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Solve for $a,b,c$: \begin{align} 2ab+a+2b=24\\ 2bc+b+c=52\\ 2ac+2c+a=74\\ \end{align}

Solving them simultaneously is leading to very difficult situation. Plz help.

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what do you mean by simultaneosly? do you just want t o solve it or to you wan to do something special ? – Dominic Michaelis Feb 22 at 9:02
yeah i mean when i am trying to solve for a, b and c, it is leading to a very complex situation. – user62947 Feb 22 at 9:03
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Also, the solution looks rather complicated wolframalpha.com/input/… – macydanim Feb 22 at 9:07
exactly @macydanim – user62947 Feb 22 at 9:09
Note: This is a problem from Brilliant.org, and MJ96 wrongly quoted the numerical value in the first line. He has been constantly posting problems from our site (4th now). – Calvin Lin Feb 22 at 14:27
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3 Answers

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At first we gonna determine $a$, at first lets say $b\neq -2$ $$ 2 ab+ a + 2b=24 \iff a(2+b)+2b =24 \iff a=\frac{24-2b}{2+b}$$ afterwards do the same on the second equation, and you will be able to express the third only with terms of $c$.

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Thank You @Dominic , that will definitely help. :)) – user62947 Feb 22 at 9:08
oh wait there is a typo in the last equaton it has to be a=... i corrected it now – Dominic Michaelis Feb 22 at 9:08
if i solve my question starting with assuming a=x^2, b=y^2, c=z^2, and then try to find x,y,z; as i have to find (abc), so then i would find (x^2)(y^2)(z^2), maybe that would be easier – user62947 Feb 22 at 9:21
i don't know what it would make easier – Dominic Michaelis Feb 22 at 9:40
up vote 2 down vote

\begin{align} (a+1)(2b+1)&=(2ab+a+2b)+1=25\\ (2b+1)(2c+1)&=2(bc+b+c)+1=105\\ (2c+1)(a+1)&=(2ca+2c+a)+1=75 \end{align} So put $u=a+1,v=2b+1,w=2c+1$ and I think you'll get the answer. (Hint: consider $u^2=(uv)(uw)/(vw)$, et cetera)

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Here is a partial answer: User the first equation and third equation (add and substract) to obtain \begin{align} c-b&=\frac{25}{a+1} \\ c+b&=\frac{49-a}{a+1} \end{align} Adding and subtracting should again give \begin{align} c=&\frac{1}{2}\frac{74-a}{a+1} \\ b=&\frac{1}{2}\frac{24-a}{a+1} \end{align} Substituting this into the second equation should give a equation in $a$ alone which looks to be quartic.

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