To solve the given problem and draw the required graphs:

1. Equation 1: $x + y + 3 = 0$ can be rearranged to: $y = -x - 3$
2. Equation 2: $3x - 2y + 4 = 0$ can be rearranged to: $y = \frac{3}{2}x + 2$

Part (i) – Plotting the Graphs:

1. For the first equation $y = -x - 3$:

We need three points to plot the graph. Choosing different values for $x$:

• When $x = 0$, $y = -3$ → Point (0, -3)
• When $x = -1$, $y = -2$ → Point (-1, -2)
• When $x = 1$, $y = -4$ → Point (1, -4)

2. For the second equation $y = \frac{3}{2}x + 2$:

We also need three points to plot the graph. Again, choosing values for $x$:

• When $x = 0$, $y = 2$ → Point (0, 2)
• When $x = -2$, $y = -1$ → Point (-2, -1)
• When $x = 2$, $y = 5$ → Point (2, 5)

Part (ii) – Finding the Intersection Point:

We can solve these two equations algebraically by setting the expressions for $y$ equal to each other: $-x - 3 = \frac{3}{2}x + 2$ Now solve for $x$:

• Multiply the entire equation by 2 to eliminate the fraction: $-2x - 6 = 3x + 4$
• Bring all $x$-terms to one side: $-2x - 3x = 4 + 6$ $-5x = 10 \quad \Rightarrow \quad x = -2$

Now substitute $x = -2$ into one of the original equations to find $y$:

• Using $y = -x - 3$: $y = -(-2) - 3 = 2 - 3 = -1$

So, the coordinates of the intersection point are (-2, -1).

---

Summary:

• Intersection point: (-2, -1)

Would you like more details or have any questions about this solution?

Relative questions:

1. How do you find the slope of a line from its equation?
2. What are other methods to solve a system of linear equations?
3. Can a system of two linear equations have more than one solution?
4. What is the geometric interpretation of parallel and intersecting lines?
5. How does the slope influence the steepness of a line?

Tip:

To quickly find the intersection of two lines graphically, ensure both lines are plotted with accurate scales and then visually locate the crossing point.