Linear Quadratic System

                    Lesson 1

 

                                                            Problem:     Solve algebraically.

 

                                                                                   x = y2 3y - 6

                                                                                   x = y + 6         

                                                                                                 

                                                            Solution      Substitute the value of x in quadratic equation.

                                                            Step 1:                         

                           

                                                            Solution      y2 -3y - 5 = y + 6

Step 2:          y2 - 4y - 12 = (y - 6)(y + 2)

                                                                                  y - 6 = 0; y = 6, y + 2 = 0; y = - 2         

                            

                                                            Solution      Substitute the values of y in linear equation.         

                                                            Step 3:  

 

                                                            Solution      y = 6,x = y + 3 = 6 + 3 = 9 ; y = -2, x = y + 3 =-2 + 3= 1

                                                            Step 4:

 

                                                            Solution     Answer is: The solution set of these equations:  (9, 6) , (1, -2)

                                                            Step 5:  

Complete      

 

                                                            Try two practice problems:  

 

A.

y = x2 - 8x + 7

y = x - 1

 

B.

y = x2 x - 7

y = x + 1

 

  

 

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