What is (2xy+2y)/(9x)*(3xy)/(2y^2+4xy^2) ?

The solution to (2xy+2y)/(9x)*(3xy)/(2y^2+4xy^2) is (x+1)/(3(1+2x))

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simplify (2xy+2y)/(9x)*(3xy)/(2y^2+4xy^2)

simplify

\frac{2 x y + 2 y}{9 x} \cdot \frac{3 x y}{2 y ^{2} + 4 x y ^{2}}

\frac{x + 1}{3 ( 1 + 2 x )}

Solution steps

\frac{2 x y + 2 y}{9 x} \cdot \frac{3 x y}{2 y ^{2} + 4 x y ^{2}}

Cancel

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

= \frac{2 x y + 2 y}{9 x} \cdot \frac{3 x}{2 y ( 1 + 2 x )}

Apply the fraction rule:

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

= \frac{( 2 x y + 2 y ) \cdot 3 x}{9 x \cdot 2 y ( 1 + 2 x )}

Cancel the common factor:

x

= \frac{( 2 x y + 2 y ) \cdot 3}{9 \cdot 2 y ( 1 + 2 x )}

Factor the number:

9 = 3 \cdot 3

= \frac{( 2 x y + 2 y ) \cdot 3}{3 \cdot 3 \cdot 2 y ( 1 + 2 x )}

Cancel the common factor:

3

= \frac{2 x y + 2 y}{3 \cdot 2 y ( 1 + 2 x )}

Factor out common term

2 y

= \frac{2 y ( x + 1 )}{3 \cdot 2 y ( 1 + 2 x )}

Cancel the common factor:

2

= \frac{y ( x + 1 )}{3 y ( 1 + 2 x )}

Cancel the common factor:

y

= \frac{x + 1}{3 ( 1 + 2 x )}

\frac{4}{3} - \frac{3}{2}

- 3 (x - 2) (x + 4)

[1 - (x + z)] [1 + (x + z)]

\frac{1}{cos ^{3} ( x )}

1 - (1)

What is (2xy+2y)/(9x)*(3xy)/(2y^2+4xy^2) ?
The solution to (2xy+2y)/(9x)*(3xy)/(2y^2+4xy^2) is (x+1)/(3(1+2x))