Precalculus help please?

Question

this is the question:

given the equations of two lines; p(x)=mx+c and q(x)=nx+d, where m>0, n<0, and c and d are real numbers, where d>c. in which quadrants will the solution to p(x)=q(x) be located. think graphically and explain your answer.

ive already figured out that it will be in quadrants 1 and 4, but i figured it out doing the guess and check method. i dont know how to explain it graphically?!!

Answer 1

Where the graphs meet:

p(x) = q(x)

mx + c = nx + d

(m - n)x = d - c

As m &gt; 0 and n &lt; 0, m - n &gt; 0.

As d &gt; c, d - c &gt; 0

x is therefore positive, ruling out quadrants 2 and 3.

The y intercept of y = nx + d is above that of y = mx + c, but they could be anywhere along the vertical axis, and so either of quadrants 1 and 4 is possible.