`3y+6=2x`
`2y-3x=6`
The system of equations above can best be described as having

Options

A. no solution
B. one solutions with the graphs intersecting at right angles in the xy-plane
C. one solution with the graphs not intersecting at right angles in the xy-plane.
D. infinitely many solutions.

Answer

To solve the system of equations given by:

1. \( 3y + 6 = 2x \)
2. \( 2y - 3x = 6 \)

we will follow these steps:

### Step 1: Rewrite the equations in slope-intercept form (y = mx + b)

**For the first equation:**

Starting with:
\[ 3y + 6 = 2x \]

Subtract 6 from both sides:
\[ 3y = 2x - 6 \]

Now, divide by 3:
\[ y = \frac{2}{3}x - 2 \]

So, the slope \( m_1 = \frac{2}{3} \).

**For the second equation:**

Starting with:
\[ 2y - 3x = 6 \]

Add \( 3x \) to both sides:
\[ 2y = 3x + 6 \]

Now, divide by 2:
\[ y = \frac{3}{2}x + 3 \]

So, the slope \( m_2 = \frac{3}{2} \).

### Step 2: Determine if the lines are perpendicular

For two lines to be perpendicular, the product of their slopes must equal -1:
\[ m_1 \cdot m_2 = \left(\frac{2}{3}\right) \cdot \left(\frac{3}{2}\right) = 1 \]

Since \( 1 \neq -1 \), the lines are not perpendicular.

### Step 3: Check if the lines intersect

To determine if the lines intersect, we can use the condition for intersection:

If the ratios of the coefficients of \( x \) and \( y \) are not equal, then the lines intersect at one point.

Rearranging the equations into standard form:

1. From \( 3y - 2x + 6 = 0 \):
- Coefficients are \( a_1 = -2, b_1 = 3, c_1 = 6 \)

2. From \( 3x - 2y + 6 = 0 \):
- Coefficients are \( a_2 = 3, b_2 = -2, c_2 = 6 \)

Now, we check the ratios:
- \( \frac{a_1}{a_2} = \frac{-2}{3} \)
- \( \frac{b_1}{b_2} = \frac{3}{-2} \)

Since \( \frac{-2}{3} \neq \frac{3}{-2} \), the lines are not parallel and thus they intersect at one point.

### Conclusion

The system of equations has one solution, and since the lines are not perpendicular, the correct description is:

**Option C: one solution with graphs not intersecting at right angles in the xy-plane.**

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