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ahora : f(u) ≡a(u-6)+6
para, u≡ 2 , tenemos : f(2)≡a(2-6) + 6 ≡-4a+b.........(1)
para u≡-3 , tenemos : f(-3) ≡a(-3 -(-6 )) ≡ -9a+b.......(2)
⇒ de (1) y(2) : -4a+b≡ -14
-9a+b≡-29
⇒ Resolviendo estas dos ecuaciones : a ≡ 3 ∧ b≡-2
∴2a - b ≡2(3) - (-2) ≡8