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If a - b = (a^2 - b^2)/(b^2 - a^2) and b^2 - a^2 ≠ 0, then b - a =

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If a - b = (a^2 - b^2)/(b^2 - a^2) and b^2 - a^2 ≠ 0, then b - a =  [#permalink]

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New post Updated on: 12 Dec 2018, 00:33
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If \(a - b = \frac{a^2 - b^2}{b^2 - a^2}\) and \(b^2 - a^2 ≠ 0\), then \(b - a =\)

A. -1
B. 0
C. 1
D. 2
E. It cannot be determined from the information given.

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Originally posted by SajjadAhmad on 11 Dec 2018, 10:21.
Last edited by Bunuel on 12 Dec 2018, 00:33, edited 1 time in total.
Renamed the topic and edited the question.
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Re: If a - b = (a^2 - b^2)/(b^2 - a^2) and b^2 - a^2 ≠ 0, then b - a =  [#permalink]

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New post 11 Dec 2018, 21:57
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\(a - b = \frac{(a^2−b^2)}{(b^2−a^2)}\) = \(\frac{(a+b)(a-b)}{(b+a)(b-a)}\)
Multiplying by -1, gives
b - a = \(\frac{(a+b)(b-a)}{(b+a)(b-a)}\) = 1.

C is the answer.
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If a - b = (a^2 - b^2)/(b^2 - a^2) and b^2 - a^2 ≠ 0, then b - a =  [#permalink]

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New post 07 Jan 2019, 19:39
Afc0892 wrote:
\(a - b = \frac{(a^2−b^2)}{(b^2−a^2)}\) = \(\frac{(a+b)(a-b)}{(b+a)(b-a)}\)
Multiplying by -1, gives
b - a = \(\frac{(a+b)(b-a)}{(b+a)(b-a)}\) = 1.

C is the answer.


Hello Afc0892

Could you please explain to me how I multiply by -1 the second term (right term)?

b - a = \(\frac{(a+b)(b-a)}{(b+a)(b-a)}\) = 1.

Kind regards!
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Re: If a - b = (a^2 - b^2)/(b^2 - a^2) and b^2 - a^2 ≠ 0, then b - a =  [#permalink]

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New post 07 Jan 2019, 19:50
jfranciscocuencag wrote:
Afc0892 wrote:
\(a - b = \frac{(a^2−b^2)}{(b^2−a^2)}\) = \(\frac{(a+b)(a-b)}{(b+a)(b-a)}\)
Multiplying by -1, gives
b - a = \(\frac{(a+b)(b-a)}{(b+a)(b-a)}\) = 1.

C is the answer.


Hello Afc0892

Could you please explain to me how I multiply by -1 the second term (right term)?

b - a = \(\frac{(a+b)(b-a)}{(b+a)(b-a)}\) = 1.

Kind regards!


Hey jfranciscocuencag, Sure I'll. :)

The question is asking for us to find the value of b-a. As you can see, we have found the value of a-b. On the RHS, a+b can be cancelled from both the numerator and the denominator. We are left with a-b/b-a. If we multiply both LHS and RHS by -1. Then RHS will be -(a-b)/(b-a)
= -a+b/b-a = b-a/b-a. Also, LHS will be -(a-b) = -a+b or b-a. Hence b-a = 1.

Hope this helps. :)

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If a - b = (a^2 - b^2)/(b^2 - a^2) and b^2 - a^2 ≠ 0, then b - a =  [#permalink]

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New post 07 Jan 2019, 19:55
1
SajjadAhmad wrote:
If \(a - b = \frac{a^2 - b^2}{b^2 - a^2}\) and \(b^2 - a^2 ≠ 0\), then \(b - a =\)

A. -1
B. 0
C. 1
D. 2
E. It cannot be determined from the information given.


So 2 things to remember,
\(a^2 - b^2\)= (a-b) (a+b)
\(b^2 - a^2\)= (b-a) (b+a)

With that in mind just substitute the values
\(a - b = \frac{a^2 - b^2}{b^2 - a^2}\)
\(a - b = \frac{(a-b) (a+b)}{(b-a) (b+a)}\)
b - a = 1

Correct Answer C
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Re: If a - b = (a^2 - b^2)/(b^2 - a^2) and b^2 - a^2 ≠ 0, then b - a =  [#permalink]

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New post 18 Jan 2019, 00:25
1
I feel like the above explanations skip steps.

\(a−b=\frac{(a−b)(a+b)}{(b−a)(b+a)}\)

(a+b)=(b+a), so we cancel each in the numerator and denominator

\(a−b=\frac{(a−b)}{(b−a)}\)

We have (a-b) on the left and in the numerator, so we can divide both sides by (a-b)

\(1=\frac{1}{(b−a)}\)

And multiply both sides by (b-a)

(b-a)=1/1=1
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Re: If a - b = (a^2 - b^2)/(b^2 - a^2) and b^2 - a^2 ≠ 0, then b - a = &nbs [#permalink] 18 Jan 2019, 00:25
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