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$ab + a + b = 250$

$bc + b + c = 300$

$ac + a + c = 216$

then find $a + b + c = ?$


MY APPROACH:

(i) * c , (ii) * a , (iii) * b then we get

$abc + ac + bc = 250c$

$abc + ab + ac = 300a$

$abc + ab + bc = 216b$

(iv)+(v)+(vi)

$3abc + 2ac +2ab + 2bc = 300a + 216b +250c$

now i cant solve it ?? how to solve??

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    $\begingroup$ $$ (a+1)(b+1) = 251, \; (b+1)(c+1)=301, \; (c+1)(a+1)= 217. $$ $\endgroup$
    – Will Jagy
    Mar 1, 2014 at 21:58
  • $\begingroup$ how??? did not understand...please explain $\endgroup$ Mar 1, 2014 at 22:00
  • $\begingroup$ You can certainly verify, by for example expanding $(a+1)(b+1)$, that the first result is correct. $\endgroup$ Mar 1, 2014 at 22:02
  • $\begingroup$ Will added $1$ to both sides of each equation, then factored each left side. $\endgroup$ Mar 1, 2014 at 23:02

1 Answer 1

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\begin{cases} ab+a+b=250 \\ bc+b+c=300 \\ ac+a+c=216 \\ \end{cases} it seems of no difficulty: a system of 3 equations and 3 variables... Let's start by adding 1 to every equations, we obtain: \begin{cases} ab+a+b+1=(a+1)(b+1)=251 \\ bc+b+c+1=(c+1)(b+1)=301 \\ ac+a+c+1=(a+1)(c+1)=217 \\ \end{cases} Now let's substitute $a+1 = x, b+1 =y, c+1 = z$, so we can reduce the amount of calculus needed, in fact to solve \begin{cases} xy=251 \\ yz=301 \\ xz=217 \\ \end{cases} you only need to find $x$ (or $y$) in the first equation, substitute it in the last (or second) equation, and then do another substitution. Then subtract 1, and you'll find the solutions:

$(a,b,c) = (-1-\sqrt{\frac{7781}{43}}, -1-\sqrt{\frac{10793}{31}}, -1-7\sqrt{\frac{1333}{251}})$

and

$(a,b,c) = (\sqrt{\frac{7781}{43}}-1, \sqrt{\frac{10793}{31}}-1, 7\sqrt{\frac{1333}{251}}-1)$

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  • $\begingroup$ Author is asking how to solve it, not what the solution is, so it seems this doesn't engage with the question. $\endgroup$ Mar 1, 2014 at 23:00
  • $\begingroup$ oook, understood! $\endgroup$
    – sirfoga
    Mar 1, 2014 at 23:04

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