1. Solving Systems of Linear Equations By Elimination

2. What is Elimination? To eliminate means to get rid of or remove. You solve equations by eliminating one of the variables (x or y) using addition and or subtraction.

3. Example 1 Solve the following system of linear equations by elimination. 2x – 3y = 15 5x + 3y = 27 (1) (2) Add equation (1) to equation (2)  7x + 0y = 42 7x = 42 x = 6  By eliminating y, we can now solve for x

4. Example 1 Substitute x= 6 into equation (1) to solve for y 2x – 3y = 15 2(6) – 3y = 15 12 – 3y = 15 – 3y = 15 – 12 – 3y = 3 y = -1 Check your solution x = 6 and y = -1 in equation (2) 5x + 3y = 27 5(6) + 3(-1) = 27 30 – 3 = 27 27 = 27 LS = RS Therefore, the solution set = {(6,-1)}

5. Example 2 5x + 4y = -28 3x + 10y = -13 (1) (2)  If we were to add these equations we would obtain 8x + 14y = -41  Even though we have only one equation now, we still have 2 variables.  We need to multiply the equations by terms that will allow us to eliminate either x or y.

6. Example 2  If we multiply equation (1) by 5 and equation (2) by -2, we be able to eliminate y. (1) x 5 (2) x -2 25x + 20y = -140 -6x – 20y = 26 (3) (4)  When you change the equations you need to renumber them. 19x = -114 Add (3) & (4)  x = -6 5x + 4y = -28 3x + 10y = -13 (1) (2)

7. Example 2 Substitute x = -6 into equation (1) 5x + 4y = -28 5(-6) + 4y = -28 -30 + 4y = -28 4y = -28 +30 4y = 2 Check your answer x = -6 and y = ½ into equation (2) 3(-6) + 10(½) = -13 -18 + 5 = -13 -13 = -13 LS = RS Therefore, the solution set = {(-6, ½)}

8. Questions? Any Questions? Homework: 1.4 # 5 – 12 Complete Homework for Monday! Have a Great Weekend