2016/107) Solve the system of equations

$xy+3y^2-x + 4y-7 = 0$

$2xy+y^2 -2x - 2y +1 = 0$

Solution
multiply the 1st one by 2 and subtract 2nd one to get
$5y^2 + 10y -15 = 0$
or $y^2+2y -3 = 0$ or $(y+3)(y-1)= 0$
y = -3 or y = 1
put y = -3 in 1st equation to get
$-3x + 27  - x - 12 -7 = 0$ or $ x= 2$
now check the 2nd one to get $ -12 + 9 -4 + 6 +1 = 0$ so 2nd one is satisfied and somution = $(x,y) = (2, -3)$
put y =1 in 1st one to get
$ -x + 3 -x + 4 -7$ or $x = 0$
put x= 0 and y = 1 in 2nd one and it is satisfied so solution $(x,y) = (0,1)$