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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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SVP
Status: Preparing GMAT
Joined: 02 Nov 2016
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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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Updated on: 12 Dec 2018, 00:33
1
1
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Difficulty:

15% (low)

Question Stats:

80% (00:35) correct 20% (00:50) wrong based on 50 sessions

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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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Last edited by Bunuel on 12 Dec 2018, 00:33, edited 1 time in total.
Renamed the topic and edited the question.
Director
Joined: 18 Jul 2018
Posts: 570
Location: India
Concentration: Finance, Marketing
WE: Engineering (Energy and Utilities)
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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11 Dec 2018, 21:57
2
$$a - b = \frac{(a^2−b^2)}{(b^2−a^2)}$$ = $$\frac{(a+b)(a-b)}{(b+a)(b-a)}$$
Multiplying by -1, gives
b - a = $$\frac{(a+b)(b-a)}{(b+a)(b-a)}$$ = 1.

_________________

If you are not badly hurt, you don't learn. If you don't learn, you don't grow. If you don't grow, you don't live. If you don't live, you don't know your worth. If you don't know your worth, then what's the point?

Intern
Joined: 12 Sep 2017
Posts: 49
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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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.

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!
Director
Joined: 18 Jul 2018
Posts: 570
Location: India
Concentration: Finance, Marketing
WE: Engineering (Energy and Utilities)
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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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.

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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_________________

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Manager
Joined: 09 Mar 2018
Posts: 196
Location: India
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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07 Jan 2019, 19:55
1
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

_________________

If you notice any discrepancy in my reasoning, please let me know. Lets improve together.

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Manager
Joined: 15 Feb 2018
Posts: 172
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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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
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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