Replacing 1, 2, 2015 with x, y, z
If x + 1 1 + y + 2 2 + z + 2 0 1 5 2 0 1 5 = 2 1 , what is x + 1 x + y + 2 y + z + 2 0 1 5 z ?
The answer is 2.5.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
8 solutions
I got it nice question
answer is 2.5 subtracting 3 from equatiom on both sides and solve the equation by taking -1 in each variables of the equation ,
THANK YOU! I AM REALLY CONVINCE BY YOUR SOLUTION.
Nice! So obvious once you see it!
x=1, y=-4, z=0
The first expression is 1 + 1 1 + − 4 + 2 2 + 0 + 2 0 1 5 2 0 1 5 = 2 1 − 1 + 1 = 2 1
The second is 1 + 1 1 + − 4 + 2 − 4 + 0 + 2 0 1 5 0 = 2 1 + − 2 + 0 = − 2 3
Any solution where z = x y − 2 x − y − 6 2 0 1 5 ( x y + 4 x + 3 y + 1 0 ) will work. There aren't just multiple solutions, there are uncountably infinite solutions.
Log in to reply
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.
X=5 Z=2015*5 Y=10 also satifies!Nice solution!
We can simply satisfy 1/6 + 1/6 + 1/6 = 1/2. 1/(x+1) = 1/6, yields x = 5. 2/(y+2) = 1/6, yields y=10. 2015/(z+2015), yields z = 5(2015). The second sum becomes 5/(5+1) + 10/(10+2) + 5(2015)/(5(2015)+2015) = 5/6 + 5/6 + 5/6 = 15/6 = 2.5.
Simplify the equation x+1/x+1 + y+2/y+2 + z+2015/z+2015 - (1/x+1 + 2/y+2 + 2015/z+2015 ) = 3 - 1/2 = 2.5
I solved it by inspection...z must equal infinity to zero out its fraction...putting x equal to 0 yields 1 and putting y equal to -6 yield -1/2 so it comes out to 1/2. Subbing those values in the second equation yields 2.5
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.
Well, if y=z=0 then x=-5/3, so x/(x+1)=5/2.
Presumably this works for any values of x , y , and z that make the first equation true, so any convenient values can be chosen for x , y , and z , such as x = 1 , y = − 4 , and z = 0 , and the desired expression computed as 2 1 + 2 + 0 .
Putting y=0,z=0 into the first equation gives x=-5/3. Then putting y=0,z=0, x=-5/3 into the second gives the result
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
Adding the two fractional expressions gives x + 1 x + 1 + y + 2 y + 2 + z + 2 0 1 5 z + 2 0 1 5 = 3 , so our unknown value must be 3 − 2 1 = 2 5 = 2 . 5 .