To solve the system of linear equations graphically and find the area of the triangle formed by the two lines and the y-axis, we will follow these steps:
### Step 1: Write the equations in slope-intercept form
The given equations are:
1. \( x - 3y = 3 \)
2. \( 2x + 3y = 6 \)
We can rearrange these equations to find the y-intercepts.
**For the first equation:**
\[
x - 3y = 3 \implies -3y = -x + 3 \implies y = \frac{1}{3}x - 1
\]
**For the second equation:**
\[
2x + 3y = 6 \implies 3y = -2x + 6 \implies y = -\frac{2}{3}x + 2
\]
### Step 2: Find the intercepts of both lines
To plot the lines, we need to find the x-intercept and y-intercept for both equations.
**For the first equation \( y = \frac{1}{3}x - 1 \):**
- **Y-intercept (when \( x = 0 \))**:
\[
y = \frac{1}{3}(0) - 1 = -1 \quad \text{(Point: (0, -1))}
\]
- **X-intercept (when \( y = 0 \))**:
\[
0 = \frac{1}{3}x - 1 \implies \frac{1}{3}x = 1 \implies x = 3 \quad \text{(Point: (3, 0))}
\]
**For the second equation \( y = -\frac{2}{3}x + 2 \):**
- **Y-intercept (when \( x = 0 \))**:
\[
y = -\frac{2}{3}(0) + 2 = 2 \quad \text{(Point: (0, 2))}
\]
- **X-intercept (when \( y = 0 \))**:
\[
0 = -\frac{2}{3}x + 2 \implies \frac{2}{3}x = 2 \implies x = 3 \quad \text{(Point: (3, 0))}
\]
### Step 3: Plot the points and draw the lines
Now we have the following points to plot:
1. From the first equation: (0, -1) and (3, 0)
2. From the second equation: (0, 2) and (3, 0)
On a graph:
- Plot the points (0, -1) and (3, 0) for the first line.
- Plot the points (0, 2) and (3, 0) for the second line.
- Draw the lines through these points.
### Step 4: Identify the triangle formed
The triangle is formed by the two lines and the y-axis. The vertices of the triangle are:
- Point A: (0, -1) (from the first line)
- Point B: (0, 2) (from the second line)
- Point C: (3, 0) (common point on both lines)
### Step 5: Calculate the area of the triangle
The area \( A \) of a triangle can be calculated using the formula:
\[
A = \frac{1}{2} \times \text{base} \times \text{height}
\]
Here, the base \( CB \) is the distance along the y-axis from (0, -1) to (0, 2), which is:
\[
\text{Base} = 2 - (-1) = 3
\]
The height \( OA \) is the distance along the x-axis from (0, -1) to (3, 0), which is:
\[
\text{Height} = 3
\]
Now, substituting these values into the area formula:
\[
A = \frac{1}{2} \times 3 \times 3 = \frac{9}{2} = 4.5 \text{ square units}
\]
### Final Answer
The area of the triangle formed by the two lines and the y-axis is \( 4.5 \) square units.
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