1. Warm up 12/6 or 7 1) Write the equation of a line that is parallel to y = -3x –5 and goes through the point (6,10). 2) Write the equation of a line that is perpendicular to y = -3x –5 and goes through the point (6,10). 3) Using elimination, solve the following system. a) x + y = 4 -3x + y = -8 a) x + y = 4 -3x + y = -8 b) 3x – 4y = -15 5x + y = -2 b) 3x – 4y = -15 5x + y = -2

2. Answers to warm up 1) y = -3x + 28 2)Y = 1/3x +8 3)A) Subtract the two lines : x + y = 4 -3x + y = -8 4x = 12 x = 3 Substitute: y = 1 (3, 1) b) (-1, 3)

3. Solving Systems by Elimination cont. Objective: To find the solution to a system of equations by elimination (addition). Remember: The solution is the point where the lines intersect.

4. 1) 2x + 3y = 6 +5x – 4y = -8 +5x – 4y = -8 4(2x + 3y) = 6(4) 8x + 12y = 24 3(5x – 4y) = -8(3) +15x -12y = -24 23 x = 0 23 x = 0 x = 0 Now plug 0 in for x into any of the 4 equations. 2(0) + 3y = 6 3y = 6 Y = 2 We will need to change both equations. We will have the y value drop out. The solution is (0, 2)

5. You Try: 1. 2x + 3y = 14 3x – 2y = -5 3x – 2y = -5 2. 5x + 2y = -8 2x – 5y = -9 2x – 5y = -9 (1, 4) (-2, 1)

6. Solve Systems by Elimination This time both equations will be multiplied but one must be multiplied by opposite sign to be canceled out. Example 2) 4x+ 2y = -8 5x + 7y = 8 5x + 7y = 8 Step 1 – Choose one variable to eliminate. Let’s start with x. One line will need to be multiplied by a negative number since all coefficients are positive. Let’s choose the first line. -5(4x+ 2y) = -8(-5) 4(5x + 7y = 8(4) -20x -10y = 40 20x + 28y = 32 18y = 72 18y = 72 y=4 y=4 Now substitute y back in to any of the four equations find x.

7. Example 2 Continued : 4x+ 2y = -8 5x + 7y = 8 Example 2 Continued : 4x+ 2y = -8 5x + 7y = 8 4x + 2(4) = -8 4x + 8 = -8 -8 -8 -8 -8 4x = -16 X = -4 The solution is (-4,4)

8. YOU TRY!! 8x+ 2 y = 0 5x + 3y = -7 Answer: (1,-4)

9. Summary: What is the solution of a system of linear equations? Homework:7.4B WS