3 2 a Solving Systems algebraically To solve

3. 2 a – Solving Systems algebraically

To solve systems using Substitution: Step #1: Check to see if one of the equations has a variable alone Step #2: Substitute what that variable equals into the other equation Solve for the other variable Step #3: Step #4: When you know what one variable is, plug it back in to find the other one. Step #5: Write the answer as an ordered pair (x, y)

Ex#1 Solve the system by substitution. Check by graphing – x + 2(-2 x + 1) = 2 – x – 4 x + 2 = 2 -5 x + 2 = 2 -2 -2 -5 x = 0 -5 -5 x=0 y = -2 x + 1 y = -2(0) + 1 y=0+1 y=1 (x, y) (0, 1)

Ex#2 Solve the system by substitution. 4 x – 3(-2 x + 13) = 11 4 x + 6 x – 39 = 11 10 x – 39 = 11 +39 10 x = 50 10 10 x=5 y = -2 x + 13 y = -2(5) + 13 y = -10 + 13 y=3 (x, y) (5, 3)

Try These: Solve the system by substitution. 3(2 y – 7) + 4 y = 9 6 y – 21 + 4 y = 9 10 y – 21 = 9 +21 10 y = 30 10 10 y=3 x = 2 y – 7 x = 2(3) – 7 x=6– 7 x = -1 (x, y) (-1, 3)

Ex#2 Solve the system by substitution. 6 x – 3(2 x – 5) = 15 6 x – 6 x + 15 = 15 Infinite solutions

Ex#2 Solve the system by substitution. 6 x + 2(-3 x + 1) = 5 6 x – 6 x + 2 = 5 2 5 No Solution

To eliminate a variable line up the variables with x first, then y. Afterward make one of the variables opposite the other. You might have to multiply one or both equations to do this.

Example #3: Find the solution to the system using elimination. 2 3 x – y = 8 x + 2 y = 5 3 + 2 y = 5 -3 -3 2 y = 2 y=1 6 x – 2 y = 16 x + 2 y = 5 7 x + 0 = 21 7 x = 21 7 7 x=3 (x, y) (3, 1)

Example #3: Find the solution to the system using elimination. -3 2 x + y = 6 4 x + 3 y = 24 -6 x – 3 y = – 18 4 x + 3 y = 24 – 2 x = 6 x=– 3 2(– 3) + y = 6 -6 + y = 6 y = 12 (x, y) (– 3, 12)

Example #3: Find the solution to the system using elimination. 3 3 x – 4 y = 16 2 5 x + 6 y = 14 5(4) + 6 y = 14 20 + 6 y = 14 -20 6 y = -6 y = -1 9 x – 12 y = 48 10 x + 12 y = 28 19 x = 76 x=4 (x, y) (4, -1)

Example #3: Find the solution to the system using elimination. 5 -3 x + 2 y = -10 3 5 x + 3 y = 4 -15 x + 10 y = -50 15 x + 9 y = 12 19 y = -38 y = -2 5 x + 3(-2) = 4 5 x – 6 = 4 5 x = 10 x=2 (x, y) (2, -2)

Example #3: Find the solution to the system using elimination. 12 x – 3 y = -9 3 -4 x + y = 3 12 x – 3 y = -9 -12 x + 3 y = 9 0= 0 Infinite solutions

Example #3: Find the solution to the system using elimination. 6 x + 15 y = -12 3 -2 x – 5 y = 9 6 x + 15 y = -12 -6 x – 15 y = 27 0 = 15 No solution