Simultaneous equation with curve and a line $$2x+5y=1;$$
$$x^2+2xy=2y^2+1.$$
The answer is $(3,-1)$, but I keep overcomplicating the solution with fractions and I just want the easiest solution possible.

Lets try like this
$$2x+5y=1\\x^2+2xy-2y^2=1$$Now square the first equation $$(2x+5y)^2=1\\x^2+2xy-2y^2=1$$Multiply the second by $4$ $$4x^2+20xy+25y^2=1\\4x^2+8xy-8y^2=4$$
Then first minus the second
$$12xy+33y^2=-3\\4xy+11y^2=-1\\y(4x+11y)=-1\\y(2(2x+5y)+y)=-1\\y(2+y)=-1\\y^2+2y+1=0$$
From there $y=-1$ then easily $2x-5=1$ implies $x=3$
