What A Mess
{ x + y + z = 2 6 x 1 + y 1 + z 1 = 3 1
The two equations above satisfy the real numbers x , y , z . What is the value of the expression below?
y x + z y + x z + z x + y z + x y
The answer is 803.
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4 solutions
Discussions for this problem are now closed
wow! I did the same.
I too did in the same way
Sometimes a simple idea could solve problems that look difficult:)
For example this as easy as
Really simple!
what if we have to find the value of (x+y)xy + (y+z)yz + (x+z)xz @paolaRamieze
x + y = 2 6 − z
x + z = 2 6 − y
z + y = 2 6 − x
You'll get into the same answer...
i am unable to get its answer. please pst how
I did it ur way as well, u just need to break it up into three terms and simplify. U have to do this with the denominator.
Interesting, I'll solve it
y/x + z/x + x/y + z/y + x/z + y/z ( y + z ) / x + ( x + z ) / y + ( x + y ) / z ( 26 - x ) / x + ( 26 - y ) / y + ( 26 - z ) / z 26/x - x/x + 26/y - y / y + 26 /z - z/z 26/x + 26/y + 26/z - 1 -1 -1 26( 1/x + 1/y + 1/z ) - 3 26 x 31 - 3 806 - 3 803
x+y+z=26
1/x + 1/y+ 1/z = 31
x/y+x/z + y/z+y/x + z/x+z/y = k
x( 1/y + 1/z) + y(1/x + 1/z) +z(1/x + 1/y) =k
1/y+ 1/z = 31-1/x 1/x+ 1/z = 31-1/y 1/y+ 1/z = 31-1/z
x( 31-1/x) + y(31-1/y) +z(31-1/z) =k 31x + 31y + 31z -3=k 31(x+y +z) -3=k 31(26) -3=k 803=k
OR
( x + y + z ) ( x 1 + y 1 + z 1 ) = 2 6 × 3 1
y x + z y + x z + z x + y z + x y + 3 = 8 0 6
y x + z y + x z + z x + y z + x y = 8 0 3