Question 418856
  <pre><font face = "Tohoma" size = 4 color = "indigo"><b>
Hi
solve each system by substitution.
y=2x^2-5x+3
y=x-2             
x-2=2x^2-5x+3
 2x^2 - 6x + 5 = 0   |using {{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}No real solution
y= -x^2+2x+1  
y= 2x+1
 2x+ 1 = -x^2 + 2x + 1
 x^2 = 0  x = 0 and y = 1  Ordered pair(0,1)is the solution for this system

solve each system by graphing: 
Shade the area in common according to the Inequalities or state the only Point in common
y> 2x^2+x+3  graphing y = 2(x+1/4)^2 + 23/8 Completing the square
y< -x^2-4x+1 graphing y = -(x+2)+5 Completing the square
{{{drawing(300,300,   -6, 6, -6, 6, grid(1),
circle(-2, 5,0.3),
circle(-.25, 23/8,0.3),
graph( 300, 300, -6, 6, -6, 6,0,-x^2-4x+1 ,2x^2+x+3))}}}
y>x^2+2x   graphing y = x^2 + 2x = (x+1)^2 -1  Completing the square
y>x^2-1    graphing y = x^2-1
{{{drawing(300,300,   -6, 6, -6, 6, grid(1),
circle(-1, -1,0.3),
circle(0, -1,0.3),
graph( 300, 300, -6, 6, -6, 6,0,x^2 + 2x,x^2-1))}}}