[tex]\it \dfrac{5a+4b}{5a+7b}=\dfrac{1}{2} \Rightarrow 10a+8b=5a+7b \Rightarrow8b-7b=5a-10a \Rightarrow\\ \\ \\ \Rightarrow b=-5a\ \ \ \ (*)\\ \\ \\ \dfrac{b^2-a^2}{2a^2-2ab}= \dfrac{(b-a)(b+a)}{-2a(b-a)}=\dfrac{b+a}{-2a}\ \stackrel{(*)}{=}\ \dfrac{-5a+a}{-2a}=\dfrac{-4a}{-2a}=2[/tex]
