Proof for mathd_algebra_513 - #197
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Proven lemmas: 1/1
The goal is to prove that if real numbers a and b satisfy the linear system 3a + 2b = 5 and a + b = 2, then necessarily a = 1 and b = 1. Mathematically, this is a straightforward 2×2 system over ℝ with a unique solution.
The statement was naturally decomposed into two subgoals: prove a = 1 and prove b = 1, since the target is a conjunction a = 1 ∧ b = 1. Both parts are linear consequences of the two given equations.
Progress is complete: 1 theorem out of 1 has been solved, and both subgoals are finished. In Lean, the proof uses constructor to split the conjunction and then linarith to solve each equality directly from the hypotheses.
There is nothing remaining to prove for this theorem. An interesting aspect is that although the automated linear arithmetic tactic finishes it immediately, there are also clear manual strategies: subtract 2(a + b = 2) from 3a + 2b = 5 to get a = 1, then substitute into a + b = 2 to get b = 1; or solve for one variable first and substitute back.