1. Systems of Linear Equations A system of linear equations consists of two or more linear equations. We will focus on only two equations at a time. The solution of a system of linear equations in two variables is any ordered pair that solves both of the linear equations. The solution to the system is the point that satisfies ALL of the equations. This point will be an ordered pair. When graphing, you will encounter three possibilities.

2. N UMBER OF S OLUTIONS OF A L INEAR S YSTEM I DENTIFYING T HE N UMBER OF S OLUTIONS y x y x Lines intersect one solution Lines are parallel no solution y x Lines coincide infinitely many solutions

3. Parallel Lines These lines never intersect! Since the lines never cross, there is NO SOLUTION! Parallel lines have the same slope with different y-intercepts.

4. Same Lines These lines are the same! Since the lines are on top of each other, there are INFINITELY MANY SOLUTIONS! Coinciding lines have the same slope and y-intercepts.

5. What is the solution of the system graphed below? 1. (2, -2) 2. (-2, 2) 3. No solution 4. Infinitely many solutions

6. Name the Solution

7. Solving a system of equations by graphing. Let's summarize! There are 3 steps to solving a system using a graph. Step 1: Graph both equations. Step 2: Do the graphs intersect? Step 3: Check your solution. Graph using slope and y – intercept or x- and y-intercepts. Be sure to use a ruler and graph paper! This is the solution! LABEL the solution! Substitute the x and y values into both equations to verify the point is a solution to both equations.

8. Determine whether the given point is a solution of the following system. point: (– 3, 1) system: x – y = – 4 and 2x + 10y = 4 Plug the values into the equations. First equation: – 3 – 1 = – 4 true Second equation: 2(– 3) + 10(1) = – 6 + 10 = 4 true Since the point ( – 3, 1) produces a true statement in both equations, it is a solution. Solution of a System Example

9. 1) Find the solution to the following system: 2x + y = 4 x - y = 2 Graph both equations. I will graph using x- and y-intercepts (plug in zeros). Graph the ordered pairs. 2x + y = 4 (0, 4) and (2, 0) x – y = 2 (0, -2) and (2, 0)

10. Graph the equations. 2x + y = 4 (0, 4) and (2, 0) x - y = 2 (0, -2) and (2, 0) Where do the lines intersect? (2, 0) 2x + y = 4 x – y = 2

11. Check your answer! To check your answer, plug the point back into both equations. 2x + y = 4 2(2) + (0) = 4 x - y = 2 (2) – (0) = 2 Nice job…let’s try another!

12. 2) Find the solution to the following system: y = 2x – 3 -2x + y = 1 Graph both equations. Put both equations in slope- intercept or standard form. I’ll do slope-intercept form on this one! y = 2x – 3 y = 2x + 1 Graph using slope and y-intercept

13. Graph the equations. y = 2x – 3 m = 2 and b = -3 y = 2x + 1 m = 2 and b = 1 Where do the lines intersect? No solution! Notice that the slopes are the same with different y-intercepts. If you recognize this early, you don’t have to graph them!

14. Practice – Solving by Graphing y – x = 1  (0,1) and (-1,0) y + x = 3  (0,3) and (3,0) Solution is probably (1,2) … Check it: 2 – 1 = 1 true 2 + 1 = 3 true therefore, (1,2) is the solution (1,2)

15. Practice – Solving by Graphing Inconsistent: no solutions y = -3x + 5  (0,5) and (3,-4) y = -3x – 2  (0,-2) and (-2,4) They look parallel: No solution Check it: m 1 = m 2 = -3 Slopes are equal therefore it’s an inconsistent system

16. Consistent: infinite sol’s 3y – 2x = 6  (0,2) and (-3,0) -12y + 8x = -24  (0,2) and (-3,0) Looks like a dependant system … Check it: divide all terms in the 2 nd equation by -4 and it becomes identical to the 1 st equation therefore, consistent, dependant system (1,2)

17. Graph the system of equations. Determine whether the system has one solution, no solution, or infinitely many solutions. If the system has one solution, determine the solution.

18. Solving a System of Linear Equations by the Substitution Method 1)Solve one of the equations for a variable. 2)Substitute the expression from step 1 into the other equation. 3)Solve the new equation. 4)Substitute the value found in step 3 into either equation containing both variables. 5)Check the proposed solution in the original equations. The Substitution Method

19. Solve the following system using the substitution method. 3x – y = 6 and – 4x + 2y = –8 Solving the first equation for y, 3x – y = 6 –y = –3x + 6 (subtract 3x from both sides) y = 3x – 6 (multiply both sides by – 1) Substitute this value for y in the second equation. –4x + 2y = –8 –4x + 2(3x – 6) = –8 (replace y with result from first equation) –4x + 6x – 12 = –8 (use the distributive property) 2x – 12 = –8 (simplify the left side) 2x = 4 (add 12 to both sides) x = 2 (divide both sides by 2) The Substitution Method Continued.

20. Substitute x = 2 into the first equation solved for y. y = 3x – 6 = 3(2) – 6 = 6 – 6 = 0 Our computations have produced the point (2, 0). Check the point in the original equations. First equation, 3x – y = 6 3(2) – 0 = 6 true Second equation, –4x + 2y = –8 –4(2) + 2(0) = –8 true The solution of the system is (2, 0). The Substitution Method

21. Solving a System of Linear Equations by the Addition or Elimination Method  Rewrite each equation in standard form, eliminating fraction coefficients.  If necessary, multiply one or both equations by a number so that the coefficients of a chosen variable are opposites.  Add the equations.  Find the value of one variable by solving equation from step 3.  Find the value of the second variable by substituting the value found in step 4 into either original equation.  Check the proposed solution in the original equations. The Elimination Method

22. Solve the following system of equations using the elimination method. 6x – 3y = –3 and 4x + 5y = –9 Multiply both sides of the first equation by 5 and the second equation by 3. First equation, 5(6x – 3y) = 5(–3) 30x – 15y = –15 (use the distributive property) Second equation, 3(4x + 5y) = 3(–9) 12x + 15y = –27 (use the distributive property) The Elimination Method Continued.

23. Combine the two resulting equations (eliminating the variable y). 30x – 15y = –15 12x + 15y = –27 42x = –42 x = –1 (divide both sides by 42) The Elimination Method Continued.

24. Substitute the value for x into one of the original equations. 6x – 3y = –3 6(–1) – 3y = –3 (replace the x value in the first equation) –6 – 3y = –3 (simplify the left side) –3y = –3 + 6 = 3 (add 6 to both sides and simplify) y = –1 (divide both sides by –3) Our computations have produced the point (–1, –1). The Elimination Method Continued.

25. Check the point in the original equations. First equation, 6x – 3y = –3 6(–1) – 3(–1) = –3 true Second equation, 4x + 5y = –9 4(–1) + 5(–1) = –9 true The solution of the system is (–1, –1). The Elimination Method