Question

solve the system of equations
1: y=3x^2-2x-1
     2x+3y=2

2: x^2+(y-2)^2=4
     x^2-2y=0

Answer 1

2x+3y = 2 so y = (2-2x)/3
y = 3x² -2x-1
(2-2x)/3 = 3x² - 2x - 1
2-2x = 9x² - 6x - 3
9x² -4x -5=0;
9x² - 9x + 5x - 5 =0
9x (x-1) + 5 (x-1) =0
(9x+5) ( x-1) = 0
x=1 or x= -5/9;
if x=1 then y= 2-2 / 3 = 0
if x=-5/9, y = (2- (-10/9) ) /3 = 28/27
Thus, (x,y) = (1,0) or (-5/9, 28/27)

2) x² -2y =0 so x² = 2y
x² + (y-2)² = 4
substituting x² = 2y we get
2y + (y-2)² = 4
2y+ y² -4y + 4 = 4
y² - 2y=0;
y(y-2) =0
y=0 or y=2
if y=0 then x= sqrt(4-4) = 0
if y=2 then x² = 4 ; So x= sqrt(4) = 2
Thus, (x,y) = (0,0) or (2 , 2)