Which of the following could be the combined age of all four people today? 
Joan, Kylie, Lillian, and Miriam all celebrate their birthdays today.
  Joan is 2 years younger than Kylie, Kylie is 3 years older than
  Lillian, and Miriam is one year older than Joan. Which of the
  following could be the combined age of all four people today?

So it took me a long time to realize that everyone's age was consecutive.
The order goes:
L -> J -> M -> K
So L is the youngest person and everyone's age can be expressed as a function of his age. L + L + 1 + L + 2 + L + 3 = 4L+ 6. So their combined age must be 6 greater than a multiple of 4.
the choices were:
- 51
- 52
- 53
- 54
- 55
So the answer is 54. 
Is there a way to solve this problem using algebra and not coming to a realization that the ages were consecutive? Why not? Or Why?
 A: Let $J,K,L,M$ have their obvious meanings. Then, $K=J+2$, $L=K−3=J−1$, and $M=J+1$. And so 
$$
J+K+L+M=J+(J+2)+(J−1)+J+1=4J+2\equiv 2\pmod{4},
$$
leading to $54$ as the answer.
A: 
Is there a way to solve this problem using algebra and not coming to a realization that the ages were consecutive?

If I understand correctly, this question asks if there's a way to determine that $J + K + L + M = 4L + 6$ without relying on the realization that $J, K, L, M$ are consecutive natural numbers.
You can show this equality algebraically. First, construct a system of equations reflecting parts of the question:

Joan is 2 years younger than Kylie

$J = K - 2$.
$\therefore 2 = K - J$.

Kylie is 3 years older than Lillian

$K = L + 3$.
$\therefore 3 = K - L$.

Miriam is one year older than Joan

$M = J + 1$.
$\therefore 1 = M - J$.
We have four unknowns in a system of 3 linear equations. Because these three equations are linearly independent, the system has one free variable ($L$). We can find $J, K, M$ in terms of $L$ by performing row reduction on the augmented matrix for the system of equations we identified above:
$\begin{bmatrix}-1&1&0&0&2\\0&1&-1&0&3\\-1&0&0&1&1\end{bmatrix}$ can be row reduced to $\begin{bmatrix}1&0&-1&0&1\\0&1&-1&0&3\\0&0&-1&1&2\end{bmatrix}$. This expresses $J, K, M$ in terms of $L$:
$J - L = 1. \therefore J = 1 + L.$
$K - L = 3. \therefore K = 3 + L.$
$M - L = 2. \therefore M = 2 + L.$
$\therefore J + K + L + M = (1+L) + (3+L) + L + (2+L) = 4L + 6.$
This is a really roundabout way to give three of our variables in terms of the fourth (and from those definitions conclude they're consecutive), but generating a system of linear equations and algebraically manipulating them (via row reduction in an augmented matrix or otherwise) is a useful general strategy for more complicated cases than this problem.
