Помогите решить 4 и 5 задание (34б)

4. а) $\frac{y+6}{4y+8} - \frac{y+2}{4y-8} + \frac{5}{y^{2} - 4} = \frac{y+6}{4(y+2)} - \frac{y+2}{4(y-2)} + \frac{5}{(y+2)(y-2)} =$ $\frac{(y+6)(y-2)-(y+2)^{2} + 5*4}{4(y+2)(y-2)} = \frac{y^{2} +4y - 12 - y^{2} -4y - 4 +20 }{4(y-2)(y+2)} = \frac{4}{4(y-2)(y+2)} = \frac{1}{y^{2} -4 }$
б) $\frac{6b^{3} +48b}{b^{3} +64} - \frac{3b^{2} }{b^{2} -4b+16} = \frac{6b(b^{2}+8) }{(b+4)(b^{2}-4b+16) } - \frac{3b^{2} }{b^{2} -4b+16} = \frac{6b^{3} +48b-3b^{2}(b+4) }{(b+4)(b^{2}-4b+16) } =$ $\frac{6b^{3} +48b - 3b^{3} -12b^{2} }{(b+4)(b^{2}-4b+16) } = \frac{3b^{3} -12b^{2}+48b }{(b+4)(b^{2} -4b+16)} = \frac{3b(b^{2}-4b+16) }{(b+4)(b^{2} -4b+16)} = \frac{3b}{b+4}$
5. $\frac{x+4y}{y} =10$
x+4y=10y
x=6y
1) $\frac{x}{y} = \frac{6y}{y} = 6$
2) $\frac{7x-y}{x} = \frac{41y}{6y} = \frac{41}{6} = 6 \frac{5}{6}$