Ax+by=1
bx-ay=a+b
solve in linear equation in 2 variables

Ax+by=1
bx-ay=a+b
The solution in attached file

\( \left\{\begin{array}{ccc}ax+by=1&/\cdot a\\bx-ay=a+b&/\cdot b\end{array}\right.\\\\+\left\{\begin{array}{ccc}a^2x+aby=a\\b^2x-aby=ab+b^2\end{array}\right.\\-\\.\ \ \ \ \ a^2x+b^2x=a+ab+b^2\\.\ \ \ \ \ \ (a^2+b^2)x=a+ab+b^2\\.\ \ \ \ \ \ \ \ \ \ \ \ x=\frac{a+ab+b^2}{a^2+b^2}\\\\a\cdot\frac{a+ab+b^2}{a^2+b^2}+by=1\\\\\frac{a^2+a^2b+ab^2}{a^2+b^2}+by=1\\\\by=1-\frac{a^2+a^2b+ab^2}{a^2+b^2} \)

\( by=\frac{a^2+b^2}{a^2+b^2}-\frac{a^2+a^2b+ab^2}{a^2+b^2}\\\\by=\frac{a^2+b^2-a^2-a^2b-ab^2}{a^2+b^2}\\\\by=\frac{b^2-a^2b-ab^2}{a^2+b^2}\\\\y=\frac{b^2-a^2b-ab^2}{a^2b+b^3} \)

\( y=\frac{b(b-a^2-ab)}{b(a^2+b^2)}\\\\y=\frac{b-a^2-ab}{a^2+b^2}\\\\Answer:\\\\x=\frac{a+ab+b^2}{a^2+b^2}\ and\ y=\frac{b-a^2-ab}{a^2+b^2} \)


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