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# If ab = x and a/b = y, and ab does not equal zero, then

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Math Expert
Joined: 02 Sep 2009
Posts: 57279
If ab = x and a/b = y, and ab does not equal zero, then  [#permalink]

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17 Nov 2014, 12:52
00:00

Difficulty:

55% (hard)

Question Stats:

66% (02:27) correct 34% (02:30) wrong based on 163 sessions

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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.

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Re: If ab = x and a/b = y, and ab does not equal zero, then  [#permalink]

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17 Nov 2014, 20:11
1

ab = x and a/b = y, and ab does not equal zero, then a^2 - b^2 =

a=x/b and a=by
x/b=by
x/y=b^2

since a = by
a^2=(b^2)(y^2)

So a^2-b^2 = (b^2)(y^2)-(x/y)
Since b^2=x/y

=(xy^2)/y-(x/y)
=x(y^2-1)/y
=x(y-1)(y+1)/y
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Re: If ab = x and a/b = y, and ab does not equal zero, then  [#permalink]

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17 Nov 2014, 23:02
1
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

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Re: If ab = x and a/b = y, and ab does not equal zero, then  [#permalink]

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17 Nov 2014, 23:13
Bunuel wrote:

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.

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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Re: If ab = x and a/b = y, and ab does not equal zero, then  [#permalink]

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18 Nov 2014, 01:09
4
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)

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

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

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

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

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Re: If ab = x and a/b = y, and ab does not equal zero, then  [#permalink]

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18 Nov 2014, 01:57
1
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

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Ammu
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Re: If ab = x and a/b = y, and ab does not equal zero, then  [#permalink]

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18 Nov 2014, 02:31
1
$$a^{2}$$ - $$b^{2}$$ = $$b^{2}$$[ $$(a/b)^{2}$$ -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

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

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Posts: 57279
Re: If ab = x and a/b = y, and ab does not equal zero, then  [#permalink]

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18 Nov 2014, 08:22
1
Official Solution:

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. $$\frac{(y + x)(y - x)}{x}$$

There are two paths to the solution: using algebra and picking numbers.

Algebra

Many algebraic manipulations are possible. However, we should be guided by the form of the answer choices, which contain only $$x$$'s and $$y$$'s. Thus, we should look for ways to express $$a$$ in terms of only $$x$$ and $$y$$ (not $$b$$); likewise, we want $$b$$ in the same terms, without $$a$$.

Again, we have $$ab = x$$ and $$\frac{a}{b} = y$$.

If we solve the first equation for b and plug into the second equation, here is what we get:
$$b = \frac{x}{a}$$

Solve first eq. for $$b$$
$$\frac{a}{(\frac{x}{a})} = y$$

Substitute $$\frac{x}{a}$$ for $$b$$ in second eq.
$$\frac{a^2}{x} = y$$

Bring the $$a$$ up to the numerator (dividing by $$\frac{x}{a}$$ is the same as multiplying by $$\frac{a}{x}$$)
$$a^2 = xy$$

Multiply both sides by $$x$$

In fact, we could have gotten this result directly if we had multiplied the two original equations together:
$$ab = x$$
$$\frac{a}{b} = y$$

Multiply each side (canceling the $$b$$'s):
$$a^2 = xy$$

Similarly, we can get $$b^2 = \frac{x}{y}$$ either by substitution (solving one equation for $$a$$ and plugging into the other equation) or by dividing the first equation by the second, canceling the $$a$$'s:
$$ab = x$$
$$\frac{a}{b} = y$$

Divide each side:
$$b^2 = \frac{x}{y}$$

(Note that $$ab$$ divided by $$\frac{a}{b}$$ is the same as $$ab$$ times $$\frac{b}{a}$$. The $$a$$'s cancel, leaving $$b^2$$.)

Now that we have $$a^2 = xy$$ and $$b^2 = \frac{x}{y}$$, we can substitute into the expression we are looking for:
$$a^2 - b^2 = xy - \frac{x}{y}$$

Substitute the expressions above.

This does not match any answer choice, so we should manipulate further:

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

Pull out the common factor $$x$$
$$= x(\frac{y^2}{y} - \frac{1}{y})$$

Make a common denominator of $$y$$
$$= \frac{x(y^2 - 1)}{y}$$

Pull out the common divisor $$y$$ (that is, $$y$$ in the denominator)
$$= \frac{x(y - 1)(y + 1)}{y}$$

Factor the special product $$y^2 - 1$$ into $$(y - 1)(y + 1)$$

This expression matches the first answer choice.

Plugging Numbers

Choose suitable numbers for $$a$$ and $$b$$. Here, we should use small numbers and make $$a$$ divisible by $$b$$. For instance, let $$a = 6$$ and $$b = 2$$.

Then, from $$ab = x$$, we have $$x = (6)(2) = 12$$.

Likewise, from $$\frac{a}{b} = y$$, we have $$y = \frac{6}{2} = 3$$.

Finally, we compute our target number:
$$a^2 - b^2 = 62 - 22 = 36 - 4 = 32.$$

Now, compute each answer choice and compare the result to the target of 32. Note that you should stop computing as soon as you see that you are not going to get 32, but you should check all the answer choices.

(A) $$\frac{x(y - 1)(y + 1)}{y} = \frac{(12)(3 - 1)(3 + 1)}{3} = \frac{(12)(2)(4)}{3} = (4)(2)(4) = 32$$ CORRECT

(B)$$xy - \frac{y}{x} = (12)(3) - \frac{3}{12} = 36 - \frac{1}{4} =$$ stop here (result is too big) INCORRECT

(C)$$x^2 - y^2 = 122 - 3^2 = 144 - 9 =$$ stop here (result is too big) INCORRECT

(D)$$y( x^2 - \frac{1}{x^2} ) = 3 ( 122 - \frac{1}{122}) = 3 (144 - \frac{1}{144} ) =$$ stop here (result is too big) INCORRECT

(E)$$\frac{(y + x)(y - x)}{x} = \frac{(3 + 12)(3 - 12)}{12} =$$ stop here (result is negative) INCORRECT

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Re: If ab = x and a/b = y, and ab does not equal zero, then  [#permalink]

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25 Apr 2019, 16:24
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Re: If ab = x and a/b = y, and ab does not equal zero, then   [#permalink] 25 Apr 2019, 16:24
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