A few minutes after I posted my question, I found that Poly.coeffs does
what I want and works for multivariate polynomials
```
lin = Poly(expr, x, y).coeffs()
solve(lin, [a,b,c,d,e])
```
with the following definitions
```
x, y = symbols("x, y")
a, b, c, d, e = symbols("a, b, c, d, e")
f = 3*x**2-2*y+1
g = x*y+x-2*y
h = x**2+2*y**2-2
j = x+3
k = x**2-x*y+2
expr = a*f + b*g + c*h + d*j + e*k - (7*x**2 + 2*x*y + 4*x - 4*y**2 - 14*y
+ 9)
```
On Monday, January 16, 2017 at 5:42:15 AM UTC+1, Aaron Meurer wrote:
>
> Perhaps solve_undetermined_coeffs() can do what you want.
>
> Aaron Meurer
>
> On Sat, Jan 14, 2017 at 5:46 AM, Michele Zaffalon
> <[email protected] <javascript:>> wrote:
> > I am struggling to find a way of solving the following equation
> (simplified
> > example)
> > ```
> > f = x+1
> > g = 2*x+1
> > expr = a*f + b*g - 1
> > solve(expr, (a, b))
> > ```
> > I would like to have `a=2`, `b=-1` which make `expr` identically zero.
> >
> > I am interested in polynomials with higher powers: at the moment, I
> derive
> > with respect to x, subs 0 for x, accumulate the result as column in a
> matrix
> > and solve the linear problem associated. This method works for
> univariate
> > polynomials, but it become impractical for multivariate linear systems,
> > which is my ultimate goal, because of the need to derive with respect to
> all
> > mixed terms x^m*y^n. (I also could not find a way of collecting the
> > coefficients of the terms x^m*y^n.)
> >
> > Is there a way to solve these linear problems?
> >
> > Thank you,
> > michele
> >
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