1. Warm up Solve for y 1. 3x + 2y = 5 2. -4x – 2y = 8 3. -6x + 3y = -15 Write an equation in slope intercept form 4. m = 4 and y int (0, 3) 5. m = -3/2 and y int (2, 4)

2. answers 1. y = -3/2x + 5/2 2. y = -2x – 4 3. y = 2x – 5 4. y = 4x + 3 5. y = -3/2x + 7

3. Key words System of Equations Two or more equations System Solution A point (x, y) that satisfies both equations

4. Part 1: Finding solutions to Systems of Equations There are ____ types of solutions for systems of equations. 3 One SolutionNo Solution Infinite Solutions

5. Types of Solutions One Solution: If there is one solution, then the lines are intersecting. They may or may not be perpendicular. The slopes will be opposite reciprocals if perpendicular. No Solution: If there is no solution, then the lines are parallel. The slopes will be the same, but “b” will be different. Infinite Solutions: If there are infinite solutions, then the lines are coinciding. Both “m” and “b” will be the same.

6. Solving Systems of Equations There are 3 ways to solve a system of equations. Graphing: Graph the lines. Where the lines intersect is the solution. Substitution: Solve one of the equations for one of the variables and substitute. Elimination: Set up the equations and combine them to eliminate one of the variables.

7. Steps for Using Elimination 1) Write both equations in standard form (Ax + By = C) so that variables and = line up 2) Multiply one or both equations by a number to make opposite coefficients for one of the variables. 3) Add equations together (one variable should cancel out) 4) Solve for remaining variable. 5) Substitute the solution back in to find other variable.

8. Example 1: 5x + y = 12 5x + y = 12 3x – y = 4 3x – y = 4 8x = 16 8x = 16 8 8 8 8 x = 2 x = 2 5(2) + y = 12 10 + y = 12 y = 2 The solution is: (2, 2) Step 1: Put both equations in standard form. Step 2: Check for opposite coefficients. Step 3: Add equations together Step 4: Solve for x Step 5: Substitute 2 in for x to solve for y (in either equation) Already Done y and –y are already opposites

9. Example 2 3x + 4y = 9 3x + 4y = 9 -x – 4y = 7 -x – 4y = 7 Answer: (8, -15/4)

10. Example 3 3x + 5y = 10 3x + 5y = 10 3x + y = 2 3x + y = 2 3x + 5y = 10 3x + 5y = 10 -1(3x + y) = -1(2) 4y = 8 4y = 8 y = 2 y = 2 Now plug (2) in for y. 3x + 2 = 2 X = 0 Solution is : (0,2) When you add these neither variable drops out SO…. We need to change 1 or both equations by multiplying the equation by a number that will create opposite coefficients. When we need to create opposite coefficients 3x + 5y = 10 -3x – y = -2 -3x – y = -2 Multiply the bottom equation by negative one to eliminate the x

11. 4) 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 x = 0 Now plug (0) in for x into any of the 4 equations. 2(0) + 3y = 6 3y = 6 y = 2 Solution is: (0, 2) We will need to change both equations. We will have the y value drop out.

12. Solving by Substitution 1. Solve one equation for either x or y 2. Substitute the expression into the other equation 3. Solve for the variable 4. Substitute the value back in and solve 5. Check your answer, is it a solution for both equations? Remember that a point consists of an “x” value and a “y” value. You have to find both to find the solution.

13. Step 1 Solve one equation for x or y y = x + 1 y = -2x - 2 y = -2x - 2 Already done!

14. Step 2 Substitute that expression into the other equation y = x + 1 y = -2x - 2 x + 1 = -2x - 2

15. Step 3 Solve for the other variable x + 1 = -2x – 2 +2x +2x 3x + 1 = -2 - 1 - 1 - 1 - 1 3x = -3 3x = -3 3 3 3 3 x = -1 x = -1

16. Step 4 Substitute the value back in and solve y = -1+ 1 y = 0 Is (-1, 0) a solution? Check to find out. 0 = -1 + 1 0= -2(-1) – 2 Solution (-1, 0)

17. Try this one Ex. y = x + 4 y = 3x + 10 Solution (-3, 1)

18. Substitution & the distributive property To use substitution you must have an equation that has been solved for one of the variables. To use substitution you must have an equation that has been solved for one of the variables. Ex. 3x – 2y = 1 y= 3 + 1 y = 4 Solution: (3, 4) 3x-2(x+1) =1 3x –2x -2 =1 x -2 = 1 x = 3 y=x+1

19. Your Turn: Solve the following systems of equations. 4. y = x +1 y = 2x – 1 y = 2x – 1 5. y = 2x 5. y = 2x 7x –y = 15 7x –y = 15 (2, 3) (3, 6)

20. You try these: Tell if lines are parallel, perpendicular, intersecting but not perpendicular, or coinciding if one solution, infinite solutions, or many solutions. Tell if lines are parallel, perpendicular, intersecting but not perpendicular, or coinciding if one solution, infinite solutions, or many solutions. 1. 3x + y = 22. -4x + 3y = 0 1. 3x + y = 22. -4x + 3y = 0 3x + y = 6 4x + y = 8 3x + y = 6 4x + y = 8 3. 5x -10y = 204. y = -1/4 x - 5 -x + 2y = -4 -4x + y = 12 -x + 2y = -4 -4x + y = 12 parallel lines, no solutionintersecting lines, one solution Coinciding lines, Infinite solution perpendicular lines, one solution

21. Summary: Draw and Fill in the table below in your notes. Parallel Lines Intersecting Not ┴ IntersectingPerpendicularCoincidingLines Graph Slope & b Ex. of what system looks like Number of Solutions

22. Practice Classwork: CW #7 Homework: WS #7