Ch. 3 Notes 3.1 – 3.3 and 3.6

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 = -2 + 12 = – 11 10 Not a solution Yes it is a solution

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)

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 -2x -2y = -2x – 8 -2 -2 y = x + 4 2x + 2y = 4 -2x -2x 2y = -2x + 4 2 2 Solution: (-1,3) y = -x + 2

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 = 5 2x – 2y = 9 2x + 4y = 12 x + 2y = 6 -x -x -2x -2x 2 2 -y = -x + 5 -2y = -2x + 9 x + 2y = 6 -1 -1 -2 -2 y = x – 9/2 y = x –5 Infinitely many solutions No solutions

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

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 -4x -2y = -4x – 8 -2 -2 y = 2x + 4 x + y = 1 -x -x y = -x + 1 Solution: (-1, 2)

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

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

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

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

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

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

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

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.

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 -2 -2 y  1/2 x – 3/2

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 -1 -1 y  x + 2

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

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)

3.6 Warm-Up

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.

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

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

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.

-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 + 20 = 18 – 10 – z = –11 -x = 18 - 20 – z = –11 + 10 -3x – 15y = 54 -x = -2 – z = –1 -11y = 44 (2, -4, 1) x = 2 z = 1 y = -4

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