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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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Updated on: 12 Dec 2018, 01:33
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78% (01:03) correct 22% (01:20) wrong based on 72 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, 01:33, 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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18 Jan 2019, 01:25
5
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 =  [#permalink]

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11 Dec 2018, 22:57
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.

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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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07 Jan 2019, 20:39
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!
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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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07 Jan 2019, 20:50
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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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, 20:55
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

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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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29 Jan 2019, 02:08
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.

Given ,

$$a - b = \frac{a^2 - b^2}{b^2 - a^2}$$

$$a - b = \frac{a^2 - b^2}{-1(a^2 - b^2)}$$

a - b = -1

b - a = 1...........Multiplying both side by -1.

Re: If a - b = (a^2 - b^2)/(b^2 - a^2) and b^2 - a^2 ≠ 0, then b - a =   [#permalink] 29 Jan 2019, 02:08
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