1. Ch. 3 Notes
3.1 – 3.3 and 3.6

3. 3.1 Graphing Systems of Equations
Objective – To be able to solve and graph systems of linear equations.
State Standard – 2.0 Students solve systems of linear equations by graphing
Example 1
Check whether (a) (1,4) and (b) (-5,0) are solutions of the following system.
x – 3y = -5
-2x + 3y = 10
(b) (-5) – 3(0) =
-2(-5) + 3(0) =
– 5 10
(1) – 3(4) =
-2(1) + 3(4) =
1 – 12 = =
– 11 10
Not a
solution
Yes it is a
solution

4. Number Of Solutions Of A Linear System
(lines that intersect at one point.)
exactly one solution
infinitely many solutions
(the graph is a single line.)
no solution
(lines are parallel)

5. Solve the system graphically. 2x – 2y = -8 2x + 2y = 4
Example 2
Solve the system graphically.
2x – 2y = -8
2x + 2y = 4 –5 –4 –3 –2 –1 1 2 5 4 3
2x – 2y = -8
-2x x
-2y = -2x – 8
y = x + 4
2x + 2y = 4
-2x x
2y = -2x + 4
Solution: (-1,3)
y = -x + 2

6. Example 3 Tell how many solutions the linear system has.
a) 2x + 4y = 12 b) x – y = 5
x + 2y = 6 2x – 2y = 9
x – y = x – 2y = 9
2x + 4y = x + 2y = 6
-x x
-2x x
-y = -x + 5
-2y = -2x + 9
x + 2y = 6
y = x – 9/2
y = x –5
Infinitely many solutions
No solutions

7. If you get the variables to cancel and you get:
0 = 0 (True Statement)
You will have:
Infinitely many solutions
If you get the variables to cancel and you get:
0 = (some #) (False Statement)
You will have:
No solutions

8. 3.2 WARM - UP Solve the system graphically. 4x – 2y = -8 x + y = 1–5 –4 –3 –2 –1 1 2 5 4 3
4x – 2y = -8
-4x x
-2y = -4x – 8
y = 2x + 4
x + y = 1
-x x
y = -x + 1
Solution: (-1, 2)

9. 3.2 Solving Systems Algebraically
Types of Other Systems
1) Solve one of the equations for a variable.
2) Substitute step 1 into the other equation.
3) Substitute the value in Step 2 into one of the original equations to get the other variable

12. 3.2 Solving Systems Algebraically
State Standard – 2.0 Students solve systems of linear equations and inequalities (in two or three variables) by substitution, linear combination, with graphs, or with matrices.
The Substitution Method
1) Solve one of the equations for a variable.
2) Substitute step 1 into the other equation.
3) Substitute the value in Step 2 into one of the original equations to get the other variable

13. ( 1 , -1 ) Example 1 Solve using the Substitution method: x – 2y = 3
3(2y + 3) + 2y = 1
6y y = 1
8y + 9 = 1
x –2y = 3
+2y y
8y = -8
x = 2y + 3
y = -1
( 1 , -1 )
x –2(-1) = 3
x + 2 = 3
x = 1

14. The Elimination Method
1) Multiply one or both of the equations by a constant.
2) Add the revised equations in order to eliminate one of the variables.
3) Substitute the value in Step 2 into one of the original equations to get the other variable

15. ( , ) Example 2 Solve using the Elimination method: 2x – 4y = 13
-2( )
2x – 4y = 13
4x – 5y = 8
2x + 24 = 13
( , )
-11 2 -6
-4x + 8y = -26
2x = -11
4x – 5y = 8
x =
-11 2
3y = -18
y = -6

16. ( , ) -2 1 Example 3 Solve using the Elimination method: 2x + 3y = -1
5( )
2x + 3y = -1
-5x + 5y = 15
2x + 3 = -1
2( )
( , ) -2 1
10x + 15y = -5
2x = -4
-10x + 10y = 30
x = -2
25y = 25
y = 1

17. If you get the variables to cancel and you get: 0 = 0 (True Statement)
You will have:
Infinitely many solutions
If you get the variables to cancel and you get: 0 = (some #) (False Statement)
You will have:
No solutions

19. 3.3 Systems of Inequalities
Objective – To be able to solve a system of linear inequalities to find the solution of the system.
California State Standard 2.0 – Students solve systems of linear equations and inequalities (in two or three variables) by substitution, with graphs, or with matrices.

20. y  1/2 x – 3/2 Extra Example 2a Graph x – 2y  3 y > 3x - 4–5 –4 –3 –2 –1 1 2 5 4 3
Put in Slope-
Intercept Form:
x – 2y  3
-x x
-2y  -x + 3
y  1/2 x – 3/2

21. Extra Example 2b y  x + 2 Graph x  0 y  0 x – y  -2 x – y  -2–5 –4 –3 –2 –1 1 2 5 4 3
Put in Slope-Intercept
Form:
x – y  -2
-x x
-y  -x – 2
y  x + 2

22. Extra Example 2c Graph x  0 y > 2x – 1 y  2x + 3 –5 –4 –3 –2 –1 1

23. Extra Example 4 Graph y ≥ 3 y < – |x + 2| + 5 (h,k) = (-2,5) –5 –4–3 –2 –1 1 2 5 4 3
(h,k) = (-2,5)

24. 3.6 Warm-Up

25. 3.6 Systems With Three Variables
Objective – To be able to solve a system of linear inequalities to find the solution of the system.
California State Standard 2.0 – Students solve systems of linear equations and inequalities (in two or three variables) by substitution, with graphs, or with matrices.

26. 3.6 Systems With Three Variables
The Substitution Method (in 3-Variable)
Substitute

27. Example 1 Solve the system algebraically. x + y – 4z = 4 2x + 3y = 2
(-2, 2, -1)
x = 2

28. 3.6 Systems With Three Variables
by Jason L. Bradbury
The Elimination Method (in 3-Variable)
Take the first two equations and eliminate one variable
Take the last two equations and eliminate the same variable.
With the two revised equations eliminate one of the two variables.

29. -x – 5y = 18 3x + 4y = -10 3x + 4y = -10 (3)(-x – 5y = 18) (2, -4, 1)
Example 2
Solve the system algebraically.
x + 3y – z = -11
2x + y + z = 1
5x – 2y + 3z = 21
(-3)(2x + y + z = 1)
5x – 2y + 3z = 21
-6x – 3y – 3z = -3
x + 3y – z = –11
5x – 2y + 3z = 21
2x + y + z = 1
-x – 5y = 18
3x + 4y = -10
3x + 4y = -10
x + 3y – z = –11
-x – 5y = 18
(2) + 3(-4) – z = –11
(3)(-x – 5y = 18)
-x – 5(-4) = 18
2 – 12 – z = –11
3x + 4y = -10
-x = 18
– 10 – z = –11
-x =
– z = –
-3x – 15y = 54
-x = -2
– z = –1
-11y = 44
(2, -4, 1)
x = 2
z = 1
y = -4

30. If you get the variables to cancel and you get:
0 = (True Statement)
You will have:
Infinitely many solutions
If you get the variables to cancel and you get:
0 = (some #) (False Statement)
You will have:
No solutions