Submitted by Soham Zemse

because an algebraic identity so have solution 0. ax + by + c -1 = 0, so ax = 0 ---> a = 0 ; by = 0 ---> b = 0 ; and c - 1 = 0 ---> c = 1. Thus, a + b + c = 0 + 0 + 1 = 1. Answer : 1

An algebraic identity is one where the equation satisfies (holds true) for any possible values of variables present in them.

For example, $(x + y)^{2} = x^{2} + 2xy + y^{2}$ is an algebraic identity.

Similarly, there are several trigonometric identities too. Like $sin^{2}(\theta) + cos^{2}(\theta) = 1$ .

Coming to our problem,

the equation $ax + by + c - 1 = 0$ holds true for any possible values of $x$ and $y$ .

Since we need $a + b + c$ , it can be obtained when we put $x = 1$ and $y = 1$ .

So, $a + b + c - 1 = 0$

$a + b + c = 1$

$\boxed{1}$ is the answer!

Misal x=1, y=1. Maka a(1)+b(1)+c-1=0. Kemudian a+b+c=1

ax + by + c - 1= 0 = 0 - 1 1 - 0= 1, since x and y can be read as 1. and the negative side transfer to the other side change the negative to positive.

It is given that it is an identity. This means it is true for all values of x and y. Put x and y as 1 in the equation. You'll get one.

Obviously $a=b=0$ otherwise it couldn't have been an identity. We are left with this equation: $c-1=0$ $\implies\qquad c=1\qquad\qquad$ $\text{W}^5$ (Which was what was wanted)

ax+by+c-1=0 hear x=1 y=1 a+b+c-1=0 a+b+c=1

Put x = y = 1 and on tramsposing 1 on R.H.S we get a+b+c=1

Substituting in $x=1$ and $y=1$ we find that $a+b+c=1$ .

as ax+by+c-1 = 0 , so we can say that a and b are zeros as x and y are non-zeros as said above. and as the answer is zero ,so we can say that c is 1. so the sum of them would be 0+0+1= ..... (you answer ;) )

if a+b+c gives 1,then only the equn will get to be equal to 0. (a+b+c)-1=0 => 1-1=0