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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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)].

(9x^2-4-5x^2+3)/(9x^2-4)=(4x^2-1)/(9x^2-4)

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

a^2-b^2=(a-b)(a+b)

(4x^2-1)/(9x^2-4)=(2x-1)(2x+1)/(3x-2)(3x+2)

Now we can apply the rule:

[(2x+1)/(3x-2)]*[(3x-2)(3x+2)/(2x-1)(2x+1)]

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)

Q(x)=quotient

R(x)=reminder

(3x+2)/(2x-1)=Q(x)+[R(x)/(2x-1)]

Q(x)=3/2

R(x)=7/2

(3x+2)/(2x-1)=3/2+7/2/(2x-1)

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**

please teacher giorgiana1976

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

regards

MAHA

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