Odpowiedź:
[tex]\frac{1}{3}(x^2+x)+\frac{2}{3}=2(\frac{2}{3}-2x)+11+\frac{1}{3}x^2\\\\\frac{1}{3}x^2+\frac{1}{3}x+\frac{2}{3}=\frac{4}{3}-4x+11+\frac{1}{3}x^2\\\\\frac{1}{3}x^2+\frac{1}{3}x+4x-\frac{1}{3}x^2=\frac{4}{3}+11-\frac{2}{3}\\\\\frac{1}{3}x+4x=\frac{4}{3}+\frac{33}{3}-\frac{2}{3}\\\\\frac{13}{3}x=\frac{35}{3}\ \ /\cdot3\\\\13x=35\ \ /:13\\\\x=\frac{35}{13}\\\\x=2\frac{9}{13}[/tex]
