try to flip the fractions on each side of the equality you are still left with 1 of the x's on the numerator and the other in the denominator. i suppose you could finish solving it if you found the common denominator to combine the terms.
there is prolly some special trick to take it from there.
the basic problem i see is that even if you try to flip the fractions on each side of the equality you are still left with 1 of the x's on the numerator and the other in the denominator. i suppose you could finish solving it if you found the common denominator to combine the terms.
I'm not sure as to how you arrived at that.
You might like to try from the beginning, multiplying both sides of the inequality by the denominator (x^2-x-6). Rearrange into the standard quadratic form (with 3x^2 +.... > 0).
Note that the denominator cannot be equal to zero so \(\displaystyle x\neq3\) and \(\displaystyle x\neq-2\) to continue down that path (these numbers come from factorising the denominator).
Factorise the resulting quadratic and test intervals.
Edit: woops you're different people!
You might like to try from the beginning, multiplying both sides of the inequality by the denominator (x^2-x-6). Rearrange into the standard quadratic form (with 3x^2 +.... > 0).
Note that the denominator cannot be equal to zero so \(\displaystyle x\neq3\) and \(\displaystyle x\neq-2\) to continue down that path (these numbers come from factorising the denominator).
Factorise the resulting quadratic and test intervals.
Edit: woops you're different people!
after getting rid of polynomial in denominator i did not "factorize"... i simply combined similar terms x^2 's , x 's , and constants ... u are left with:
3x^2/2 - 4x - 1/2 < 0
x ( 3x/2 - 4) < 1/2
move 1/2 to other side and factor out the x and then remove it by dividing both sides by x gives u
(3x/2) - (3x/2) - 4 < 1/2x - (3x/2)
and u get
-4 < 1/2x - 3x/2
3x^2/2 - 4x - 1/2 < 0
x ( 3x/2 - 4) < 1/2
move 1/2 to other side and factor out the x and then remove it by dividing both sides by x gives u
(3x/2) - (3x/2) - 4 < 1/2x - (3x/2)
and u get
-4 < 1/2x - 3x/2
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Inequalities always have to be "(this) compared to (zero)". So first you have to move the 3/2 over to the left-hand side.forsure15 said:
. . . . .(3x<sup>2</sup> + 8x + 1)/(x<sup>2</sup> - x - 6) - 3/2 < 0
Convert to a common denominator, and combine:
. . . . .(6x<sup>2</sup> + 16x + 2)/(2x<sup>2</sup> - 2x - 12) - (3x<sup>2</sup> - 3x - 18)/(2x<sup>2</sup> - 2x - 12) < 0
. . . . .[(3x + 4)(x + 5)] / [2(x - 3)(x + 2)] < 0
Find the zeroes of all the factors. Use these to divide the number-line into intervals. Then test the sign of the rational on each interval. The solution will be the intervals for which you get a "minus" sign on the answer.
Eliz.