How do I solve this?

[uneven hull]

The straight line y = kx + m passes through the point (4, 3) and intersects the line 3y + 5x - 15 = 0 at a point where both the x and y coordinates are positive numbers. Investigate what values of k are possible.

[alpine lance]

!status

[robust quail]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[uneven hull]

Number 2. I think I should start converting 3y + 5x - 15 = 0 to y = 5 - (5/3)x but not sure what to do next

[alpine lance]

That would be fine. If you have two line equations $y=l_1(x)$ and $y=l_2(x)$, then their graphs interect where their $y$ values become equal. That is, they interesct when $l_1(x)=l_2(x)$. From that, you'll need to alegbarically solve for $x$. That result will be the $x$-coordinate of the intersection. You can find the $y$ coordinate simply by plugging in your $x$ answer into either $l_1(x)$ or $l_2(x)$ (we set them equal, so it doesn't matter which you choose).

[mortal sigil]

SWR

[alpine lance]

You also know the line $y=kx+m$ passes through the point $(x=4, y=3)$, so $k$ and $m$ must also satisfy $3=k(4)+m$

[mortal sigil]

SWR

[uneven hull]

So what's the next step for a solution?

[alpine lance]

Exactly what I just all described to you

[uneven hull]

Oh wait my bad. And what is done after that?

[alpine lance]

You have solutions for x and y in terms of k. So you want to find which values of k will keep both positive.

[uneven hull]

And do I just test my way to find that?