How do you divide $\frac{6 x^{3} y^{2}}{8 x^{4} y^{5}} \div \frac{2 x}{12 x y^{2}}$ $\frac { 6x ^ { 3} y ^ { 2} } { 8x ^ { 4} y ^ { 5} } \div \frac { 2x } { 12x y ^ { 2} }$ ?

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How do you divide $\frac{6 x^{3} y^{2}}{8 x^{4} y^{5}} \div \frac{2 x}{12 x y^{2}}$ $\frac { 6x ^ { 3} y ^ { 2} } { 8x ^ { 4} y ^ { 5} } \div \frac { 2x } { 12x y ^ { 2} }$ ?

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Answer 1 · 2016-12-01T02:26:06

$\frac{9}{2 x y}$ $(9)/(2xy)$

Re-write into multiplication format, multiply together like terms, divide away like terms on the numerator and denominator.

Explanation:

Dividing something by a fraction is the same as multiplying that same 'something' by the inverse of the fraction (swap the denominator and numerator locations).

$\frac{6 x^{3} y^{2}}{8 x^{4} y^{5}} \cdot \frac{12 x y^{2}}{2 x}$ $(6x^3y^2)/(8x^4y^5) * (12xy^2)/(2x)$

From here, multiply like terms together.

$\frac{72 x^{4} y^{4}}{16 x^{5} y^{5}}$ $(72x^4y^4)/(16x^5y^5)$

Divide like terms away.

$\frac{9}{2 x y}$ $(9)/(2xy)$

This can be done because $x^{5}$ $x^5$ and $y^{5}$ $y^5$ can be re-written as $x^{4} \cdot x$ $x^4 * x$ and $y^{4} \cdot y$ $y^4 * y$ respectively. Thus, the variable being raised to the fourth on both the numerator and denominator effectively cancels each other out, leaving you with just one $x$ $x$ and one $y$ $y$ on the denominator of the end result.

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How do you divide $\frac{6 x^{3} y^{2}}{8 x^{4} y^{5}} \div \frac{2 x}{12 x y^{2}}$ $\frac { 6x ^ { 3} y ^ { 2} } { 8x ^ { 4} y ^ { 5} } \div \frac { 2x } { 12x y ^ { 2} }$ ?

1 Answer

Explanation:

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How do you divide 6x3y28x4y5÷2x12xy2\frac { 6x ^ { 3} y ^ { 2} } { 8x ^ { 4} y ^ { 5} } \div \frac { 2x } { 12x y ^ { 2} } ?

1 Answer

Explanation:

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How do you divide $\frac{6 x^{3} y^{2}}{8 x^{4} y^{5}} \div \frac{2 x}{12 x y^{2}}$ $\frac { 6x ^ { 3} y ^ { 2} } { 8x ^ { 4} y ^ { 5} } \div \frac { 2x } { 12x y ^ { 2} }$ ?