Respuesta :

caylus
Hello,


Let's assume k the ratio.

[tex] \left \{ {{3x=k*(4x+4)} \atop {4x+4=k*(7x-2)}} \right. \\\\ \left \{ {{3x-4kx=4k} \atop {4x-7kx=-2k-4}} \right. \\\\ \left \{ {{x(3-4k)=4k} \atop {x(4-7k)=-2k-4}} \right. \\\\ \left \{ {{x=\dfrac{4k}{3-4k} \atop {x=\dfrac{-2k-4}{4-7k} \right. \\\\ \dfrac{4k}{3-4k}=\dfrac{-2k-4}{4-7k}\\\\ 36k^2-6k-12=0\\ 6k^2-k-2=0\\ \boxed{k= \dfrac{2}{3} \ or\ k= -\dfrac{1}{2}}\\\\ x=\dfrac{4*\dfrac{2}{3} }{3-4*\dfrac{2}{3} }=8\\ x=\dfrac{4*\dfrac{-1}{2} }{3-4*\dfrac{-1}{2} }=-\dfrac{2}{5}\\ [/tex]

[tex] \boxed{x= 8 \ or\ x= -\dfrac{2}{5}}\\\\ [/tex]





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