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ax+by=1bx+ay=(a+b)²/a²+b²-1 |
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Answer» ax + by = 1---(1) bx + ay = \(\frac{(a+b)^2}{a^2+b^2-1}\)---(2) ∴ a2x + aby = a - b2x\(\pm\)aby = \(-\frac{(a+b)^2b}{a^2+b^2-1}\) (a2 - b2)x = a \(-\frac{(a+b)^2b}{a^2+b^2-1}\) = \(\frac{a^3+ab^2-a-a^2b-b^3-2ab^2}{a^2+b^2-1}\) = \(\frac{a^3-ab^2-a^2b-b^3-a}{a^2+b^2-1}\) ∴ x = \(\frac{a(a^2-b^2)-b(a^2+b^2)-a}{(a^2-b^2)(a^2+b^2-1)}\) ∴ y = \(\frac{1-ax}b\) = \(\frac1b(1-\frac{a^4-a^2b^2-a^3b-ab^3-a^2}{(a^2-b^2)(a^2+b^2-1)})\) = \(\frac1b(\frac{a^4+a^2b^2-a^2-a^2b^2-b^4+b^2-a^4+a^2b^2+a^3b+ab^3+a^2}{(a^2-b^2)(a^2+b^2-1)})\) = \(\frac{-b^4+ab^3+a^2b^2+a^3b+b^2}{(a^2-b^2)b(a^2+b^2-1)}\) Hence, x = \(\frac{a^3-ab^2-a^2b-b^3-a}{(a^2-b^2)(a^2+b^2-1)}\) and y = \(\frac{-b^4+ab^3+a^2b^2+a^3b+b^2}{(a^2-b^2)b(a^2+b^2-1)}\) |
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