1. Solving Systems by Elimination 5.4 NOTES, DATE ____________

2. 1. Line up like terms 2. Multiply one or both equations by a constant to get the coefficients of one variable to be opposites. Note: Sometimes this step is not needed. 3. Add the two equations together to eliminate a variable, then solve. 4. Substitute this value back into one of the original equations and solve. 5. Write your answer as an ordered pair.

3. Example 1: a - 3b = -1 2a + 3b = 16 + 3a = 15 a = 5 a - 3b = -1 5 - 3b = -1 -5 -3b = -6 b = 2 (5, 2)

4. Example 2: 8q +12 r = 20 5q +12 r = -1 ( )-1 -8q -12 r = -20 5q +12 r = -1 -3q = -21 q = 7 5(7) +12r = -1 35 +12r = -1 - 35 -35 12r = -36 r = -3 (7, -3) + 5q +12 r = -1

5. Example 3: 6x + y = 6 3x + 2y =9 ( )-2 -9x = -3 x = 1/3 3(1/3) + 2y = 9 - 1 -1 2y = 8 y = 4 (1/3, 4) -12x – 2y = -12 3x + 2y = 9 1 + 2y = 9 + 3x + 2y =9

6. Example 4: 2x + 3y = 35 5x = 7 + 4y ( )3 5(7) = 7 + 4y -7 -7 28 = 4y 7 = y (7, 7) 5x – 4y = 7 2x + 3y = 35 35 = 7 + 4y -4y ( )4 15x – 12y = 21 8x + 12y = 140 + 23x = 161 x = 7 5x = 7 + 4y

7. Example 5: 5a + 35b = 45 a + 7b = 9 ( )-5 5a + 35b = 45 -5a - 35b = -45 0 = 0 + True No Solution

8. Example 6: 2x + 3y = 8 4x + 6y = 4 ( )2 -8x – 12y = -32 8x + 12y = 8 + ( )-4 0 = -24 False Infinitely Many Solutions