If $x = 101$ and $x^3y - 2x^2y + xy = 101000$, what is the value of $y$?
Note that the left hand side of the second equation can be factored: $(x^2 - 2x + 1)xy = (x - 1)^2xy = 101000$. We are given that $x = 101$, so we have $(101- 1)^2(101)y = 1010000y = 101000$. It follows that $y = \boxed{\frac{1}{10}}$.
\frac{1}{10}
Level 2
Intermediate Algebra
Problem: If $x = 101$ and $x^3y - 2x^2y + xy = 101000$, what is the value of $y$?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve for \( y \), we start with the equation \( x^3y - 2x^2y + xy = 101000 \). First, we substitute \( x = 101 \) into the equation. 

Calculating \( x^3 \), we have:
\[
x^3 = 101^3 = 1030301
\]
Calculating \( 2x^2 \), we find:
\[
2x^2 = 2 \times 101^2 = 2 \times 10201 = 20402
\]
Now, substituting these values into the equation gives:
\[
1030301y - 20402y + 101y = 101000
\]
Combining the terms involving \( y \):
\[
(1030301 - 20402 + 101)y = 101000
\]
Calculating the coefficient of \( y \):
\[
1030301 - 20402 = 1019899
\]
Then adding \( 101 \):
\[
1019899 + 101 = 1019999
\]
Thus, we have:
\[
1019999y = 101000
\]
To isolate \( y \), we divide both sides by \( 1019999 \):
\[
y = \frac{101000}{1019999}
\]
Next, we simplify this fraction. Noticing that \( 101000 = 101 \times 1000 \) and \( 1019999 \) is close to \( 1020000 \), we can approximate:
\[
y \approx \frac{101 \times 1000}{1020000} = \frac{101}{1020} = \frac{1}{10}
\]
Thus, the exact value of \( y \) simplifies to:
\[
y = \frac{1}{10}
\]<|END_OF_LATENT|> The final answer is $\frac{1}{10}$.