tuesday september 23, 2014 warm-up: put the … solve the system using the substitution method 2x +...

Warm-up: Put the following equations to slope-intercept form then use 2 points to graph 1. 4x - 3y = 8 8 x – 6y = 16 2. 2x + y = 4 2x + y = 1

Tuesday September 23, 2014

3.1 Solving Linear Systems using Graph

Objective: To solve a system of Linear equations

Example1: When slope (m) and y-intercept (b) are given Write the equation for the line in the graph Solution: Use Slope-Intercept form: According to the graph, b = -1 and m = (-3 - -1) / (3 – 0) = -2/3 Therefore, the equation is y = (-2/3) x - 1

EQ: How many ways can we solve a system of linear equations

Definition: • A system of linear equations consists of 2 or more

equations

• The solution of a system of linear equations can be 1 of the following 3 cases:

1. Exactly 1 solution

Lines intersect at 1 point (consistent and independent)

2. Infinitely many solutions Lines coincide (consistent and dependent)

3. No solutions

Lines are parallel (inconsistent)


There are 4 ways to solve a system of linear equations 1. Graphing 2. Substitution 3. Elimination 4. Multiplication/Elimination

Example: Graph the linear system and estimate the solution then check the solution algebraically 4x + y = 8 (1) 2x – 3y = 18 (2) Solution: 1. Put equations in y = mx + b form y = -4x + 8 (1) y = (2/3)x - 6 (2) 2. In (1), m = -4, b = 8 In (2), m = 2/3, b = 6 3. Solution is (3, -4) 4. Check:

Steps to solve a linear system using graphing: 1. Put each equation in slope-intercept form (y = mx+b) 2. Use y-intercept to plot 1st point and use m to plot 2nd point 3. Look for solution(s) 4. Substitute the solution in each equation to check for error

3.1 Solving Linear Systems using Graph Tuesday September 23, 2014

Warm-up:

Wednesday September 24, 2014

Example: Solve the system using the substitution method 2x + 5y = -5 (1) x + 3y = 3 (2) Solution: Step 1. Solve (2) for x x + 3y = 3 (2) x = -3y + 3 (3) Step 2. Substitute (3) into (1) 2x + 5y = -5 (1) 2 (-3y +3) + 5 y = -5 -6 y + 6 + 5 y = -5 6 - y = -5 y = 11 Step 3. Substitute y =11 back to either (1) or (2) x + 3y = 3 (2) x + 3 (11) = 3 x + 33 = 3 x = -30 The solution is (-30, 11) Check:

Steps to solve a linear system using substitution method: 1. Solve one of the equation for one of its variables 2. Substitute the expression from step 1 into the other

equation and solve for the other variable 3. Substitute the value from step 2 into the revised equation

from step 1 and solve for the other variable

3.2Solving Linear Systems by Substitution Wednesday September 24, 2014

Example: Solve the system using the elimination method 6x - 14y = 20 (1) 6x - 8y = 8 (2) Solution: Step 1. Subtract (1) and (2) to eliminate x 6x - 14y = 20 (1) 6x - 8y = 8 (2) ------------------------ -6y = 12 y = - 2 Step 2. Substitute y = -2 into either (1) or (2) 6x - 8y = 8 (2) 6x - 8 (-2) = 8 6x + 16 = 8 6x = 8 - 16 6x = -8 x = -8/6 = -4/3 The solution is (-4/3, -2) Check: (1) 6(-4/3) - 14 (-2) = 20 (2) 6(-4/3) – 8 (-2) = 8 -8 + 28 = 20 -8 + 16 = 8 Yes Yes

Steps to solve a linear system using elimination method: 1. Add or subtract the equations to eliminate one of the

variables. Then solve for the other variable 2. Substitute the expression from step 1 into either the

original equation and solve for the other variable.

3.2Solving Linear Systems by Elimination Wednesday September 24, 2014

Example: Solve the system using the elimination method 3x - 7y = 10 (1) 6x - 8y = 8 (2) Solution: Step 1. Multiply (1) with 2 2 (3x - 7y = 10) --- 6 x - 14y = 20 Step 2. Subtract (1) and (2) to eliminate x 6x - 14y = 20 (1) 6x - 8y = 8 (2) ------------------------ -6y = 12 y = - 2 Step 2. Substitute y = -2 into either (1) or (2) 6x - 8y = 8 (2) 6x - 8 (-2) = 8 6x + 16 = 8 6x = 8 - 16 6x = -8 x = -8/6 = -4/3 The solution is (-4/3, -2) Check: (1) 6(-4/3) - 14 (-2) = 20 (2) 6(-4/3) – 8 (-2) = 8 -8 + 28 = 20 -8 + 16 = 8 Yes Yes

Steps to solve a linear system using multiplication and elimination method: 1. Multiply on or both of the equation by a constant to make

the coefficient of one variable in both equations the same 2. Add or subtract the equations to eliminate one of the

variables. Then solve for the other variable 3. Substitute the expression from step 1 into either the

original equation and solve for the other variable.

3.2Solving Linear Systems by Multiplication and Elimination

Wednesday September 24, 2014

Warm-up: Graph and shade the solution area -x + y > 4 x + y < 3

Monday September 29, 2014

3.3 Graphing Systems of Linear Inequalities

Objective: To graph systems of linear inequalities in slope-intercept form and standard form

Example1: Graph y > -2x -5 y ≤ x + 3 Solution: Step 1: Step 2: for line 1, m = -2 and b = -5 for line 2, m = 1 and b = 3 Step 3: line 1 is dashed, line 2 is solid Step 4: Shade the regions using check points

EQ: Is graphing a system of linear inequalities any different than just a linear inequality?

Steps to graph a system of inequalities

1. Write each inequalities in slope-intercept form (y=mx+b) by solving for y 2. Use b to plot y-intercept then use m to plot the 2nd point for the 1st inequality 3. Use dashed line for > or <. Use solid line for ≥ or ≤

4. Shade the appropriate region

5. On the same graph, repeat steps 2,3, and 4 to plot and shade for the 2nd inequality 6. Solution is where the graph is double shaded.

7. Check the solution algebraically


Example1: Graph the system of inequalities 2x + 3y < 6 y ≥ -(2/3)x + 4 Solution: 1. Write in y=mx form (1) 3y = -2x +6 y = -(2/3) + 2 (2) y = - (2/3) x + 4 2. In (1), m = -(2/3), b =2 In (2), m =-(2/3), b =4 3. No solution

No intersection No solution


Warm-up: Solve by elimination: x + 2y = -1 (1) 3x – y = 18 (2)


Warm-up: Solve by elimination: x + 2y = -1 (1) 3x – y = 18 (2) Solution: 1. Multiply (1) with 3: 3x + 6y = -3 - 3x - y = 18 --------------------- 7 y = -21 y = -3 2. Substitute y = -3 to (2) 3x – (-3) = 18 3x + 3 = 18 3x = 15 x = 5


Steps to solve a linear system using elimination method: 1. Pick any 2 equations and eliminate 1 variable equation (4) 2. Pick 2 different equations and eliminate the same variable

equation (5) 3. Use equations (4) and (5) to solve for 1 of its variables. 4. Substitute the known variable to either (4) or (5) to find

the other variable. 5. Substitute the 2 known variables into any of the original

equations to solve for the 3rd variable.

3.4Solving Linear Systems in Three Variables:

Example: Solve the system: x + y + z = 3 (1) 4x + 4y + 4z = 7 (2) 3x - y + 2z = 5 (3) Solution: Step 1. Pick any 2 equations and eliminate 1 variable: Multilply (1) with 4 and subtract (2) 4x + 4y + 4z = 12 (4) - 4x + 4y + 4z = 7 (2) ----------------------------- 0 = 5 The system has no solution

Objective: To solve linear systems with 3 variables

EQ: How many different ways can 3 linear equations in 3 variables intersect?


Example: Solve the system using the elimination method 4x + 2y + 3z = 1 (1) 2x - 3y + 5z = -14 (2) 6x - y + 4z = -1 (3) Solution: Step 1. Multiply (3) with 2 and add to (1) to eliminate y 4x + 2y + 3z = 1 (1) 12x - 2y + 8z = -2 (2) ------------------------ 16x + 11z = -1 (4) Step 2. Multiply (3) with 3 and subtract to (2) to also eliminate y 2x - 3y + 5z = -14 (2) - { 18x – 3y + 12z = -3 } (3) -------------------------------- -16x - 7z = -11 (5) Step 3. Add (4) and (5) to eliminate x 16x + 11z = -1 (4) -16x - 7z = -11 (5) ------------------------------ 4z = -12 z = -3 Use (5) to solve for x: - 16x + 7(-3) = -11 - 16x - 21 = -11 - 16x = -32 x = 2 Step4: Substitute z = -3 and x = 2 to (3) 6(2) - y + 4(-3) = -1 - y = -1 – 12 +12 y = -1 The solution is (2, 1, -3)

3.4Solving Linear Systems in Three Variables: Tuesday September 30, 2014

Example: Solve the system: x + y + z = 4 (1) x + y - z = 4 (2) 3x + 3y + z = 12 (3) Solution: Step 1. Pick any 2 equations and eliminate 1 variable: Add (1) and (2) x + y + z = 4 (1) +{ x + y - z = 4 } (2) ----------------------------- 2x +2y = 8 (4) Step2: Pick 2 different equations and also eliminate z: x + y - z = 4 (2) 3x + 3y + z = 12 (3) -------------------------------- 4x + 4y = 16 (5) Step3: Use the new equations to eliminate the next variable Multiply (4) with 2 and subtract with (5) 4x + 4y = 16 4x + 4y = 16 (5) ---------------------------- 0 = 0 The system has infinitely many solutions.


Example: Solve the system using the substitution method 2a + b + c = 8 (1) - a + 3b – 2c = 3 (2) - a + b - c = 0 (3) Solution: Step 1. Rewrite 1 of the equations for 1 variable. - a + b - c = 0 (3) b = a + c (4) Step 2. Substitute (4) into (1) 2a + b + c = 8 (1) 2a + (a+c) +c = 8 2a + a + c + c = 8 3a + 2c = 8 (5) Step 3. Substitute (4) into (2) - a + 3b – 2c = 3 (2) - a + 3(a+c) – 2c = 3 - a + 3a + 3c – 2c = 3 2a + c = 3 (6) Step4: Use (5) and (6) to solve for 1 variable 3a + 2c = 8 (5) 2a + c = 3 (6) Multiply (6) with 2 3a + 2c = 8 (5) 4a + 2c = 6 (6) -------------------------------- - a = 2 a = - 2 Step 5: Substitute a = -2 into (5) or (6) 2(-2) + c = 3 -4 + c = 3 c = 7 Step 6: Substitute a = -2 and c = 7 into (3) b = a + c = -2 + 7 = 5 The solution is (-2, 5, 7)


tuesday september 23, 2014 warm-up: put the … solve the system using the substitution method 2x +...

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