Question

Use method of elimination to find the solution of $9x-4y=14$ and $7x-3y=11$

Answer 1

$xy\frac{dy}{dx}=\frac{1+y^2}{1+x^2}(1+x+x^2)$

or, $\frac{ydy}{1+y^2}=\frac{1+x+x^2}{x+x^3}dx$

Integrating both sides,

$\int \frac{ydy}{1+y^2}=\int \frac{1+x+x^2}{x+x^3}$

$9x-4y=14---\left(1\right)\times 3$

$7x-3y=11---\left(2\right)\times 4$

$\frac{27x-12y=4228x-12y=44-+-}{-x=-2}$

$\therefore x=2$

Putting $x=2$ in equation $\left(2\right)$

$7\left(2\right)-3y=11$

or, $14-3y=11$

or, $3y=3$

$\therefore y=1$

Ans: $x=2,y=1$