If the line y = - 2x-1 intersects with the line y = 3x + m at a point in the third quadrant, what is the value range of M

If the line y = - 2x-1 intersects with the line y = 3x + m at a point in the third quadrant, what is the value range of M

(-1,3/2)
Two straight lines are set up simultaneously, and the coordinates of the intersection point are calculated, which are respectively less than 0

Given that y = - 4-2x intersects with straight y = 3x + B in the third quadrant, then the value range of B is--
I have to finish it now

-4

If the line y = - 2x-1 intersects with the line y = 3x + m in the third quadrant, then the value range of M is

The intersection of two straight lines is p ((- m-1) / 5, (2m-3) / 5)
P is in the third quadrant X

It is known that the intersection point of the line y = 2x + 1 and y = 3x + B is in the third quadrant
Otherwise I don't understand

When two points intersect, X and Y in y = 2x + 1 and y = 3x + B are equal
That is, 2x + 1 = 3x + B solves x = 1-B, then y = 3-2b is obtained by y = 2x + 1
Intersection of lines (1-B, 3-2b)
The intersection is in the third quadrant, so the coordinates of the intersection are less than 0
So 1-B