Question

If the x-coordinate of the point of intersection of the lines x - y + 3 = 0 and mx + 2y + 1 = 0 is integer, then the number of all possible values of m are

• A. 2
• B. 4
• C. 5
• D. 3

Answer 1

Answer: (B)

4

Answer 2

We are given two lines:

1. $x - y + 3 = 0$
2. $mx + 2y + 1 = 0$

From the first equation, we can express $y$ in terms of $x$: $y = x + 3$

Substitute this expression for $y$ into the second equation: $mx + 2(x + 3) + 1 = 0$ $mx + 2x + 6 + 1 = 0$ $x(m + 2) + 7 = 0$

For the lines to intersect, they must not be parallel. The slope of the first line is 1, and the slope of the second line is $-m/2$. They are parallel if $1 = -m/2$, which means $m = -2$. If $m = -2$, the equation becomes $x(0) + 7 = 0$, which is $7 = 0$, a contradiction. Thus, $m \neq -2$ for an intersection point to exist.

Solving for $x$, we get: $x(m + 2) = -7$ $x = \frac{-7}{m + 2}$

The problem states that the x-coordinate of the point of intersection is an integer. Therefore, $\frac{-7}{m + 2}$ must be an integer. This implies that $(m + 2)$ must be an integer divisor of $-7$.

The integer divisors of $-7$ are $1, -1, 7, -7$.

The question asks for "the number of all possible values of m". Given the finite integer options, it is implied that $m$ must be an integer. If $m$ is an integer, then $m+2$ is also an integer.

We set $m+2$ equal to each of the divisors of $-7$:

1. $m + 2 = 1 \implies m = 1 - 2 = -1$
2. $m + 2 = -1 \implies m = -1 - 2 = -3$
3. $m + 2 = 7 \implies m = 7 - 2 = 5$
4. $m + 2 = -7 \implies m = -7 - 2 = -9$

All these values of $m$ ($-1, -3, 5, -9$) are integers, and none of them is equal to $-2$, ensuring that the lines intersect. Each of these values of $m$ results in an integer value for $x$:

• If $m=-1$, $x = -7/(-1+2) = -7/1 = -7$.
• If $m=-3$, $x = -7/(-3+2) = -7/(-1) = 7$.
• If $m=5$, $x = -7/(5+2) = -7/7 = -1$.
• If $m=-9$, $x = -7/(-9+2) = -7/(-7) = 1$.

There are 4 distinct integer values for $m$.

The final answer is $\boxed{4}$.