Question

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

Answer 1

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

Answer 2

`\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=(a+ab+b^2)/(a^2+b^2)\\\\a\cdot(a+ab+b^2)/(a^2+b^2)+by=1\\\\(a^2+a^2b+ab^2)/(a^2+b^2)+by=1\\\\by=1-(a^2+a^2b+ab^2)/(a^2+b^2)`


`by=(a^2+b^2)/(a^2+b^2)-(a^2+a^2b+ab^2)/(a^2+b^2)\\\\by=(a^2+b^2-a^2-a^2b-ab^2)/(a^2+b^2)\\\\by=(b^2-a^2b-ab^2)/(a^2+b^2)\\\\y=(b^2-a^2b-ab^2)/(a^2b+b^3)`


`y=(b(b-a^2-ab))/(b(a^2+b^2))\\\\y=(b-a^2-ab)/(a^2+b^2)\\\\Answer:\\\\x=(a+ab+b^2)/(a^2+b^2)\ and\ y=(b-a^2-ab)/(a^2+b^2)`