Question

. The prime numbers p, q, r satisfy the simultaneous equations

Answer 1

$first\ equation :\\pq+pr=80\\pq=80-pr..........[1]\\\\secon\ equation :\\pq+qr=425..........[2]$

$Now\ substituting\ equation\ 1\ into\ 2:\\\\pq+qr=425\\(80-pr)+qr=425\\-pr+qr=425-80\\-pr+qr=345\\r(q-p)=345\\\\we\ have :\\ p(q+r)=80=2^{4} \times 5,\\q(p+r)=425=5^{2} \times 17\\r(q-p)=345=3 \times 5 \times 23\\\\so\ : p = 2\ or\ 5, ~q=5\ or\ 17,~ r = 3,5,or\ 23.$

$From\ : p(q+r)=80, if~p=5~then\ q+r=16, which\ has\ no\ solution\\So~~p=2, then\ ~q+r=40, hence~~q=17~~and~~r=23.\\\\\therefore~~ p+q+r~~~is~~~\boxed{2+17+23=42}$

Answer 2

Pq + pr = 80  => p (q+r) = 80
pq + qr = 425 => q (p+r ) = 425

Factorize 80 with having one factor as prime number
80 =  2 * 40 = 5 *16 

Factrorize 425 with having one factor as prime number
425 = 5 * 85 = 17 * 25...

Therfore, p has to be 2.
q = 17 and r= 23...
p + q +r = 2 + 17+ 23 = 42