SOLUTION: by using quadratic system of equations solve each system by substitution. {y=2x^2-5x+3 {y=x-2 both are one problem solve each system by graphing: {y>2x^2+x+3

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Question 418856: by using quadratic system of equations solve each system by substitution.
{y=2x^2-5x+3
{y=x-2 both are one problem

solve each system by graphing:
{y>2x^2+x+3
{y<-x^2-4x+1


{y=-x^2+2x+1 same as before
{y=2x+1
solve each system by graphing:
{y>x^2+2x
{y>x^2-1

Answer by ewatrrr(24785)   (Show Source): You can put this solution on YOUR website!

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 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

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


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