1. Basic algebra question

let say if we have
x(a)+x(b)+x(c) and a+b+c=d
we can write it as=x(d)

but what if we have
x(a)+y(b)+z(c) and a+b+c=d
how can we write it then?

2. Re: Basic algebra question

if all we are given is $a + b + c = d$

then there is no simplification of

$a x + b y + c z$

3. Re: Basic algebra question

I think I rewrite my question

(a + b ) + (c + d ) + (e + f ) = S ------------>Equation 1

alpha(a + b ) + beta(c + d ) + gemma(e + f )=??? ..............>equation 2

Is there any way to rewrite this equation semiller to the first one

4. Re: Basic algebra question

Originally Posted by gevni
I think I rewrite my question
(a + b ) + (c + d ) + (e + f ) = S ------------>Equation 1

alpha(a + b ) + beta(c + d ) + gemma(e + f )=??? ..............>equation 2
I think that you are confusing function notation and simple multiplication.
$\alpha(a + b )$ is function notation meaning that alpha is a function with argument $a+b$.
As such $\alpha(a + b )\ne\alpha(a )+\alpha(b )$
If that is not what you mean then what do you mean?

5. Re: Basic algebra question

Originally Posted by gevni
I think I rewrite my question

(a + b ) + (c + d ) + (e + f ) = S ------------>Equation 1

alpha(a + b ) + beta(c + d ) + gemma(e + f )=??? ..............>equation 2

Is there any way to rewrite this equation semiller to the first one
Suppose $\beta = \gamma = 0$. Then you have for equation 2:
$\alpha (a+b)+0(c+d)+0(e+f) = \alpha (a+b)$
That has very little relation left to $S$. You can write:
$(a+b) = S-(c+d)-(e+f) = S-(c+d+e+f)$
which gives:
$\alpha (a+b) + \beta (c+d) + \gamma (e+f) = \alpha (S-c-d-e-f) + \beta (c+d) + \gamma (e+f)$, but this is probably not what you are looking for.

Alternately, you can have:

$\alpha (a+b) + \beta (c+d) + \gamma (e+f) = \alpha S + (\beta - \alpha)(c+d) + (\gamma - \alpha)(e+f)$