Algebra C A Extra Examples Lesson 13-3 by tsu23398

VIEWS: 0 PAGES: 3

									Algebra Concepts and Applications                                                                 Chapter 13


Lesson 13-3


Examples                 Use substitution to solve each system of equations.
                     1   y = -3x
                         x + 2y = 10
                         The first equation tells you that y is equal to -3x. So, substitute -3x for y in the
                         second equation. Then solve for x.
                             x + 2y = 10
                         x + 2(-3x) = 10      Replace y with -3x.
                              x - 6x = 10
                                 -5x = 10
                                -5x 10
                                 -5
                                    = -5       Divide each side by -5.
                                  x = -2

                         Now substitute -2 for x in either equation and solve for y.
                         Choose the equation that is easier for you to solve.
                         y = -3x
                         y = -3(-2)   Replace x with -2.
                         y=6

                         The solution of this system of
                         equations is (-2, 6). You can see
                         from the graph that the solution
                         is correct. You can also check by
                         substituting (-2, 6) into each of
                         the original equations.

                     2   x - y = -4
                         x = 3y + 5
                         Substitute 3y + 5 for x in the first equation. Then solve for y.
                                x - y = -4
                          3y + 5 - y = -4         Replace x with 3y + 5.
                              2y + 5 = -4
                         2y + 5 - 5 = -4 - 5      Subtract 5 from each side.
                                  2y = -9
                                 2y -9
                                 2
                                    =2            Divide each side by 2.
                                       9
                                  y = -2
Algebra Concepts and Applications                                                              Chapter 13


                                             9
                         Now substitute -2 for y into either equation and solve for x.
                         x = 3y + 5
                                  9                              9
                         x = 3-2 + 5           Replace y with -2.
                             27
                         x=-2 +5
                             17
                         x=-2
                                                                         17 9
                         The solution of this system of equations is - 2 , -2. Check by replacing (x, y)
                                  17 9
                         with - 2 , -2 in each equation.


Example              3   -x + y = 7
                         3x - y = -5
                         Solve the first equation for y since the coefficient of y is 1.
                         -x + y = 7  y = x + 7

                         Next, find the value of x by                   Now substitute 1 for x in either
                         substituting x + 7 for y in the second         equation and solve for y.
                         equation.                                          -x + y = 7
                               3x - y = -5                                  -1 + y = 7       Replace x with 1.
                         3x - (x + 7) = -5                              -1 + y + 1 = 7 + 1
                           3x - x - 7 = -5                                       y =8
                               2x - 7 = -5
                          2x - 7 + 7 = -5 + 7
                                  2x = 2
                                      2x 2
                                      2
                                        =2
                                   x =1
                         The solution is (1, 8).


Examples                 Use substitution to solve each system of equations.
                     4   y = -5x + 1
                         5x + y = -2
                         Find the value of x by substituting -5x + 1 for y in the second equation.
                                  5x + y = -2
                          5x + (-5x + 1) = -2    Replace y with -5x + 1.
                                       1 = -2

                         The statement 1 = -2 is false. This means
                         that there are no ordered pairs that are
                         solutions to both equations. Compare the
                         slope-intercept forms of the equations,
                         y = -5x + 1 and y = -5x - 2. Notice that the
                         graphs of these equations have the same
                         slope but different y-intercepts. Thus, the
                         lines are parallel, and the system has no solution.
Algebra Concepts and Applications                                                             Chapter 13


                     5   y = -4x + 1
                         8x + 2y = 2
                                 8x + 2y    =2
                         8x + 2(-4x + 1)    =2    Replace y with -4x + 1.
                              8x - 8x + 2   =2    Distributive Property
                                        2   =2

                         The statement 2 = 2 is true. This means that an ordered pair for a point on either
                         line is a solution to both equations. The system has infinitely many solutions.


Example              6   The Garden Center sells bags of wild flower seed that contain part seed and
Gardening Link           part sand to aid in planting. One seed mix contains 35% seed and a second
                         type contains 55% seed. How much of each type of mix should be added to
                         make 500 pounds of a 45% seed mix?
                         Explore Let a = the number of pounds of the 35% seed.
                                   Let b = the number of pounds of the 55% seed.

                                                                35% Seed        55% Seed        45% Seed
                                     Total Pounds                   a               b              500
                                     Pounds of Seed               0.35a           0.55b         0.45(500)

                         Plan       Write two equations to represent the information.
                                    a + b = 500                     total pounds
                                    0.35a + 0.55b = 0.45(500)       total pounds of seed

                         Solve      Use substitution to solve this system. Since a + b = 500, a = 500 - b.
                                            0.35a + 0.55b = 0.45(500)
                                    0.35(500 - b) + 0.55b = 0.45(500)       Replace a with 500 - b.
                                      175 - 0.35b + 0.55b = 225             Distributive Property
                                        175 + 0.2b - 175 = 225 - 175        Subtract 175 from each side.
                                                     0.2b = 50
                                                      0.2b 50
                                                       0.2
                                                           = 0.2            Divide each side by 0.2.
                                                         b = 250

                                    Now substitute 250 for b in either equation and solve for a.
                                             a + b = 500
                                           a + 250 = 500            Replace b with 250.
                                     a + 250 - 250 = 500 - 250      Subtract 250 from each side.
                                                 a = 250
                                    So, 250 pounds of 35% seed mix and 250 pounds of 55% seed mix
                                    should be used.

                         Examine Check by substituting (250, 250) into the original equations. The
                                 solution is correct.

								
To top