3 equations, 3 variables! part 4

$\begin{cases} 5a + 5b + 2ab = 92 &...(1) \\ 5b + 5c + 2bc = 136 &...(2) \\ 5c + 5a + 2ca = 244 &...(3) \end{cases}$

$\begin{aligned} (1): \quad 5a + 5b + 2ab & = 92 \\ \Rightarrow \frac{5}{2} (a+b) + ab & = 46 \\ ab & = 46 - \frac{5}{2}(a+b) \end{aligned}$

Since $a$ and $b$ are positive integers, then $a+b$ is even.

\(\begin{array} {} \text{If } a + b = 2 & \Rightarrow ab = 1 & \Rightarrow 46 - \frac{5}{2}(a+b) = 41 & \color{red}{\text{No solutions for } a \text{ and } b} \\ \text{If } a + b = 4 & \Rightarrow ab = \{3,4 \} & \Rightarrow 46 - \frac{5}{2}(a+b) = 36 & \color{red}{\text{No solutions for } a \text{ and } b} \\ \text{If } a + b = 6 & \Rightarrow ab = \{6,8,9 \} & \Rightarrow 46 - \frac{5}{2}(a+b) = 31 & \color{red}{\text{No solutions for } a \text{ and } b} \\ \text{If } a + b = 8 & \Rightarrow ab = \{7,12,15,16 \} & \Rightarrow 46 - \frac{5}{2}(a+b) = 26 & \color{red}{\text{No solutions for } a \text{ and } b} \\ \text{If } a + b = 10 & \Rightarrow ab = \{9,16, \color{blue}{21}, 24, 25 \} & \Rightarrow 46 - \frac{5}{2}(a+b) = \color{blue}{21} & \color{blue}{\Rightarrow a, b = 3, 7} \end{array} \)

Comparing $(2)$ and $(3)$ , we note that $b < a$ , therefore, $b = 3$ and $a = 7$ .

$\Rightarrow \begin{cases} (2): &15 + 5c + 6c = 136 & \Rightarrow c = 11 \\ (3): & 5c + 35 + 14c = 244 & \Rightarrow c = 11 \end{cases}$

Therefore, $7a+8b+9c = \boxed{172}$

Quadratic Diophantine Equations - Solve by Simon's Favorite Factoring Trick is sufficient to solve this problem.

Consider the expansion, $(5+2a)(5+2b) = 25 + 10b + 10a + 4ab = 25 + 2(5a + 5b + 2ab) = 25 + 2(92) = 209 .$ Analogously, we can have the following system of equations:

$\begin{aligned} (5+2a)(5+2b) &=& 25 + 2(92) = 209 \\ (5 + 2b)(5 + 2c) &=& 25 + 2(136) = 297 \\ (5 + 2a)(5 + 2c) &=&25 + 2(244) = 513. \end{aligned}$

Multiplying all these equations and take its positive square root:

$(5+ 2a)(5+2b)(5+2c) = 5643$

Solving for $a,b$ and $c$ :

$\begin{aligned} 5 + 2c &=& 5643 \div 209 = 27 \Rightarrow c = 11 \\ 5 + 2a &=& 5643 \div 297 = 19 \Rightarrow a = 7 \\ 5 + 2b &=& 5643 \div 513 = 11 \Rightarrow b = 3. \end{aligned}$

The desired answer is $7a + 8b+ 9c = 7(7) + 8(3) + 9(11) = \boxed{172}$ .

5a+5b+2ab=92 ........》 $\boxed 1$ . 5c+5b+2bc=136 ........》 $\boxed 2$ .5c+5a+2ca=244 ........》 $\boxed 3$ . On subtracting $\boxed 1$ from $\boxed 2$ We get . 5c+5b+2bc-(5a+5b+2ab)=136-92 Or 5(c-a)+2b(c-a)=44
Or (5+2b)(c-a)=11×4 (by factorizing 44 into suitable factors) therefore b=3 & c-a=4 or c=a+4》 $\boxed 4$ _
_
Similarly on subtracting $\boxed 2$ from $\boxed 3$ and factorizing gives
(5+2c)(a-b)=108 . Now on substituting the value of $\boxed 4$ in the above equation we get 2 $a^{2}$ + 7a -147=0
Or (a-7)(2a+21)=0 _for positive value a=7 and therefore c=11(from 4) & required answer of the given expression is $\boxed {172}$ _ $\boxed {Note}$ In the first equation 11×4 is the only expansion that would satisfy the requirements of the question which demands for positive integral value.

$(\sqrt{2}a+5\sqrt{2}/2)(\sqrt{2}b+5\sqrt{2}/2)=92+25/2$

$(\sqrt{2}b+5\sqrt{2}/2)(\sqrt{2}c+5\sqrt{2}/2)=136+25/2$

$(\sqrt{2}c+5\sqrt{2}/2)(\sqrt{2}a+5\sqrt{2}/2)=244+25/2$

$a'b'=209/2$

$b'c'=297/2$

$c'a'=513/2$

$b'=297/(2c')$

$c'=513/(2a')$

$a'297/(2c')=209/2$

$a'297/(2*513/(2a'))=209/2$

$a'297/(513/a')=209/2$

$a'^2=361/2$

$a'=\sqrt{361/2}=19\sqrt{2}/2$

$b'=209/(2*(19\sqrt{2}/2))=209/(19\sqrt{2})=11\sqrt{2}/2$

$c'=513/(2*(19\sqrt{2}/2))=513/(19\sqrt{2})=27\sqrt{2}/2$

$\sqrt{2}a+5\sqrt{2}/2=19\sqrt{2}/2$

$\sqrt{2}b+5\sqrt{2}/2=11\sqrt{2}/2$

$\sqrt{2}c+5\sqrt{2}/2=27\sqrt{2}/2$

$a=7$

$b=3$

$c=11$

$7a+8b+9c=172$