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Comment on The Something Method
Brilliant!!
Awesome!
This is the best technique so
Wow! Thank You.
The technique seems promising
Hi Sajid,
Hi Sajid,
How did you conclude that (y/x) = 1?
IF it were true that y/x = 1, then it would also be true that x/y = 1.
However, cannot conclude that y/x = 1 (as you suggest)
For example, one solution to the equation is x = 1/2 and y = 2
Notice that these values don't satisfy the equation y/x = 1
Cheers,
Brent
This is awesome. I solved by
Hi brent
In the above question, I have one more technique which is even more faster. that is to blindly assume 1/x=1 and y=1 and substitute in the given eqn. After substitution it seems good. So obviously 1/x=y => xy=1. Though this approach avoids the deep analysis of the numbers in the given eqn, this may not work for all the questions.
Hi Vineet,
Hi Vineet,
Good idea.
The main problem with your approach is that it relies on assuming values that satisfy the given equation.
You randomly chose 1/x = 1 and y = 1.
These values just happen to satisfy the given equation.
If you had chosen different values like 1/x = 2 and y = 1, then those values would NOT satisfy the equation, in which case you wouldn't be able to identify the correct answer.
Cheers,
Brent
Amazing.
I love this method, but I'm
Great question!
Great question!
I'd say that it's useful when there is a some algebraic expression (e.g., 3xy or 5z or 4w²-19, etc) sitting the place where we typically see just one number.
For example, 13.6^(4w²-19) = 1
Here, we have the expression 4w²-19 representing the exponent of 13.6
So, we might look at this as: 13.6^(something) = 1 and go from there.
Does that help?
Cheers,
Brent
Yes, that's helpful, thank