Linear Quadratic Systems Worksheet

                    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:  

1.

y = x^2 - 8x + 7 y = x - 1

2.

y = x^2 – x - 7 y = x + 1

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