### Instruction: Let $A,$ $B,$ and $C$ be constants such that the equation \[\frac{(x+B)(Ax+28)}{(x+C)(x+7)} = 2\]has infinitely many solutions for $x.$ For these values of $A,$ $B,$ and $C,$ it turns out that there are only finitely many values of $x$ which are not solutions to the equation. Find the sum of these values of $x.$.### Response: I notice that the equation has rational functions on both sides, so I can try to cross-multiply and simplify it.
Multiplying both sides by $(x+C)(x+7),$ I get \[(x+B)(Ax+28) = 2(x+C)(x+7).\]
Expanding the brackets, I get \[A x^2 + (A B + 28) x + 28 B = 2 x^2 + (14 + 2 C) x + 14 C.\]
Rearranging the terms, I get \[(A - 2) x^2 + (A B + 14 - 2 C) x + (28 B - 14 C) = 0.\]
This is a quadratic equation in $x,$ so it has at most two solutions.
However, the problem says that the original equation has infinitely many solutions for $x.$
That means that this quadratic equation must be identically zero for all values of $x.$
That can only happen if the coefficients of $x^2,$ $x,$ and the constant term are all zero.
So I have the following system of equations: \[\begin{aligned} A - 2 &= 0 \\ A B + 14 - 2 C &= 0 \\ 28 B - 14 C &= 0 \end{aligned}\]
Solving for $A,$ I get $A = 2.$
Plugging that into the second equation, I get $2 B + 14 - 2 C = 0,$ or $B + 7 - C = 0.$
Plugging that into the third equation, I get $28 B - 14 C = 0,$ or $2 B - C = 0.$
Subtracting the second equation from the third, I get $B = 7.$
Plugging that into the third equation, I get $C = 14.$
So I have found the values of $A,$ $B,$ and $C$ that make the equation have infinitely many solutions: $A = 2,$ $B = 7,$ and $C = 14.$
Now I need to find the values of $x$ that are not solutions to the equation.
Since the equation is equivalent to the quadratic equation \[0 x^2 + 0 x + 0 = 0,\]the only values of $x$ that are not solutions are those that make the denominator of the original equation zero.
The denominator is $(x+C)(x+7),$ so the values of $x$ that make it zero are $x = -C$ and $x = -7.$
Therefore, the sum of these values of $x$ is $-C + (-7) = -14 + (-7) = -21.$