Draw the graph of equation `x+2y-3=0`.
From the graph find
`x_(1)` the value of x, wheny=3
(ii) `x_(2)`, the value of x, when y=-2

Answer

To solve the problem, we will follow these steps:

### Step 1: Rewrite the equation in slope-intercept form
The given equation is:
\[ x + 2y - 3 = 0 \]
We can rearrange this to express \( y \) in terms of \( x \):
\[ 2y = -x + 3 \]
\[ y = -\frac{1}{2}x + \frac{3}{2} \]

### Step 2: Determine points on the graph
To draw the graph, we need to find some points. We can choose different values for \( x \) and calculate the corresponding \( y \) values.

1. **When \( x = 0 \)**:
\[ y = -\frac{1}{2}(0) + \frac{3}{2} = \frac{3}{2} \]
Point: \( (0, \frac{3}{2}) \)

2. **When \( x = 2 \)**:
\[ y = -\frac{1}{2}(2) + \frac{3}{2} = -1 + \frac{3}{2} = \frac{1}{2} \]
Point: \( (2, \frac{1}{2}) \)

3. **When \( x = 4 \)**:
\[ y = -\frac{1}{2}(4) + \frac{3}{2} = -2 + \frac{3}{2} = -\frac{1}{2} \]
Point: \( (4, -\frac{1}{2}) \)

4. **When \( x = 6 \)**:
\[ y = -\frac{1}{2}(6) + \frac{3}{2} = -3 + \frac{3}{2} = -\frac{3}{2} \]
Point: \( (6, -\frac{3}{2}) \)

### Step 3: Plot the points on the graph
Now we can plot the points \( (0, \frac{3}{2}) \), \( (2, \frac{1}{2}) \), \( (4, -\frac{1}{2}) \), and \( (6, -\frac{3}{2}) \) on a Cartesian plane and draw a straight line through them.

### Step 4: Find \( x_1 \) when \( y = 3 \)
To find \( x_1 \) when \( y = 3 \), we substitute \( y = 3 \) into the original equation:
\[ x + 2(3) - 3 = 0 \]
\[ x + 6 - 3 = 0 \]
\[ x + 3 = 0 \]
\[ x = -3 \]
So, \( x_1 = -3 \).

### Step 5: Find \( x_2 \) when \( y = -2 \)
To find \( x_2 \) when \( y = -2 \), we substitute \( y = -2 \) into the original equation:
\[ x + 2(-2) - 3 = 0 \]
\[ x - 4 - 3 = 0 \]
\[ x - 7 = 0 \]
\[ x = 7 \]
So, \( x_2 = 7 \).

### Final Answers:
- \( x_1 = -3 \) when \( y = 3 \)
- \( x_2 = 7 \) when \( y = -2 \)

---