The number of integral value of m so that the abscissa of point of intersection of lines 3x+4y=9 and y=mx+1 is also an integers is

Answer

To solve the problem, we need to find the number of integral values of \( m \) such that the abscissa (x-coordinate) of the point of intersection of the lines \( 3x + 4y = 9 \) and \( y = mx + 1 \) is an integer.

### Step-by-Step Solution:

1. **Set the equations equal to find the intersection**:
We have the two equations:
\[ 3x + 4y = 9 \quad \text{(1)} \]
\[ y = mx + 1 \quad \text{(2)} \]
Substitute equation (2) into equation (1):
\[ 3x + 4(mx + 1) = 9 \]

2. **Simplify the equation**:
Expanding the left-hand side:
\[ 3x + 4mx + 4 = 9 \]
Combine like terms:
\[ (3 + 4m)x + 4 = 9 \]
Rearranging gives:
\[ (3 + 4m)x = 5 \]
Therefore, we can express \( x \) as:
\[ x = \frac{5}{3 + 4m} \quad \text{(3)} \]

3. **Determine when \( x \) is an integer**:
For \( x \) to be an integer, \( 3 + 4m \) must be a divisor of 5. The divisors of 5 are \( \pm 1, \pm 5 \).

4. **Set up equations for each divisor**:
- Case 1: \( 3 + 4m = 1 \)
\[ 4m = 1 - 3 \implies 4m = -2 \implies m = -\frac{1}{2} \]
- Case 2: \( 3 + 4m = -1 \)
\[ 4m = -1 - 3 \implies 4m = -4 \implies m = -1 \]
- Case 3: \( 3 + 4m = 5 \)
\[ 4m = 5 - 3 \implies 4m = 2 \implies m = \frac{1}{2} \]
- Case 4: \( 3 + 4m = -5 \)
\[ 4m = -5 - 3 \implies 4m = -8 \implies m = -2 \]

5. **List the integral values of \( m \)**:
From the cases above, we find:
- \( m = -\frac{1}{2} \) (not an integer)
- \( m = -1 \) (integer)
- \( m = \frac{1}{2} \) (not an integer)
- \( m = -2 \) (integer)

The integral values of \( m \) are \( -1 \) and \( -2 \).

6. **Count the integral values**:
Thus, the total number of integral values of \( m \) is:
\[ \text{Total integral values} = 2 \]

### Final Answer:
The number of integral values of \( m \) such that the abscissa of the point of intersection is an integer is **2**.