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x, y = var('x, y') solve([x+y==6, x-y==4], x, y)
[[x == 5, y == 1]]
t_a,t_b=var('t_a,t_b') a,b,c,d=var('a,b,c,d') eq1=a+b*t_a+c*t_a**2+d*t_a**3==0 eq2=a+b*t_b+c*t_b**2+d*t_b**3==1 eq3=b+2*c*t_a+3*d*t_a**2==0 eq4=b+2*c*t_b+3*d*t_b**2==0 solve([eq1,eq2,eq3,eq4],a,b,c,d)
[[a == (t_a^3 - 3*t_a^2*t_b)/(t_a^3 - 3*t_a^2*t_b + 3*t_a*t_b^2 - t_b^3), b == 6*t_a*t_b/(t_a^3 - 3*t_a^2*t_b + 3*t_a*t_b^2 - t_b^3), c == -3*(t_a + t_b)/(t_a^3 - 3*t_a^2*t_b + 3*t_a*t_b^2 - t_b^3), d == 2/(t_a^3 - 3*t_a^2*t_b + 3*t_a*t_b^2 - t_b^3)]]
show(solve([eq1,eq2,eq3,eq4],a,b,c,d))
\newcommand{\Bold}[1]{\mathbf{#1}}\left[\left[a = \frac{t_{a}^{3} - 3 \, t_{a}^{2} t_{b}}{t_{a}^{3} - 3 \, t_{a}^{2} t_{b} + 3 \, t_{a} t_{b}^{2} - t_{b}^{3}}, b = \frac{6 \, t_{a} t_{b}}{t_{a}^{3} - 3 \, t_{a}^{2} t_{b} + 3 \, t_{a} t_{b}^{2} - t_{b}^{3}}, c = -\frac{3 \, {\left(t_{a} + t_{b}\right)}}{t_{a}^{3} - 3 \, t_{a}^{2} t_{b} + 3 \, t_{a} t_{b}^{2} - t_{b}^{3}}, d = \frac{2}{t_{a}^{3} - 3 \, t_{a}^{2} t_{b} + 3 \, t_{a} t_{b}^{2} - t_{b}^{3}}\right]\right]
solve([eq1,eq2,eq3,eq4,t_a==0,t_b==1],a,b,c,d,t_a,t_b)
[[a == 1, b == 0, c == -3, d == 2, t_a == 0, t_b == 1]]