Reduce the following equations into intercept form and find their intercepts on the axes.
`(i)3x+2y-12=0`
`(ii) 6x+3y-5=0`
`(iii) 3y+2=0`

Answer

To reduce the given equations into intercept form and find their intercepts on the axes, we will follow the steps below:

### Step-by-Step Solution

**(i) For the equation \(3x + 2y - 12 = 0\)**

1. **Rearranging the equation**:
\[ 3x + 2y = 12 \]

2. **Dividing the entire equation by 12**:
\[ \frac{3x}{12} + \frac{2y}{12} = 1 \]

3. **Simplifying the fractions**:
\[ \frac{x}{4} + \frac{y}{6} = 1 \]

4. **Identifying the intercepts**:
- The x-intercept \(a\) is 4 (when \(y = 0\)).
- The y-intercept \(b\) is 6 (when \(x = 0\)).

**Intercepts**:
- x-intercept = 4
- y-intercept = 6

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**(ii) For the equation \(6x + 3y - 5 = 0\)**

1. **Rearranging the equation**:
\[ 6x + 3y = 5 \]

2. **Dividing the entire equation by 5**:
\[ \frac{6x}{5} + \frac{3y}{5} = 1 \]

3. **Simplifying the fractions**:
\[ \frac{x}{\frac{5}{6}} + \frac{y}{\frac{5}{3}} = 1 \]

4. **Identifying the intercepts**:
- The x-intercept \(a\) is \(\frac{5}{6}\) (when \(y = 0\)).
- The y-intercept \(b\) is \(\frac{5}{3}\) (when \(x = 0\)).

**Intercepts**:
- x-intercept = \(\frac{5}{6}\)
- y-intercept = \(\frac{5}{3}\)

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**(iii) For the equation \(3y + 2 = 0\)**

1. **Rearranging the equation**:
\[ 3y = -2 \]

2. **Solving for \(y\)**:
\[ y = -\frac{2}{3} \]

3. **Identifying the intercepts**:
- Since there is no \(x\) term, the x-intercept is 0 (the line is vertical).
- The y-intercept \(b\) is \(-\frac{2}{3}\) (the point where the line crosses the y-axis).

**Intercepts**:
- x-intercept = 0
- y-intercept = \(-\frac{2}{3}\)

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### Summary of Intercepts

1. For \(3x + 2y - 12 = 0\):
- x-intercept = 4
- y-intercept = 6

2. For \(6x + 3y - 5 = 0\):
- x-intercept = \(\frac{5}{6}\)
- y-intercept = \(\frac{5}{3}\)

3. For \(3y + 2 = 0\):
- x-intercept = 0
- y-intercept = \(-\frac{2}{3}\)

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