21b-32 +7b-20=0 then b=?
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2mn - (m2 - nZ) = 7 can be rewritten (n + m)2 - 2m 2 = 7, and ... Since m is thus the largest of a Pythagorean triple, we can let m = y (a2 + f32), ... b were odd, then ( a - 21b)2 + b2 == 3 (mod 4), again an impossibility.
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21b - 32 + 7b - 20 = 0
21b+7b-32-20 = 0
28b-52 = 0
28b = 52
b = 52/28 ✓✓✓[
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