What is the slope and the y -intercept of the equation 9+27 over 2x = −3 over 2y. Write the y -intercept as an ordered pair.

To find the slope and the y-intercept of the given equation \( \frac{9 + 27}{2x} = -\frac{3}{2}y \), we first simplify the equation.

First, we can rewrite the left side:

\[
\frac{36}{2x} = -\frac{3}{2}y
\]

This simplifies to:

\[
\frac{18}{x} = -\frac{3}{2}y
\]

Next, we can multiply both sides by \(2x\) to eliminate the fraction:

\[
36 = -3xy
\]

To express \(y\) in terms of \(x\), we can rearrange this equation:

\[
3xy = -36
\]

Now, divide by \(3x\):

\[
y = -\frac{36}{3x} = -\frac{12}{x}
\]

However, the equation is still in a complicated form. We will rearrange this into the slope-intercept form, \(y = mx + b\). To do that, we can multiply both sides by \(x\) to make it easier to identify:

\[
xy = -12
\]

This implies

\[
y = -\frac{12}{x}
\]

Rearranging doesn't really help us in identifying slope. Instead, let's try plugging in values to see how the equation behaves.

Alternatively, we can rewrite \(y = -\frac{12}{x}\) as:

\[
y = -12 \cdot \frac{1}{x}
\]

Now, this isn't a linear function, but if we consider the equation from yet another perspective, rearranging the original provided equation directly into \(y\) terms:

Starting from \( \frac{36}{2x} = -\frac{3}{2}y \):

Multiplying it through by \(2x\):

\[
36 = -3xy
\]

And moving all terms involving \(y\) to one side gives us:

\[
xy = -36
\]

This means it can be seen as inversely proportional (not a straight line in standard slope-intercept form).

However, if we analyze around specific coordinates, and find the y-intercept, we observe that:

When \(x=0\), \(y\) has no defined value (since division by zero is undefined); thus the transformation does not lead to a visual line.

Thus, the nature of this equation is hyperbolic rather than linear, meaning it does not have a slope and an intercept we can describe.

Conclusively, the situation is as such: the equation does not yield a slope or an ordered pair as a y-intercept due to its nature, which is critical for linear equations. It forms a hyperbola. Therefore, traditional linear analysis does not apply.

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