Problem No. 43

The system of equations can be rewritten as

$(a + 1)(b + 1) = 16, (b + 1)(c + 1) = 100, (a + 1)(c + 1) = 400.$

Thus $\dfrac{(a + 1)(c + 1)}{(a + 1)(b + 1)} = \dfrac{400}{16} = 25 \Longrightarrow (c + 1) = 25(b + 1).$

So then $(b + 1)(c + 1) = 25(b + 1)^{2} = 100 \Longrightarrow (b + 1)^{2} = 4 \Longrightarrow b = 1,$

since we want $a,b,c$ positive. (The other solution is $b = -3$ .)

So with $b = 1$ we have $(a + 1)(1 + 1) = 16 \Longrightarrow a = 7,$ and

$c + 1 = 25(1 + 1) = 50 \Longrightarrow c = 49.$

Thus $a + b + c + bc = 7 + 1 + 49 + 1*49 = \boxed{106}.$

a+b+ab=15 has two solutions i.e 1st) considering the a=3, and b=3 but that won't satisfies the other two equations 2nd) considering a=7 and b=1 which will surely satisfies both equations so the value of c=49 and the answer will be '' 106 ''.