Animated Solution for Mathematics - Straight Lines: The number of integer values of $m$, for which the x-coordinate of the point of intersection of the lines $3x+4y=9$ and $y=mx+1$ is also an integer, is

Imagine you are standing on a coordinate plane, watching two lines dance. The first line, $3x+4y=9$, is a rigid, unmoving structure.

The second line, $y=mx+1$, is a dynamic, shifting entity. Because its $y$-intercept is fixed at $1$, this line acts like a gate, always anchored at the point $(0,1)$, but swinging its slope $m$ like a pendulum.

We want to know how many integer slopes $m$ will cause this swinging line to intersect our fixed line at a point where the $x$-coordinate is a clean, whole number. This is not just algebra; it is a search for harmony between two geometric objects.

To find where these lines meet, we must force them to agree. We have $3x+4y=9$ and $y=mx+1$.

Since we are hunting for the $x$-coordinate, let us eliminate $y$ by substituting the second equation into the first. Replacing $y$ with $(mx+1)$, we get:

Expanding this, we distribute the $4$ to get $3x+4mx+4=9$. Now, we group the $x$ terms: $x(3+4m)=9-4$, which simplifies to $x(3+4m)=5$.

Finally, we isolate $x$ to see the relationship clearly:

This is our master equation. It tells us exactly how the $x$-coordinate behaves based on the slope $m$.

Here is the crux of the problem: we need $x$ to be an integer. For the fraction $\frac{5}{3+4m}$ to be an integer, the denominator $(3+4m)$ must be a perfect divisor of the numerator $5$.

This is the gatekeeper of our solution. Because $5$ is a prime number, its only possible integer divisors are $1,-1,5,$ and $-5$.

We must test each of these four possibilities to see if they yield an integer value for $m$.

Let us test our cases one by one:

Case 1: $3+4m=1$. This leads to $4m=-2$, or $m=-1/2$. Since $-1/2$ is not an integer, we must reject this.

Case 2: $3+4m=-1$. This gives $4m=-4$, so $m=-1$. This is a valid integer! We have found our first winner.

Case 3: $3+4m=5$. This results in $4m=2$, or $m=1/2$. Again, not an integer, so we reject it.

Case 4: $3+4m=-5$. This gives $4m=-8$, so $m=-2$. This is our second valid integer!

We have systematically tested all possible divisors of $5$. We found that only $m=-1$ and $m=-2$ satisfy the condition that $m$ must be an integer.

The beauty of this problem lies in how a simple geometric intersection constraint forces us into the realm of number theory. We started with lines on a graph and ended by analyzing the divisors of a prime number.

There are exactly $2$ such integer values of $m$. You have successfully navigated the trap and found the truth hidden in the algebra.