Question 1074078
We can write two equations for the common ratio:
r = 6/(3-x)  [1]
r = (7-5x)/6 [2]
We have two equations in two unknowns. Let's solve for x first.
6/(3-x) = (7-5x)/6 -> 36 = 21 - 22x + 5x^2 -> 5x^2 - 22x - 15 = 0
This can be factored as (5x+3)(x-5) = 0.  The two solutions are x = -3/5, x = 5.
But, x=5 gives r<0, so we choose the 1st solution, x = -3/5.
Substitute in [1] to get the value for r:
r = 6/(3--3/5) = 6/(18/5) = 5/3.
Ans: r = 5/3, x = -3/5