Proof for mathd_algebra_513 - #64
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Proven lemmas: 1/1
The goal is to prove that if real numbers a and b satisfy 3a + 2b = 5 and a + b = 2, then necessarily a = 1 and b = 1. In other words, this is a small linear-system uniqueness argument over ℝ.
The proof was naturally decomposed into two subgoals because the conclusion is a conjunction: first show a = 1, then show b = 1. Both subgoals have already been solved.
So the progress is complete: 2 out of 2 sub-problems are finished, and the theorem is proved. The key idea is to use the two linear equations together: 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 can be handled very cleanly by splitting the conjunction and letting linarith solve each linear arithmetic goal. There do not appear to be any remaining obstacles; this is a straightforward but tidy application of linear equation solving.