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given that a=by &ab=x
so dividing 1 with 2 results in 1/b=by/x
b^2=x/y

so a^2-b^2=(by)^2-(x/a)^2
=((aby)^2-x^2)/a^2
=((xy)^2-x^2)/a^2 ;from equation 2
=(x^2)(y^2-1)/a^2
=(ab^2)(y-1)(y+1)/a^2 ;from equation 2
=b^2(y-1)(y+1)
=x(y-1)(y+1)/y ; since b^2=x/y

Answer is A
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Bunuel

Tough and Tricky questions: Algebra.



If \(ab = x\) and \(\frac{a}{b} = y\), and \(ab\) does not equal zero, then \(a^2 - b^2 =\)

A. \(\frac{x(y - 1)(y + 1)}{y}\)
B. \(xy - \frac{y}{x}\)
C. \(x^2 - y^2\)
D. \(y(x^2 - \frac{1}{x^2})\)
E. \(\frac{(y + x)(y - x)}{x}\)

Kudos for a correct solution.

Answer = A.
ab=x a/b=y.
Multiply the two and a^2 = xy.
Divide the two and b^2 = x/y.
a^2-b^2 = xy-x/y = x*(y^2-1)*(1/y) = x*(y+1)*(y-1)/y = Choice A.
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ab = x ............. (1)

\(\frac{a}{b} = y\) ............. (2)

(1) * (2)

\(a^2 = xy\) .............. (3)

a = by ............ From (2)

by*b = x ........... from (1)

\(b^2 = \frac{x}{y}\) ................... (4)

(3) - (4)

\(a^2 - b^2 = xy - \frac{x}{y}\)

\(= x(y - \frac{1}{y})\)

\(= x(\frac{y^2 - 1}{y})\)

\(= \frac{x(y+1)(y-1)}{y}\)

Answer = A
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Question is a^2-b^2 = ?

Solution:

As per question stem,
ab= x -- equation (1)
a= by --equation (2)

Let us substitute eq (2) int0 eq(1), we will get

by*b= x

(b^2)*y=x

hence, b^2 = x/y ---eq (3)

Now solve equation 3 & 2.
we will get a^2 = xy -- eq 4


using eq 3 & 4, we can solve for a^2 - b^2

a^2 - b^2 = xy- x/y = (xy^2-x)/y = x(y^2-1)/y = x(y-1)(y+1)/y

Answer is A

Regards,
Ammu
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\(a^{2}\) - \(b^{2}\) = \(b^{2}\)[ \((a/b)^{2}\) -1] ....1)

As given --> \(\frac{a}{b}\) = y, ab = x

a = by => by*b = x =>\(b^{2}\) = x/y

Let's put these values in equation 1

\(\frac{x[(y^{2} - 1)]}{y}\) => \(\frac{x(y - 1)(y + 1)}{y}\)

Answer A
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