What is (3y+4/(y+1))/(2y-3/y) ?

The solution to (3y+4/(y+1))/(2y-3/y) is ((3y^2+3y+4)y)/((y+1)(2y^2-3))

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simplify (3y+4/(y+1))/(2y-3/y)

simplify

\frac{3 y + \frac{4}{y + 1}}{2 y - \frac{3}{y}}

\frac{( 3 y ^{2} + 3 y + 4 ) y}{( y + 1 ) ( 2 y ^{2} - 3 )}

Solution steps

\frac{3 y + \frac{4}{y + 1}}{2 y - \frac{3}{y}}

Join

3 y + \frac{4}{y + 1} : \frac{3 y ^{2} + 3 y + 4}{y + 1}

= \frac{\frac{3 y ^{2} + 3 y + 4}{y + 1}}{2 y - \frac{3}{y}}

Join

2 y - \frac{3}{y} : \frac{2 y ^{2} - 3}{y}

= \frac{\frac{3 y ^{2} + 3 y + 4}{y + 1}}{\frac{2 y ^{2} - 3}{y}}

Apply the fraction rule:

\frac{\frac{a}{b}}{\frac{c}{d}} = \frac{a \cdot d}{b \cdot c}

= \frac{( 3 y ^{2} + 3 y + 4 ) y}{( y + 1 ) ( 2 y ^{2} - 3 )}

5 + \frac{1}{75} (x - 125)

2^{5} \cdot 3^{2} \cdot 5^{2}

\frac{3}{4} + \frac{4}{8}

- \frac{1}{4} + \frac{5}{4}

e^{\frac{3}{2} l n (x)}

What is (3y+4/(y+1))/(2y-3/y) ?
The solution to (3y+4/(y+1))/(2y-3/y) is ((3y^2+3y+4)y)/((y+1)(2y^2-3))