Proof for mathd_algebra_513

[aleph-prover-test]

Proven lemmas: 1/1

The goal is to prove that if real numbers a, b ∈ ℝ satisfy 3a + 2b = 5 and a + b = 2, then necessarily a = 1 and b = 1. The proof naturally splits into two subgoals because the conclusion is a conjunction: first prove a = 1, then prove b = 1.

At this point, both subgoals are solved, so progress is 2 out of 2 complete. The key idea is that each equality is a direct linear consequence of the two given equations. For example, subtracting 2(a + b = 2) from 3a + 2b = 5 gives a = 1, and then substituting into a + b = 2 yields b = 1.

In Lean, this is handled very cleanly by splitting the conjunction and letting linear arithmetic solve both parts automatically. There do not appear to be any remaining gaps or active difficulties: the theorem is fully proved. The main strategy is simply recognizing the system as a pair of linear equations with a unique solution.