The graphs of $2y + x + 3 = 0$  and $3y + ax + 2 = 0$ are perpendicular. Solve for $a.$
Solving $2y + x + 3 = 0$ for $y$ gives $ y = \frac{-1}{2}x - \frac{3}{2},$ so the slope of this line is $-\frac{1}{2}.$

Solving $3y + ax + 2 = 0$ for $y$ gives $ y = \frac{-a}{3}x - \frac{2}{3},$ so the slope of this line is $- \frac{a}{3}.$

In order for these lines to be perpendicular, we must have $$\left(-\frac{1}{2}\right)\left(-\frac{a}{3}\right) = -1.$$Solving for $a$ gives $a = \boxed{-6}.$
-6
Level 4
Algebra
Problem: The graphs of $2y + x + 3 = 0$  and $3y + ax + 2 = 0$ are perpendicular. Solve for $a.$
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To determine the value of \( a \) such that the lines represented by the equations are perpendicular, we first need to find the slopes of both lines. 

The first equation is \( 2y + x + 3 = 0 \). Rearranging this into slope-intercept form \( y = mx + b \) gives:
1. \( 2y = -x - 3 \)
2. \( y = -\frac{1}{2}x - \frac{3}{2} \)

From this, we identify the slope \( m_1 = -\frac{1}{2} \).

Next, we analyze the second equation \( 3y + ax + 2 = 0 \). Rearranging it similarly yields:
1. \( 3y = -ax - 2 \)
2. \( y = -\frac{a}{3}x - \frac{2}{3} \)

Here, the slope is \( m_2 = -\frac{a}{3} \).

For the lines to be perpendicular, the product of their slopes must equal \(-1\):
\[
m_1 \cdot m_2 = -1
\]
Substituting the slopes we found:
\[
\left(-\frac{1}{2}\right) \left(-\frac{a}{3}\right) = -1
\]
This simplifies to:
\[
\frac{a}{6} = -1
\]
To isolate \( a \), we multiply both sides by 6:
\[
a = -6
\]

Thus, the value of \( a \) that makes the lines perpendicular is \( -6 \).<|END_OF_LATENT|> The final answer is $-6$.