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

The solution to (1/(2x)+1/(3y))/(4/x-3/(4y)) is (2(3y+2x))/(3(16y-3x))

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

simplify

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

\frac{2 ( 3 y + 2 x )}{3 ( 16 y - 3 x )}

Solution steps

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

Join

\frac{1}{2 x} + \frac{1}{3 y} : \frac{3 y + 2 x}{6 x y}

= \frac{\frac{3 y + 2 x}{6 x y}}{\frac{4}{x} - \frac{3}{4 y}}

Join

\frac{4}{x} - \frac{3}{4 y} : \frac{16 y - 3 x}{4 x y}

= \frac{\frac{3 y + 2 x}{6 x y}}{\frac{16 y - 3 x}{4 x y}}

Apply the fraction rule:

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

= \frac{( 3 y + 2 x ) \cdot 4 x y}{6 x y ( 16 y - 3 x )}

Cancel

\frac{( 3 y + 2 x ) \cdot 4 x y}{6 x y ( 16 y - 3 x )} : \frac{( 3 y + 2 x ) \cdot 2}{3 ( 16 y - 3 x )}

= \frac{( 3 y + 2 x ) \cdot 2}{3 ( 16 y - 3 x )}

= \frac{2 ( 3 y + 2 x )}{3 ( 16 y - 3 x )}

\frac{\frac{π}{6}}{10 π}

\frac{8}{x - 3} > \frac{6}{x - 1}

∣ 3 v - 6 ∣ = - 3

48 - 16 g + g^{2}

lo g_{3} (19683)

What is (1/(2x)+1/(3y))/(4/x-3/(4y)) ?
The solution to (1/(2x)+1/(3y))/(4/x-3/(4y)) is (2(3y+2x))/(3(16y-3x))