Question

ex 8 c si d (nu sunt sigura daca le am facut corect)

Answer 1

$\it c)\ \dfrac{x}{2}= \dfrac{y}{5} = k\Rightarrow \begin{cases}\it x=2k\\ \\ \it y=5k\end{cases}\ \ \ \ \ (*)\\ \\ \\ (*) \Rightarrow \dfrac{7x-2y}{x+4y}= \dfrac{7\cdot2k-2\cdot5k}{2k+4\cdot5k} = \dfrac{14k-10k}{2k+20k}= \dfrac{\ 4k^{(k}}{22k}= \dfrac{\ 4^{(2}}{22}= \dfrac{2}{11}$

$\it d)\ \dfrac{x}{3}= \dfrac{y}{11} = k\Rightarrow \begin{cases}\it x=3k\\ \\ \it y=11k\end{cases}\ \ \ \ \ (*)\\ \\ \\ (*) \Rightarrow \dfrac{10x-y}{x+y}= \dfrac{10\cdot3k-11k}{3k+11k} = \dfrac{30k-11k}{3k+11k}= \dfrac{\ 19k^{(k}}{14k}= \dfrac{19}{14}$

Answer 2

$\dfrac{x}{3}=\dfrac{y}{11}=>x = 3y \div11$

înlocuim pe x în celalalt raport

$\dfrac{10 \cdot\dfrac{3y}{11} - y}{ \dfrac{3y}{11} + y} = \dfrac{\dfrac{30y - 11y}{11}}{ \dfrac{3y +11y}{11}} =$

$\dfrac{19y}{11} \div \dfrac{14y}{11} = \dfrac{19y}{11} \cdot\dfrac{11}{19y} =$

$\dfrac{19 \not y}{ \not 11} \cdot\dfrac{14 \not y}{ \not 11} = \boxed{\dfrac{19}{14}}$