1. Lesson 1

2.  Example 1. Use either elimination or the substitution method to solve each system of equations.  3x -2y = 7 & 2x +5y = 9  A. Using substitution method  3x -2y = 7 ( 1 st eq.)  2x +5y = 9 ( 2 nd eq.)

3.  1. Choose an equation to change either in y form or x form.  For example, I choose eq. 1 & change it to x form.  3x -2y = 7 ( 1 st eq.)  3x = 2y + 7   x = 3 rd equation

4.  2. Substitute the value of x in the 2 nd equation.  2( ) +5y = 9 ( 2 nd eq.)  Use distributive property  Multiply both sides by 3  4y + 14 + 15y = 27

5.  Multiply both sides by 3  4y + 14 + 15y = 27  by simplifying,  19y = 13  using division,  y =

6.  3. Find x by substituting the value of y in the 1 st equation or 3 rd eq. 3 rd equation By simplifying Get the LCD & simplify

7.  Example 1. Use either elimination or the substitution method to solve each system of equations.  3x -2y = 7 ( 1 st eq.)  2x +5y = 9 ( 2 nd eq.)  The solution set is

8.  A. Using elimination method  3x -2y = 7 ( 1 st eq.)  2x +5y = 9 ( 2 nd eq.)

9.  1. Choose a variable to eliminate.  For example, I choose x.  3x -2y = 7 ( 1 st eq.)  2x +5y = 9 ( 2 nd eq.)  To eliminate x, multiply the 1 st eq. by -2 and multiply the 2 nd eq. by 3. then add the 1 st eq. to the 2 nd eq..  -2(3x -2y = 7) ( 1 st eq.)  3(2x +5y = 9) ( 2 nd eq.)

10.  -6x + 4y = -14  + 6x +15y = 27  19y = 13  y = 13  19  2. Now substitute the value of y in either equation to find the value of x.

11.  I choose the 2 nd equation.  2x +5y = 9) ( 2 nd eq.)  Multiply both sides by 19  38 x + 65 = 171  38x = 171- 65 (subtract both sides by 65)

12.  38 x = 106 (by subtraction prop.)   X = 106 (by division prop.)  38  X = 53 (by simplifying)  19