How do you simplify $\frac{3 x^{3} y^{2}}{2 x^{2} y^{3}}$ $\frac { 3x ^ { 3} y ^ { 2} } { 2x ^ { 2} y ^ { 3} }$ ?

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Answer 1 · 2018-07-04T11:45:27

Here.

Start by crossing out the $x$ $x$ of the numerator with that of the denominator using indices rules.

$\frac{x^{3}}{x^{2}} = x$ $x^3/x^2 = x$

That's because the rule of indices states that when powers are divided, they get subtracted.

Do the same for the $y$ $y$ .

You get:

$\frac{y^{2}}{y^{3}} = \frac{1}{y}$ $y^2/ y^3= 1/y$

Now since you can't simplify the 2 and the 3, you can leave them:

$\frac{3 x}{2 y}$ $(3x)/(2y)$

Answer 2 · 2018-07-04T13:48:43

$\frac{3 x}{2 y}$ $(3x)/(2y)$ or I prefer it as $1.5 x y^{- 1} = \frac{1.5 x}{y}$ $1.5xy^-1=(1.5x)/y$ .

Given: $\frac{3 x^{3} y^{2}}{2 x^{2} y^{3}}$ $(3x^3y^2)/(2x^2y^3)$ .

Separate into:

$= \frac{3}{2} \cdot \frac{x^{3}}{x^{2}} \cdot \frac{y^{2}}{y^{3}}$ $=3/2*x^3/x^2*y^2/y^3$

$= 1.5 \cdot x \cdot \frac{1}{y}$ $=1.5*x*1/y$

$= \frac{1.5 x}{y}$ $=(1.5x)/y$

$= 1.5 x y^{- 1}$ $=1.5xy^-1$

How do you simplify 3x3y22x2y3\frac { 3x ^ { 3} y ^ { 2} } { 2x ^ { 2} y ^ { 3} }?