Let's solve the equation you provided, \( 2y = -3xy - 5 \), with the condition \( x = 1 \).
- Substitute \( x = 1 \) into the equation:
\[ 2y = -3(1)y - 5 \]
- Simplifying the equation:
\[ 2y = -3y - 5 \]
- Now, let's add \( 3y \) to both sides to isolate \( y \):
\[ 2y + 3y = -5 \]
\[ 5y = -5 \]
- Divide by 5:
\[ y = -1 \]
So the solution for \( y \) when \( x = 1 \) is \( y = -1 \).
To check the linear equation \( y = -\frac{5}{3}x + 2 \), we need to substitute the value of \( x = 1 \):
\[ y = -\frac{5}{3}(1) + 2 \]
\[ y = -\frac{5}{3} + 2 \]
\[ y = -\frac{5}{3} + \frac{6}{3} \]
\[ y = \frac{1}{3} \]
The line \( y = -\frac{5}{3}x + 2 \) gives \( y = \frac{1}{3} \) when \( x = 1 \), which does not equal \( -1 \).
The answer \( y = -\frac{5}{3}x + 2 \) is indeed a linear equation, but for \( x = 1 \) it results in \( y = \frac{1}{3} \), not \( -1 \).
Thus, the correct conclusion from our original equation is that when \( x = 1 \), \( y = -1 \). The provided linear equation does not yield the same value for \( y \) at that point.