Verified answer

[tex]\displaystyle \sf \frac{3y}{y-4}=3+\frac{2}{y} \\\\\\ \frac{3y}{y-4}=\frac{3y+2}{y} \\\\\\ 3y\cdot y = (y-4)\cdot (3y+2) \\\\ 3y^2=3y^2+2y-12y-8 \\\\ 3y^2-3y^2-10y- 8 = 0 \\\\ 10 y = -8 \\\\ y = \frac{-8}{10} \\\\\ \huge\boxed{\sf \ y = \frac{-4}{5}\ }\checkmark[/tex]

Após resolver os cálculos, concluímos que o valor de "y" para que as expressões sejam iguais é [tex]\sf y=-\dfrac{4}{5}\,\cdot\\[/tex]

Para resolver esse exercício, basta iguala as expressões, veja:

[tex]\sf\boxed{\sf\dfrac{3y}{y-4}}\quad e\quad\boxed{\sf3+\dfrac{2}{y}}~, com~y\neq0;\,y\neq4[/tex]

[tex] \mathsf{\dfrac{3y}{y-4}=3+\dfrac{2}{y} } [/tex]

[tex] \mathsf{\dfrac{3y}{y-4} -\dfrac{2}{y}=3} [/tex]

[tex] \mathsf{\dfrac{3y^2-2(y-4)}{y(y-4)}=3 } [/tex]

[tex] \mathsf{ \dfrac{ 3y^2 -2y+8}{y(y-4)}=3} [/tex]

[tex] \mathsf{3y^2-2y+8=3y(y-4) } [/tex]

[tex] \mathsf{\diagdown\!\!\!\!\!\!3y^2-2y+8=\diagdown\!\!\!\!\!\!3y^2-12y } [/tex]

[tex] \mathsf{-2y+8=-12y } [/tex]

[tex] \mathsf{ -2y+12y= -8} [/tex]

[tex] \mathsf{ 10y= -8} [/tex]

[tex] \mathsf{ y=-\dfrac{8}{10}} [/tex]

[tex]\boxed{\boxed{\sf y=-\dfrac{4}{5}}}[/tex]

Logo, o valor de "y" para que as expressões sejam iguais é [tex]\sf y=-\dfrac{4}{5}\,\cdot\\[/tex]

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