Question

Resolva o sistema:
Log x - Log y = Log y
3x + 2y = 33

Answer 1

$\begin{cases}\mathrm{\log{x}-\log{y}=\log{y}\ \mathbf{(I)}}\\ \mathrm{3x+2y=33\ \mathbf{(II)}}\\ \end{cases}$

$\mathbf{*\ De\ I,\ vem:}\\\\ \mathrm{\log{x}=2\log{y}\ \to\ \log{x}=\log{y^2}\ \to\ \boxed{\mathrm{x=y^2}}}\\\\ \mathbf{*\ De\ II,\ vem:}\\\\ \mathrm{3y^2+2y=33\ \to\ 3y^2+2y-33=0}\\\\ \mathrm{y=\dfrac{-2\pm\sqrt{2^2-4.3.(-33)}}{2.3}=\dfrac{-2\pm\sqrt{4+396}}{6}=}\\\\ \mathrm{=\dfrac{-2\pm\sqrt{400}}{6}=\dfrac{-2\pm20}{6}\ \to\ \boxed{\mathrm{y=3}}\ \ ou\ \ \boxed{\mathrm{y=-\dfrac{11}{3}}}}$

$\mathbf{*\ Para\ y=3}\ \to\ \mathrm{x=3^2\ \to\ \boxed{\mathrm{x=9}}}\\\\ \mathbf{*\ Para\ y=-\dfrac{11}{3}}\ \to\ \mathrm{x=\bigg(\dfrac{-11}{3}\bigg)^2\ \to\ \boxed{\mathrm{x=\dfrac{121}{9}}}}\\\\\\ \boxed{\boxed{\mathbf{\mathbb{S}=\bigg\{(x,y)\in\mathbb{R}^2\ |\ (x,y)=(9,3)\ \ ou\ \ (x,y)=\bigg(\dfrac{121}{9},-\dfrac{11}{3}\bigg)\bigg\}}}}$