Anonymous

Ans. There are 2 sets of solution: x=-2; y=-5 x=1; y=4

Anonymous

y = -x² + 2x + 3 y = 2x² + 5x - 3 2x² + 5x - 3 = -x² + 2x + 3 3x² + 3x - 6 = 0 Quadratic formula x = [-3 ± √(3² – 4·3(-6))] / [2·3]  = [-3 ± √81] / 6  = [-3 ± 9] /6  = -2, 1 (-2,-5) and (1,4)

Anonymous

2x^2 + 5x - 3 = 3 + 2x - x^2 3x^2 + 3x - 6 = 0 x^2 + x - 2 = 0 [ x + 2 ] [ x - 1 ] = 0 x = - 2 , x = 1 y = - 5 , y = 4

Anonymous

y = y therefore 3 + 2x -x^2 = 2x^2 +5x -3 Combining like terms 3x^2 +3x -6 = 0 Factoring 3(x +2)(x -1) = 0 x = -2 ; 1

Anonymous

Y=3+2x-x^2 =2x^2+5x-3 --->3x^2+3x-6=0 solve for x then find y

Anonymous

Y = Y.... so 3 + 2x - x^2 = 2x^2 + 5x - 3 ==> 3x^2 + 3x - 6 = 0 ==> x^2 + x - 2 = 0 ==> (x+2)(x-1) = 0 ==> x = -2 or x = 1 etc.