Решите...

Г) $\frac{2y+1}{y-4}= \frac{3y-1}{y-3} \\ (2y+1)*(y-3)=(3y-1)*(y-4) \\ 2y^2-6y+y-3=3y^2-12y-y+4 \\ y^2-8y+7=0 \\ D=64-4*7=36 \\y_{1,2}= \frac{8+-6}{2} \\ y_1=7;y_2=1$
д) $\frac{3p+1}{p-1} = \frac{2p+3}{4p-7} \\ (3p+1)*(4p-7)=(p-1)*(2p+3) \\ 12p^2-21p+4p-7=2p^2+3p-2p-3 \\ 12p^2-2p^2-17p-p-7+3=0 \\ 10p^2-18p-4=0 \\ 5p^2-9p-2=0 \\ D=81-4*(-2*5)=81+40=121 \\ p_{1,2}= \frac{9+-11}{10} \\ p_1=2; p_2=-0,2$