The answer is 2.5.

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I got it nice question

Yash Gupta
- 5 years, 8 months ago

answer is 2.5 subtracting 3 from equatiom on both sides and solve the equation by taking -1 in each variables of the equation ,

Abhishek Kumar Verma
- 5 years, 8 months ago

THANK YOU! I AM REALLY CONVINCE BY YOUR SOLUTION.

jonathan dapadap
- 5 years, 8 months ago

Nice! So obvious once you see it!

Dana Rubin
- 5 years, 8 months ago

x=1, y=-4, z=0

The first expression is $\frac {1}{1+1}+\frac {2}{-4+2}+\frac {2015}{0+2015} = \frac {1}{2}-1+1 = \frac {1}{2}$

The second is $\frac {1}{1+1}+\frac {-4}{-4+2}+\frac {0}{0+2015} = \frac {1}{2}+-2+0 = -\frac {3}{2}$

Any solution where $z=\frac {2015(xy+4x+3y+10)}{xy-2x-y-6}$ will work. There aren't just multiple solutions, there are uncountably infinite solutions.

Justin Eberlein
- 5 years, 8 months ago

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There may be infinite number of possible values for x, y, and z satisfying the first equation. But inserting the values of x, y and z (that satisfy the first equation) on the second equation will always yield to 2.5 as proven by Saunak. Your second expression has quite a mistake in basic division. The second fraction should be 2, not -2. Therefore, 0.5 + 2 + 0 = 2.5.

Edrian Laki
- 5 years, 8 months ago

X=5 Z=2015*5 Y=10 also satifies!Nice solution!

Govind Rathi
- 5 years, 8 months ago

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Why do you assume that x must be zero - it is a lucky guess in this case, but the logic of how you got there escapes me - especially given your first statement that x must be infinity.

Tony Flury
- 5 years, 8 months ago

Well, if y=z=0 then x=-5/3, so x/(x+1)=5/2.

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1-(1/(1+x)+1-2/(y+2)+1-z/(z+2015)=3-(1/1+x)+2/(y+2)+z/(z+2015))=3-(1/2)=2.5

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Adding the two fractional expressions gives $\frac{x+1}{x+1}+\frac{y+2}{y+2} +\frac{z+2015}{z+2015} = 3,$ so our unknown value must be $3-\frac{1}{2} = \frac{5}{2} = 2.5.$