Find the intercepts cuts on `X`-axis and `Y` -axis from the following lines :
`(i) 3x+4y=12` `(ii) 2x-5y=8`
`(iii) x+2y+3=0` `(iv) 2x-y+3=0`

Answer

To find the intercepts on the x-axis and y-axis for the given lines, we will follow these steps for each equation:

### (i) For the line \(3x + 4y = 12\):

1. **Find the x-intercept**:
- Set \(y = 0\) in the equation.
- \(3x + 4(0) = 12\)
- \(3x = 12\)
- \(x = \frac{12}{3} = 4\)
- **X-intercept**: \( (4, 0) \)

2. **Find the y-intercept**:
- Set \(x = 0\) in the equation.
- \(3(0) + 4y = 12\)
- \(4y = 12\)
- \(y = \frac{12}{4} = 3\)
- **Y-intercept**: \( (0, 3) \)

### (ii) For the line \(2x - 5y = 8\):

1. **Find the x-intercept**:
- Set \(y = 0\) in the equation.
- \(2x - 5(0) = 8\)
- \(2x = 8\)
- \(x = \frac{8}{2} = 4\)
- **X-intercept**: \( (4, 0) \)

2. **Find the y-intercept**:
- Set \(x = 0\) in the equation.
- \(2(0) - 5y = 8\)
- \(-5y = 8\)
- \(y = \frac{8}{-5} = -\frac{8}{5}\)
- **Y-intercept**: \( (0, -\frac{8}{5}) \)

### (iii) For the line \(x + 2y + 3 = 0\):

1. **Find the x-intercept**:
- Set \(y = 0\) in the equation.
- \(x + 2(0) + 3 = 0\)
- \(x + 3 = 0\)
- \(x = -3\)
- **X-intercept**: \( (-3, 0) \)

2. **Find the y-intercept**:
- Set \(x = 0\) in the equation.
- \(0 + 2y + 3 = 0\)
- \(2y + 3 = 0\)
- \(2y = -3\)
- \(y = -\frac{3}{2}\)
- **Y-intercept**: \( (0, -\frac{3}{2}) \)

### (iv) For the line \(2x - y + 3 = 0\):

1. **Find the x-intercept**:
- Set \(y = 0\) in the equation.
- \(2x - 0 + 3 = 0\)
- \(2x + 3 = 0\)
- \(2x = -3\)
- \(x = -\frac{3}{2}\)
- **X-intercept**: \( (-\frac{3}{2}, 0) \)

2. **Find the y-intercept**:
- Set \(x = 0\) in the equation.
- \(2(0) - y + 3 = 0\)
- \(-y + 3 = 0\)
- \(y = 3\)
- **Y-intercept**: \( (0, 3) \)

### Summary of Intercepts:
1. For \(3x + 4y = 12\): X-intercept = 4, Y-intercept = 3
2. For \(2x - 5y = 8\): X-intercept = 4, Y-intercept = -\(\frac{8}{5}\)
3. For \(x + 2y + 3 = 0\): X-intercept = -3, Y-intercept = -\(\frac{3}{2}\)
4. For \(2x - y + 3 = 0\): X-intercept = -\(\frac{3}{2}\), Y-intercept = 3