$x=-\frac{3}{4}y=\frac{1}{3}$  

 

$\text{a)}\frac{2x^2-4y}{15xy}$  

$\frac{2\cdot {\left(-\frac{3}{4}\right)}^2-4\cdot \frac{1}{3}}{15\cdot \left(-\frac{3}{4}\right)\cdot \frac{1}{3}}=\frac{{\sout{2}}^1\cdot \frac{9}{{\sout{16}}^8}-\frac{4}{3}}{15\cdot \left(-\frac{{\sout{3}}^1}{4}\right)\cdot \frac{1}{{\sout{3}}^1}}=\frac{\frac{27}{24}-\frac{32}{24}}{-\frac{15}{4}}=-\frac{5}{24}:\left(-\frac{15}{4}\right)=-\frac{{\sout{5}}^1}{{\sout{24}}^6}\cdot \left(-\frac{{\sout{4}}^1}{{\sout{15}}^3}\right)=\frac{1}{18}$ $\frac{2\cdot {\left(-\frac{3}{4}\right)}^2-4\cdot \frac{1}{3}}{15\cdot \left(-\frac{3}{4}\right)\cdot \frac{1}{3}}=\frac{{\sout{2}}^1\cdot \frac{9}{{\sout{16}}^8}-\frac{4}{3}}{15\cdot \left(-\frac{{\sout{3}}^1}{4}\right)\cdot \frac{1}{{\sout{3}}^1}}=\frac{\frac{27}{24}-\frac{32}{24}}{-\frac{15}{4}}=-\frac{5}{24}:\left(-\frac{15}{4}\right)=-\frac{{\sout{5}}^1}{{\sout{24}}^6}\cdot \left(-\frac{{\sout{4}}^1}{{\sout{15}}^3}\right)=\frac{1}{18}$  

$\underline{}$

$\text{b)}\frac{15xy-2x^2}{3y^2}$

$\frac{15\cdot \left(-\frac{3}{4}\right)\cdot \frac{1}{3}-2\cdot {\left(-\frac{3}{4}\right)}^2}{3\cdot {\left(\frac{1}{3}\right)}^2}=\frac{-\frac{{\sout{45}}^{15}}{4}\cdot \frac{1}{{\sout{3}}^1}-{\sout{2}}^1\cdot \frac{9}{{\sout{16}}^8}}{{\sout{3}}^1\cdot \frac{1}{{\sout{9}}^3}}=\frac{-\frac{15}{4}\cdot \frac{1}{1}-\frac{9}{8}}{\frac{1}{3}}=\frac{-\frac{15}{4}-\frac{9}{8}}{\frac{1}{3}}=\frac{-\frac{30}{8}-\frac{9}{8}}{\frac{1}{3}}=$  

$=\frac{-\frac{39}{8}}{\frac{1}{3}}=-\frac{39}{8}\cdot 3=-\frac{117}{8}$  

$\underline{}$  

 

$\text{c)}\frac{2x+7y-xy}{8x^2}$

$<br>$

$\frac{{\sout{2}}^1\cdot \left(-\frac{3}{{\sout{4}}^2}\right)+7\cdot \frac{1}{3}-\left(-\frac{{\sout{3}}^1}{4}\right)\cdot \frac{1}{{\sout{3}}^1}}{8\cdot {\left(-\frac{3}{4}\right)}^2}=\frac{\left(-\frac{3}{2}\right)+\frac{7}{3}+\frac{1}{4}}{{\sout{8}}^1\cdot \frac{9}{{\sout{16}}^2}}=\frac{-\frac{18}{12}+\frac{28}{12}+\frac{3}{12}}{\frac{9}{2}}=\frac{\frac{13}{12}}{\frac{9}{2}}=\frac{13}{{\sout{12}}^6}\cdot \frac{{\sout{2}}^1}{9}=\frac{13}{54}$