Hi, hope this helps:
\(2a+3b=52 \) (1)
\(3b+c=41\) (2)
\(bc=60\) (3)
Consider (2) and (3):
From (3): \(c=\dfrac{60}{b}\), which when we substitute into (2) gives:
\(3b+\dfrac{60}{b}=41\) Multiply by b
\(3b^2-41b+60=0\)
This can be factorised or you can use the Quadratic formula.
\(3b^2-41b+60=0 \iff (3b-5)(b-12)=0\)
Given that a,b,c are integers, then we only accept \(b=12\).
Now from (3): \(12*c=60 \implies c=5\)
Then from (1): \(2a+3(12)=52 \implies a=8\)
So, \(a+b+c=8+12+5=25\)
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