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Comment on Strange Operators
If xθy=(x+1)/(y−1) for all y
A. b−2
B. b−1
C. b
D. b+1
E. b+2
I did (a+1)/(b-1) = 1
I don't understand what should I do to take out a
Thanks in advance
Let's take it from (a+1)/(b-1
Let's take it from (a+1)/(b-1) = 1
Multiply both sides of the equation by (b-1) to get: (a+1) = 1(b-1)
Expand right side: a + 1 = b - 1
Subtract 1 from both sides: a = b - 2
Answer: A
Hi Brent, for the below
(√5 + √5)² - (√5 - √5)² = using the property a² - b² = (a+b)(a-b) I got, (√5 + √5) (√5 - √5) = (2√5)(0) = 0
If x ¤ y = (x + y)² - (x - y)². Then √5 ¤ √5 =
A. 0
B. 5
C. 10
D. 15
E. 20
Question link: https:/
Question link: https://gmatclub.com/forum/if-x-y-x-y-x-y-then-218476.html
You made a small mistake in your solution.
You're using the property that says a² - b² = (a + b)(a - b)
So, in the expression, (√5 + √5)² - (√5 - √5)², what is a, and what is b?
Well, a = (√5 + √5) and b = (√5 - √5)
This means (a + b)(a - b) = [(√5 + √5) + (√5 - √5)][(√5 + √5) - (√5 - √5)]
= [(√5 + √5) + 0][(√5 + √5) - 0]
= (√5 + √5)(√5 + √5)
= (2√5)(2√5)
= 4√25
= 20
I have two different solutions posted here: https://gmatclub.com/forum/if-x-y-x-y-x-y-then-218476.html#p1727422
Does that help?
Cheers,
Brent
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