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The solutions of the equations 3x + 5Y = m + 2,2x + 3Y = m are suitable for the equations x + y = 2, and the values of M & sup2; - 2m + 1 are obtained
I'm not sure, right···
Let 3x + 5Y = m + 2 be: 1
2X + 3Y = m is: 2
① - 2, x + 2Y = 2 is: 3
Let x + y = 2 be 4
③ - 4, we get that y = 0 x = 2, y = 0 is the solution of the original equation
Substituting x = 2, y = 0 into the original equation, M = 4 is obtained
When m = 4
The original formula is M & sup2; - 2m + 1
=9
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