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

Algebra ->  Quadratic Equations and Parabolas -> 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       Log On


   

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) About Me  (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 x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+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