To solve for \( y \) in the equation \( 2y = -3xy - 5 \), we can start by isolating \( y \).
-
Start with the equation:
\[ 2y = -3xy - 5 \]
-
Rearrange the equation to group the \( y \) terms:
\[ 2y + 3xy = -5 \]
-
Factor out \( y \) on the left side:
\[ y(2 + 3x) = -5 \]
-
Now, solve for \( y \) by dividing both sides by \( (2 + 3x) \):
\[ y = \frac{-5}{2 + 3x} \]
This is the expression for \( y \) in terms of \( x \).
Now, substituting \( x = 1 \) to find the specific value of \( y \):
\[ y = \frac{-5}{2 + 3(1)} = \frac{-5}{2 + 3} = \frac{-5}{5} = -1 \]
So the answer is \( y = -1 \) when \( x = 1 \).
If you meant to express the equation in slope-intercept form (like \( y = mx + b \)), notice that \( \frac{-5}{2 + 3x} \) can't be rewritten into the form you mentioned, as it represents a rational function.
However, if you want to represent this in a linear form involving \( x \) and \( y \), you'd need to appropriately analyze the structure given the \( x \) value.
The conclusion is:
- The derived formula for \( y \) is \( y = \frac{-5}{2 + 3x} \).
- For \( x = 1 \), \( y = -1 \).
If you're looking for another interpretation or simplification, please clarify!