Question - Solving a Fraction Expression
Solution:
The image is a bit blurry, but it looks like the expression is a fraction where the numerator is $$ \frac{1}{7x^2y^3} $$ and the denominator is $$ \frac{1}{3x^3y} $$.To solve this expression, you can simply invert the denominator and multiply it with the numerator:$$ (\frac{1}{7x^2y^3}) \times (\frac{3x^3y}{1}) $$When you multiply fractions, you multiply the numerators together and the denominators together:$$ \frac{1 \times 3x^3y}{7x^2y^3 \times 1} $$$$ \frac{3x^3y}{7x^2y^3} $$Now simplify by canceling out the common terms:The $$ x^2 $$ in the denominator cancels out with two of the $$ x $$s in the numerator, leaving you with one $$ x $$ in the numerator. Similarly, the $$ y $$ in the numerator and one $$ y $$ in the denominator will cancel out, leaving you with $$ y^2 $$ in the denominator.So, you're left with:$$ \frac{3x}{7y^2} $$This is the simplified form of the given expression.
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