F.2 Maths (Urgent, 20 Points)

p, q, r, and s are positive numbers.
The only requirement is that p must be 1, p = 1

Say p =1, q = 0.3, r =0.4, s = 1.3

To satisfy the first equation: 2p + q + r + s = 4
2(1) + 0.3 + 0.4 + 1.3 = 4
2 + 0.3 + 0.4 + 1.3 = 4
4 = 4

To satisfy the second equation: p + q + r + s = 3
1 + 0.3 + 0.4 + 1.3 = 3
3 = 3

The answer is that it is possible provided that

q + r + s = 2 and p = 1, and 0 < q < 2, 0 < r < 2, 0 < s < 2


Another alternative p =1, q = 0.5, r = 0.8 and s = 0.7
There are almost infinite possibilities.

I may interpret the question wrong.

2P+Q+R+S-4=0
P+Q+R+S-3=0

2P+Q+R+S-4 = P+Q+R+S-3

2P-P=1

P=1