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E(x)=[(2x+1)/(3x-2)]/[1-(5x^2-3)/(9x^2-4)].Is it E(x)=(3x+2)/(2x-1)? Find x integer so...

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fardelimits | Student, Grade 9 | eNotes Newbie

Posted June 1, 2009 at 11:38 PM via web

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E(x)=[(2x+1)/(3x-2)]/[1-(5x^2-3)/(9x^2-4)].Is it E(x)=(3x+2)/(2x-1)? Find x integer so that E(x) to be natural.

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted June 2, 2009 at 3:26 AM (Answer #1)

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First of all, when a fraction is divided by the other, in fact we'll multiply one of them with the inverse of the other one. For example:

(a/b)/(c/d)= (a/b)*(d/c)

In our expression E(x), before applying the rule above, we have to have the same denominator in the expression [1-(5x^2-3)/(9x^2-4)].


We've noticed that at the numerator and at the denominator, we have difference of squares, like:



Now we can apply the rule:


We can reduce the similar terms (2x+1) and (3x-2).

E(x)=(3x+2)/(2x-1) q.e.d.

For finding x integer, in order to obtain E(x) natural, (3x+2) has to be divided by (2x-1).

We'll write the ule of division with reminder:

(3x+2)= (2x-1)*Q(x)+R(x)







If we'd have

E(x)= (3x+2)/(2x-1)=3/2+7/2=10/2=5, which is a natural number

For this, 2x-1=1, 2x=2, x=1, which is integer


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mahag | Student, Undergraduate | eNotes Newbie

Posted August 4, 2009 at 9:31 PM (Answer #2)

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please teacher giorgiana1976

how did u obtain 3/2 & 7/2 ?






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