b) The question must be : 7x = 3y.We solve it as shown :

\(7x=3y\Leftrightarrow\dfrac{x}{3}=\dfrac{y}{7}\).Applying the property of sequence of equivalent ratios,we have :

\(\dfrac{x}{3}=\dfrac{y}{7}=\dfrac{x-y}{3-7}=\dfrac{-16}{-4}=4\Rightarrow\left\{{}\begin{matrix}x=12\\y=28\end{matrix}\right.\)

d) Kayasari Ryuunosuke,you should write the roots as shown :

(x ; y) = (6 ; 8) ; (-6 ; 8) ; (6 ; -8) ; (-6 ; -8)

a) Apply the properties of the ratios are equal , we have :

\(\dfrac{x}{7}=\dfrac{y}{13}=\dfrac{x+y}{7+13}=\dfrac{40}{20}=2\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=14\\y=26\end{matrix}\right.\)

Similar with b,c and d , we have :

b) \(\dfrac{2x}{38}=\dfrac{y}{21}=\dfrac{2x-y}{38-21}=\dfrac{34}{17}=2\)

\(\Rightarrow\left\{{}\begin{matrix}x=....\\y=.....\end{matrix}\right.\)

c) It's wrong

d) \(\dfrac{x^2}{9}=\dfrac{y^2}{16}=\dfrac{x^2+y^2}{9+16}=\dfrac{100}{25}=4\)

\(\Rightarrow\left\{{}\begin{matrix}x^2=36\\y^2=64\end{matrix}\right.\)

\(\Rightarrow x=\left[{}\begin{matrix}6\\-6\end{matrix}\right.\) and \(y=\left[{}\begin{matrix}8\\-8\end{matrix}\right.\)