sjb is correct, your answer is wrong.

However, I do not like the two equation approach for this because it puts more emphasis on math and not thinking about what is going on behind it, in my opinion. But yes, math is required.

You have a mixture of two components, producing the same thing. Think about it at a molar level first.

1 mol Na_{2}CO_{3} + 1 mol MgCO_{3} -> 2 mol CO_{2}

You know the value for the total mols of CO2 produced (and I would use more significant figures in your calculations and do not round until the end, it is off a decent amount).

So you know

x mols Na_{2}CO_{3} + y mols MgCO_{3} -> 0.084 mols CO_{2}

Correct?

But you only know the mass, but how can you relate mass to mols?

Molecular Weight!

So, call one arbitrarily x and the other 7.63-x (because you know you have 7.63 grams total)

So now you have

x g Na_{2}CO_{3} + 7.63 -x MgCO_{3} = 0.084

105.99 g Na_{2}CO_{3} /mol 84.32 g / MgCO_{3}

Which the units cancel and leave mols + mols = 2 mols, which is what your balanced equation said.

You use the molecular weight to relate the mass to the amount of mols, as dictated by stoichmetry, directly into the balanced equation with the appropriate coefficient. It is no longer a system of equations and just a math problem, but the math is now displaying the chemical relationships.

Remember, balanced equation relates mols, not mass!

Now it is just the basic algebra and solving for x, and once you know x you know the other as well.

Hope that makes sense and helps.