Question

2-) Considere as proporções $\frac{2x}{x-3} = \frac{4}{3} e \frac{2}{y} = \frac{6}{15}$ .  Nessas condições, calcule o valor de $x^{2} + y^{2}$

Answer 1

$\Large\textsf{Primeira equa\c{c}\~ao, descobrindo o valor de x:}$


$\mathsf{\dfrac{2x}{x-3}=\dfrac{4}{3}}\\\\\\\\\mathsf{3}{\begin{pmatrix}\mathsf{\dfrac{2x}{x-3}}\end{pmatrix}}\mathsf{=\dfrac{4}{3\hspace{-7}\diagup}\cdot 3\hspace{-7}\diagup}\\\\\\\mathsf{\dfrac{6x}{x-3}=4}\\\\\\\mathsf{(x\hspace{-7}\diagup-3\hspace{-7}\diagup)}{\begin{pmatrix} \mathsf{\dfrac{6x}{x\hspace{-7}\diagup-3\hspace{-7}\diagup}} \end{pmatrix}\mathsf{=4(x-3)}}\\\\\\\mathsf{6x=4x-12}\\\\\\\mathsf{2x=-12}\\\\\\\mathsf{x=\dfrac{-12}{2}}\\\\\\\fbox{ \mathsf{x=-6} }$


$\Large\textsf{Segunda equa\c{c}\~ao, descobrindo o valor de y:}$



$\mathsf{\dfrac{2}{y}=\dfrac{6}{15}}\\\\\\\\\mathsf{15}\begin{pmatrix}\mathsf{\dfrac{2}{y}} \end{pmatrix}\mathsf{=\dfrac{6}{15\hspace{-9}\diagup}\cdot 15\hspace{-9}\diagup}\\\\\\\mathsf{\dfrac{30}{y}=6}\\\\\\\mathsf{y=\dfrac{30}{6}}\\\\\\\fbox{ \mathsf{y=5} }$


$\Large\begin{array}{l}\mathsf{Descobrindo~o~resultado~da~express\~ao~x^2+y^2:}\end{array}$



$\mathsf{x^2+y^2}\\\\\\\mathsf{=(-6)^2+5^2}\\\\\\\mathsf{=36+25}\\\\\\\mathsf{=61}\\\\\\\fbox{ \mathsf{x^2+y^2=61} }$