Respuesta :

dalendrk
[tex]\dfrac{y^2}{4}+\dfrac{y}{3}+\dfrac{y^2}{3}-\dfrac{y}{5}=\dfrac{3y^2}{3\cdot4}+\dfrac{4y^2}{3\cdot4}+\dfrac{5y}{3\cdot5}-\dfrac{3y}{3\cdot5}\\\\=\dfrac{3y^2}{12}+\dfrac{4y^2}{12}+\dfrac{5y}{15}-\dfrac{3y}{15}=\boxed{\dfrac{7y^2}{12}+\dfrac{2y}{15}}\\\\or\\\\\dfrac{7y^2}{12}+\dfrac{2y}{15}=\dfrac{7y^2\cdot5}{12\cdot5}+\dfrac{2y\cdot4}{15\cdot4}}=\dfrac{35y^2}{60}+\dfrac{8y}{60}=\boxed{\frac{35y^2+8y}{60}}[/tex]

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