Question

For all real numbers
x
and
y such that,

x-y = 3
. The following identity holds:

axy +bx +cy +9=0
.What is

a+b+c ?​

Answer 1

Given :  For all real numbers x and y such that x - y = 3, the following identity hold  axy +bx+cy +9 = 0.

To find : Value of a + b + c

Solution:

For all real numbers x and y such that x - y = 3,

the following identity hold

axy +bx+cy +9 = 0

as  above identity holds for all real numbers x & y

so we can take any values of x & y satisfying x - y = 3

Let say  y = 0  => x = 3  as x  - y = 3

putting these values in  axy + bx + cy + 9 = 0

=> a(3)(0) + b(3) + c(0) + 9 = 0

=> 3b = -9

=> b = - 3

now let say  x = 0 => y =  -3   as x  - y = 3

putting these values in axy + bx + cy + 9 = 0

=> a(0)(-3) + b(0) + c(-3) + 9 = 0

=> 3c =  9

=> c = 3

axy  + bx + cy  + 9 =  0

=> axy  -3x  + 3(x - 3) + 9 = 0

=> axy  = 0

=> a = 0    

a = 0  , b = -3  , c = 3

=> a + b + c  = 0 - 3 + 3  =0

Hence a+b + c = 0

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