[tex]-8\dfrac{3}{4}+\left(-11\dfrac{1}{3}\right)=-\left(8\dfrac{3\cdot3}{4\cdot3}+11\dfrac{1\cdot4}{3\cdot4}\right)\\\\=-\left(8\dfrac{9}{12}+11\dfrac{4}{12}\right)=-\left[(8+11)+\left(\dfrac{9}{12}+\dfrac{4}{12}\right)\right]\\\\=-\left(19+\dfrac{13}{12}\right)=-\left(19+1\dfrac{1}{12}\right)=-20\dfrac{1}{12}[/tex]
