$\{\begin{array}{l}6u\left(x\right)+10v\left(x\right)=x^2+1\\ 2u\left(x\right)+5v\left(x\right)=x^3-x\mid \cdot \left(-2\right)\end{array}$  

$\{\begin{array}{l}6u\left(x\right)+10v\left(x\right)=x^2+1\\ -4u\left(x\right)-10v\left(x\right)=-2x^3+2x\end{array}$  

Dodajemy równania stronami.

$2u\left(x\right)=-2x^3+x^2+2x+1\mid \cdot \frac{1}{2}$  

$u\left(x\right)=-x^3+\frac{1}{2}{x}^2+x+\frac{1}{2}$  

Podstawiamy wyznaczoną wartość u(x) do dowolnego równania w układzie i wyznaczamy v(x):

$\{\begin{array}{l}u\left(x\right)=-x^3+\frac{1}{2}{x}^2+x+\frac{1}{2}\\ 2u\left(x\right)+5v\left(x\right)=x^3-x\end{array}$  

$\{\begin{array}{l}u\left(x\right)=-x^3+\frac{1}{2}{x}^2+x+\frac{1}{2}\\ -2x^3+x^2+2x+1+5v\left(x\right)=x^3-x\end{array}$  

$\{\begin{array}{l}u\left(x\right)=-x^3+\frac{1}{2}{x}^2+x+\frac{1}{2}\\ 5v\left(x\right)=3x^3-x^2-3x-1\mid :5\end{array}$  

$\{\begin{array}{l}u\left(x\right)=-x^3+\frac{1}{2}{x}^2+x+\frac{1}{2}\\ v\left(x\right)=\frac{3}{5}{x}^3-\frac{1}{5}{x}^2-\frac{3}{5}x-\frac{1}{5}\end{array}$