Question 388218
Okay this is what I got:
let a = articles/hr done by A
b = articles/hr done by B.
Let x = # hrs done by A in the morning BEFORE B joined him.
Then:
ax = 1500  <------(1)
(a+b)(4-x) = 4000 <------(2) 
4a + 2b = 5200, which is the same as    2a + b = 2600<---(3)
We got a system of non-linear equations.
From eq'n (3),  b = 2600 - 2a
From eq'n(1), x = 1500/a.  Plugging both of these into (2), we get
{{{(a+2600-2a)(4 - 1500/a) = 4000}}};
{{{(2600 - a)(4-1500/a) = 4000}}};
{{{10400- 3900000/a - 4a + 1500 = 4000}}};
{{{-3900000/a - 4a = -7900}}};
{{{975000/a+a = 1975}}}.  This is equivalent to 
{{{a^2- 1975a+975000 = 0}}} after multiplying both sides by a and transposition.
(a - 1000)(a - 975) = 0----> a = 1000, or 975.

If a = 1000, b = 600, and x = 1.5 hrs.
If a = 975, b = 650, and x = 20/13 hrs.

So we actually have 2 possible scenarios here, both being viable.
B worked for 4 - x = 4 - 1.5 = 2.5 hours in the morning, OR, equally possible,
4 - 20/13 = 32/13 hours in the morning.