Best

$b+c =8$ ; $a+c = 9$

$\rightarrow (b+c)+(a+c)=8+9$

But $a+b=7$

$\rightarrow 7+2c=17 \rightarrow c=5$

Let's use Cramer's Rule to solve for $c$ in this 3x3 system:

$D = \begin{vmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{vmatrix} = 2$ ;

$D_{c} = \begin{vmatrix} 1 & 1 & 7 \\ 0 & 1 & 8\\ 1 & 0 & 9 \end{vmatrix} = 10$ ;

Thus, $c = \frac{D_{c}}{D} = \boxed{5}.$

Now I add the three equations.

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

2*(a+b+c)=24

Divide the equation by 2.

Equation A a+b+c=12

Subtract Equation 1 from Equation A.

c=12-7

c=5

[a+b] - [c+b]=7-8, a-c = -1, a=-1+c Putting the value of a in c+a=9, -1+c + c =9, 2c-1 =9, 2c =10, c=5

here, c+a=9 so, a=9-c & from c+b=8 , b=8-c so, put this value in a+b=7 we get c=5.