# Solving system of linear congruences (3 pairs)

So we have the following:

$$2x \equiv 3\pmod {5} \\ 3x \equiv 4\pmod {7} \\ 5x \equiv 7\pmod {11}$$

which reduces to:

$$x \equiv 4\pmod {5} \\ x \equiv 6\pmod {7} \\ x \equiv 8\pmod {11}$$

Now the confusion begins here. At this point, I choose the first two pairs of congruences and equate them, giving:

$$5k+4= 7l +6 \\ \\$$

But I'm not sure what to do past this point. I know in essence I need to solve this and pair this new equation with the last one and re-do the steps. It's just past this point I don't know how to solve.