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Algebra Difficulty 4.8 AIME Find the answer

Solve the system of equations p+3q+r=3p+3q+r=3, p+2q+3r=3p+2q+3r=3, p+q+r=2p+q+r=2 for the ordered triple (p,q,r)(p, q, r).

A number or a short expression. Spacing and $ signs are ignored.

Solution

We can rewrite the equation in terms of ln⁡2,ln⁡3,ln⁡5\ln 2, \ln 3, \ln 5, to get 3ln⁡2+3ln⁡3+2ln⁡5=ln⁡5400=px+qy+rz=(p+3q+r)ln⁡2+(p+2q+3r)ln⁡3+(p+q+r)ln⁡53 \ln 2+3 \ln 3+2 \ln 5=\ln 5400=p x+q y+r z=(p+3 q+r) \ln 2+(p+2 q+3 r) \ln 3+(p+q+r) \ln 5. Consequently, since p,q,rp, q, r are rational we want to solve the system of equations p+3q+r=3,p+2q+3r=3,p+q+r=2p+3 q+r=3, p+2 q+3 r=3, p+q+r=2, which results in the ordered triple (54,12,14)\left(\frac{5}{4}, \frac{1}{2}, \frac{1}{4}\right).

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