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Answer :
We start with the expression
[tex]$$
\frac{-4x^9y + 48x^3 - 35xy^2}{-32x^3y}.
$$[/tex]
Our goal is to simplify it by breaking it up into separate terms and reducing each one individually. We can write the expression as a sum of three fractions:
[tex]$$
\frac{-4x^9y}{-32x^3y} + \frac{48x^3}{-32x^3y} + \frac{-35xy^2}{-32x^3y}.
$$[/tex]
Let’s simplify these one by one.
1. First Term:
Consider
[tex]$$
\frac{-4x^9y}{-32x^3y}.
$$[/tex]
Notice that the negatives cancel:
[tex]$$
\frac{-4}{-32} = \frac{4}{32} = \frac{1}{8}.
$$[/tex]
For the variables, cancel the common factors:
[tex]$$
\frac{x^9}{x^3} = x^{9-3} = x^6, \quad \frac{y}{y} = 1.
$$[/tex]
Thus, the first term simplifies to:
[tex]$$
\frac{1}{8}x^6 = \frac{x^6}{8}.
$$[/tex]
2. Second Term:
Consider
[tex]$$
\frac{48x^3}{-32x^3y}.
$$[/tex]
Simplify the constants first:
[tex]$$
\frac{48}{-32} = -\frac{48}{32} = -\frac{3}{2}.
$$[/tex]
Cancel the common factor [tex]$x^3$[/tex]:
[tex]$$
\frac{x^3}{x^3} = 1.
$$[/tex]
With the variable [tex]$y$[/tex] in the denominator, the term becomes:
[tex]$$
-\frac{3}{2} \cdot \frac{1}{y} = -\frac{3}{2y}.
$$[/tex]
3. Third Term:
Consider
[tex]$$
\frac{-35xy^2}{-32x^3y}.
$$[/tex]
The negatives cancel:
[tex]$$
\frac{-35}{-32} = \frac{35}{32}.
$$[/tex]
Now, simplify the variables:
- For [tex]$x$[/tex]: [tex]$$\frac{x}{x^3} = \frac{1}{x^2}.$$[/tex]
- For [tex]$y$[/tex]: [tex]$$\frac{y^2}{y} = y.$$[/tex]
So the third term simplifies to:
[tex]$$
\frac{35}{32} \cdot \frac{1}{x^2} \cdot y = \frac{35y}{32x^2}.
$$[/tex]
4. Combine the Terms:
Putting these results together, we arrive at the simplified expression:
[tex]$$
\frac{x^6}{8} - \frac{3}{2y} + \frac{35y}{32x^2}.
$$[/tex]
Thus, the final simplified expression is
[tex]$$
\boxed{\frac{x^6}{8} - \frac{3}{2y} + \frac{35y}{32x^2}}.
$$[/tex]
[tex]$$
\frac{-4x^9y + 48x^3 - 35xy^2}{-32x^3y}.
$$[/tex]
Our goal is to simplify it by breaking it up into separate terms and reducing each one individually. We can write the expression as a sum of three fractions:
[tex]$$
\frac{-4x^9y}{-32x^3y} + \frac{48x^3}{-32x^3y} + \frac{-35xy^2}{-32x^3y}.
$$[/tex]
Let’s simplify these one by one.
1. First Term:
Consider
[tex]$$
\frac{-4x^9y}{-32x^3y}.
$$[/tex]
Notice that the negatives cancel:
[tex]$$
\frac{-4}{-32} = \frac{4}{32} = \frac{1}{8}.
$$[/tex]
For the variables, cancel the common factors:
[tex]$$
\frac{x^9}{x^3} = x^{9-3} = x^6, \quad \frac{y}{y} = 1.
$$[/tex]
Thus, the first term simplifies to:
[tex]$$
\frac{1}{8}x^6 = \frac{x^6}{8}.
$$[/tex]
2. Second Term:
Consider
[tex]$$
\frac{48x^3}{-32x^3y}.
$$[/tex]
Simplify the constants first:
[tex]$$
\frac{48}{-32} = -\frac{48}{32} = -\frac{3}{2}.
$$[/tex]
Cancel the common factor [tex]$x^3$[/tex]:
[tex]$$
\frac{x^3}{x^3} = 1.
$$[/tex]
With the variable [tex]$y$[/tex] in the denominator, the term becomes:
[tex]$$
-\frac{3}{2} \cdot \frac{1}{y} = -\frac{3}{2y}.
$$[/tex]
3. Third Term:
Consider
[tex]$$
\frac{-35xy^2}{-32x^3y}.
$$[/tex]
The negatives cancel:
[tex]$$
\frac{-35}{-32} = \frac{35}{32}.
$$[/tex]
Now, simplify the variables:
- For [tex]$x$[/tex]: [tex]$$\frac{x}{x^3} = \frac{1}{x^2}.$$[/tex]
- For [tex]$y$[/tex]: [tex]$$\frac{y^2}{y} = y.$$[/tex]
So the third term simplifies to:
[tex]$$
\frac{35}{32} \cdot \frac{1}{x^2} \cdot y = \frac{35y}{32x^2}.
$$[/tex]
4. Combine the Terms:
Putting these results together, we arrive at the simplified expression:
[tex]$$
\frac{x^6}{8} - \frac{3}{2y} + \frac{35y}{32x^2}.
$$[/tex]
Thus, the final simplified expression is
[tex]$$
\boxed{\frac{x^6}{8} - \frac{3}{2y} + \frac{35y}{32x^2}}.
$$[/tex]
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