7.1 solving linear systems by graphing · 7.1 solving linear systems by graphing objectives: •...

7.1 Solving Linear Systems by Graphing Objectives:

• Learn how to solve a system of linear equations by graphing

• Learn how to model a real-life situation using a system of linear equations

With an equation, any point on the line (x, y) is called a solution to the equation. With 2 or more equations, any point that is true for both equations is also a solution to the system. Is (2, –1) the solution to the system?

3x + 2y = 4

– x + 3y = –5

Plug in the (x, y) values

3(2) + 2(–1) = 4 – (2) + 3(–1) = –5

6 – 2 = 4 – 2 – 3 = –5

4 = 4 – 5 = – 5 Yes Yes

Check by graphing: write in slope-intercept form first

3x + 2y = 4 – x + 3y = –5

2y = –3x + 4 3y = x – 5

y = − 32 x + 2 y = 1

3 x – 53

Mr. Noyes, Akimel A-al Middle School 1 Heath Algebra 1 - An Integrated Approach

(2, –1)4–4–8 x

2

4

6

8

–2

–4

–6

–8

y

7.1 Solving Linear Systems by Graphing Graph and check:

y = x + 1 y = – x + 5

Check: Substitute solution

into each equation 3 = 2 + 1 3 = – (2) + 5

3 = 3 3 = 3 Yes Yes

Solve: 2 4

2x y

x y+ =− =

Rewrite 2x + y = 4 in slope-intercept form: y = – 2x + 4 Rewrite x – y = 2 in slope-intercept form: y = x – 2

check: Substitute (2, 0) into each equation

y = – 2x + 4

0 = –2(2) + 4

0 = 0

y = x – 2

0 = 2 – 2

0 = 0

(2, 3)

1 2 3 4 5 6–1–2–3–4–5–6 x

123456

–1–2–3–4–5–6

y

(2, 0)

1 2 3 4 5 6–1–2–3–4–5–6 x

123456

–1–2–3–4–5–6

y

Solution (2, 0)

Solution


7.1 Solving Linear Systems by Graphing Examples: You invest $9,000 in two different accounts. One account pays 5% interest and the other pays 6% interest. If you earn $510 in total interest, how much did you invest in each account?

Equation #1: .05x + .06y = 510 Equation #2: x + y = 9,000 Solve by graphing (find the x-, y-intercepts) When x = 0 When y = 0 x + y = 9,000 x + y = 9,000 y = 9,000 x = 9,000 .06y = 510 .05x = 510

y = 8,500 x = 10,200 Graph is upper right quadrant, crossing at (3000, 6000)

4000 8000 x

4000

8000

y

Answer: $3,000 is invested at 5% and $6,000 is invested at 6%.

Check answer to make sure!


7.2 Solving Linear Systems by Substitution Objectives:

• Learn how to use substitution to solve a linear equation

• Learn how to model real-life situations using a system of linear equations

Basic Steps: 1) Solve one of the equations for one of its variables

2) Substitute that expression into the other equation and solve for the other variable

3) Substitute this value into the revised first equation and solve

4) Check the solution pair in each of the original equations

Example:

– x + y = 1

2x + y = –2

1) Solve Equation #1 for y:

Equation #1: – x + y = 1

y = x + 1 (call this Revised Equation #1)

2) Substitute the expression found for y from equation #1 into equation #2:

Equation #2: 2x + y = –2

2x + (x + 1) = –2

2x + x + 1 = –2

3x + 1 = –2 – 1 – 1 3x = –3 3 3

x = – 1


7.2 Solving Linear Systems by Substitution

3) Substitute the solution found for x back into Revised Equation #1 and solve for y:

Revised Equation #1: y = x + 1

y = (– 1) + 1

y = 0

solution: (– 1, 0) 4) Check by substituting the solution into both equations:

Equation #1 Equation #2

– x + y = 1 2x + y = –2

– (– 1) + (0) = 1 2(– 1) + (0) = –2

1 + 0 = 1 – 2 + 0 = – 2

1 = 1 – 2 = – 2 Since the statements both equations make are true when substituting the values (–1, 0) in each, we know point (–1, 0) is a solution

Graph to check:

(–1, 0)1 2 3 4 5 6 7–1–2–3–4–5–6–7–8 x

12345678

–1–2–3–4–5–6–7–8

y


7.2 Solving Linear Systems by Substitution Real-Life Example: Dinner at China Buffet

Adult Dinner price: $11.95 Child Dinner price: $6.95

If your bill for 6 people is $61.70, how many adults and children were in your dinner party? Verbal Model:

EQ #1: # of Adults + # of Children = Total # People in Dinner Party

EQ #2: Price per Adult · # of Adults + Price per Child · # of Children = Total Bill

Labels:

Let A = # of adults

C = # of children

Adult Dinner price = $11.95

Child Dinner price = $6.95

Total Bill = $61.70

Total # People in Dinner Party = 6

Algebraic Model:

EQ #1: A + C = 6

EQ #2: 11.95 · A + 6.95 · C = 61.70

Step 1: Solve Equation #1 for A

EQ #1: A + C = 6

A = – C + 6 (Revised Equation #1)


7.2 Solving Linear Systems by Substitution


Step 2: Substitute the expression found for A into Equation #2 and solve for C

EQ #2: 11.95 · A + 6.95 · C = 61.70

11.95 · (– C + 6) + 6.95 · C = 61.70

71.70 – 11.95C + 6.95C = 61.70

71.70 – 5C = 61.70

–5C = –10

C = 2

Step 3: Substitute the solution found for C into Revised Equation #1 and solve for A

Revised EQ #1: A = – C + 6

A = – (2) + 6

A = 4 Step 4: Check by substituting the solution into both equations:

Equation #1 Equation #2

A + C = 6 11.95A + 6.95C = 61.70

(4) + (2) = 6 11.95(4) + 6.95(2) = 61.70

6 = 6 47.80 + 13.90 = 61.70

61.70 = 61.70

7.3 Solving Linear Systems by Linear Combinations Objectives:

• Learn to use linear combinations to solve a linear system

• Learn how to model a real-life situation using a system of linear equations

Linear Combinations: A Third way of Solving a Linear System of Equations

Basic Steps:

1) Arrange equations with like terms in columns.

2) Study the x or y coefficients. Multiply one or both equations by an appropriate number to obtain new coefficients for x or y that are opposites.

3) Add the two equations together. The sum should have eliminated one of the variables. Solve for the remaining variable

4) Substitute the value in Step 3 into either of the original equations and solve for the other variable. You now have an ordered pair

5) Check the solution in each original equation. If both are true, you have a solution


7.3 Solving Linear Systems by Linear Combinations Example #1: Equation #1: 4x + 3y = 1 Equation #2: 2x – 3y = 1 (since 3y and – 3y are opposites, add equations) 6x = 2 6 6

x = 13

x = 13 (substitute 13 back into one of the equations and solve for y)

Equation #1: 4x + 3y = 1 (Substituting into Equation #1)

4( )13 + 3y = 1

43 + 3y = 1

3y = –13

y = – 19

Solution: ( )1 1,3 9−

Check: Substitute the solution into the other Equation

Equation #2: 2x – 3y = 1

2( )13 – 3( )1

9− = 1

23 + 3

9 = 1

1 = 1


7.3 Solving Linear Systems by Linear Combinations Example #2:

Equation #1: 3x + 5y = 6 Equation #2: – 4x + 2y = 5 (won’t combine as currently written)

Try to get rid of either x or y

Multiply top equation by 4 and bottom equation by 3 to end up with 12x and – 12x. When added together, the x’s will cancel each other out.

4(3x + 5y) = 4(6) 12x + 20y = 24 3(– 4x + 2y) = 3(5) –12x + 6y = 15 26y = 39 y = 3

2

Solve for x: Substitute the solution found for y into either one of the equations and solve for x:

Equation #1: 3x + 5y = 6

3x + 5( )32 = 6

3x + 152 = 6

3x = – 32

x = – 12

Solution: ( )31 ,2 2−


7.3 Solving Linear Systems by Linear Combinations

Check: Substitute the solution into the other Equation

Equation #2: – 4x + 2y = 5

– 4( )12− + 2( )3

2 = 5

2 + 3 = 5

5 = 5 Helpful Hint: When there are fractions in the equations, clear the fractions first. Choosing when to use each method:

Graph: to approximate a solution, check a solution or show visual. Substitution or combination: for exact solution Substitution: when one variable has 1 or –1 for the coefficient Combination: when coefficients are larger


7.4 Problem Solving Using Linear Systems Objective:

• Learn how to write and use a linear system as a real-life model

Use any of the 3 Methods for Solving Linear Systems of Equations: 1. Graph 2. Substitute 3. Combine

Example 1:

Sam sells gloves at a department store. At the end of the day, Sam sold 20 pair of

gloves. Of the 3 types, Sam only sold knit gloves at $7.95 and cloth gloves at $11.50.

None of the leather gloves at $27.50 sold. If Sam sold a total of $212.25 for the day,

how many gloves of each type did Sam sell?

k = knit gloves c = cloth gloves

EQ #1: k + c = 20

EQ #2: 7.95k + 11.50c = 212.25 Use the Substitution Method for this problem

Revised EQ #1: k = 20 – c

EQ #2: 7.95(20 – c) + 11.50c = 212.25

159 – 7.95c + 11.50c = 212.25

159 + 3.55c = 212.25

3.55c = 53.25

c = 15


7.4 Problem Solving Using Linear Systems

Revised EQ #1: k = 20 – c

k = 20 – (15)

k = 5 You sold 15 cloth gloves and 5 knit gloves.

Check:

EQ #2: 7.95k + 11.50c = 212.25

7.95(5) + 11.50(15) = 212.25

39.75 + 172.50 = 212.25

212.25 = 212.25

Example 2: Two planes leave the same airport but fly in opposite directions. If

plane #2 flies 50 mph faster than plane #1, but starts one-half hour later, how fast is

each plane flying when after 2 hours they are 1,825 miles apart?

*remember D = rt

Verbal Model:

EQ #1: Plane #2 speed = Plane #1 speed + 50 mph

EQ #2: Plane #1 speed · time + Plane #2 speed · time = Distance Apart

Labels:

plane #1’s speed = x

plane #2’s speed = x + 50

plane #1’s time = t = 2 hours

plane #2’s time = 2 – 0.5 = 1.5 hours (remember, 30 minutes is 0.5 hours)


7.4 Problem Solving Using Linear Systems Algebraic Model:

EQ #1: plane #2 speed = x + 50

EQ #2: 2x + 1.5(x + 50) = 1825

2x + 1.5x + 75 = 1825

3.5x + 75 = 1825

3.5x = 1750

x = 500

x + 50 = 550 Plane 1’s speed is 500 mph

Plane 2’s speed is 550 mph

Check:

2x + 1.5(x + 50) = 1825

2(500) + 1.5(500 + 50) = 1825

1000 + 1.5(550) = 1825

1000 + 825 = 1825

1825 = 1825


7.4 Problem Solving Using Linear Systems Example 3: (See problem #22 on page 372 of textbook)

x + 5y = 45 x = –5y + 45 2x – y = 2

2(–5y + 45) – y = 2

– 10y + 90 – y = 2 – 90 – 90 – 11y = – 88 11 11

y = 8

x + 5y = 45

x + 5(8) = 45

x + 40 = 45 – 40 – 40

x = 5

Solution: (5, 8)

Check:

x + 5y = 45 2x – y = 2 (5) + 5(8) = 45 2(5) – (8) = 2 5 + 40 = 45 10 – 8 = 2 45 = 45 2 = 2 Graph: y = mx + b

x + 5y = 45 2x – y = 2 – x – x – 2x – 2x 5y = – x + 45 – y = – 2x + 2 5 5 5 – 1 – 1 – 1

y = – 15 x + 9 y = 2x – 2

(5, 8)

2 4 6 8 10 12–2–4–6–8–10–12 x

2468

1012

–2–4–6–8

–10–12

y


7.5 Special Types of Linear Systems Objectives:

• Learn how to visualize the solutions possible for linear systems

• Learn how to identify a linear system as having many solutions

A farmer keeps track of his cows and chickens by counting legs and heads. If he counts 78 legs and 35 heads, how many cows and chickens does he have? How many ways can you solve this? Solve by using systems of equations, substitution method. a = chickens c = cows a + c = 35 ⇒ c = 35 – a 2a + 4c = 78 2a + 4(35 – a) = 78 2a + 140 – 4a = 78 – 2a = – 62 a = 31 31 + c = 35 c = 4 There are 4 cows and 31 chickens.


7.5 Special Types of Linear Systems Special Linear Systems:

Intersecting Parallel Same Line one solution

(an ordered pair) no solution

(a false statement) Many (infinite) solutions

(the same line) (x, y) 0 = 2 0 = 0

6 = 6

x

y

x

y

x

y

When you solve each system, you either get an ordered pair, a false statement or both sides are equal. Example #1: Solve by substitution or combination then graph.

3x – 2y = 3

– 6x + 4y = – 6

Try combination -- Multiply top equation by 2

6x – 4y = 6 – 6x – 4y = – 6 0 = 0

What does this mean? It means that there are many solutions because each equation describes the same line.


7.5 Special Types of Linear Systems

Graph to check: Change to y = mx + b EQ #1: y = 3

2 x – 32

EQ #2: y = 32 x – 3

2

What do you notice?

• same slope

• same intercept

• same line!

• MANY SOLUTIONS

Example #2: Solve by substitution or combination then graph.

2 4 6–2–4–6 x

2

4

6

–2

–4

–6

y

x + y = 8

x + y = –1

Try substitution: Revised EQ #1: y = – x + 8

EQ #2: x + (– x + 8) = –1

x – x + 8 = –1

8 = –1 (This statement clearly isn’t true!)

No solution, this system of equations describe parallel lines.


7.5 Special Types of Linear Systems Change each to slope-intercept form:

Revised EQ #1: y = – x + 8

Revised EQ #2: y = – x – 1

What do you notice?

• same slope

• different y-intercept

• parallel lines!

• NO SOLUTIONS

Given a graph, write a system of equations to match

y = 2x – 1

2y = 4x – 2

y = 2x + 2

y = 2x – 2 y = 2

3 x + 2

x = 2

2 4 6–2–4–6 x

2

4

6

–2

–4

–6

y

2 4 6–2–4–6 x

2

4

6

–2

–4

–6

y

2 4 6–2–4–6 x

2

4

6

–2

–4

–6

y

• same slope

• same y-intercept

• same line!

• MANY SOLUTIONS

• same slope

• different y-intercept

• parallel lines!

• NO SOLUTIONS

• different slope

• different intercepts

• intersecting lines!

• ONE SOLUTION

2 4 6 8–2–4–6–8 x

2

4

6

8

–2

–4

–6

–8

y


7.6 Solving Systems of Linear Inequalities Objectives:

• Learn how to solve a system of linear inequalities by graphing

• Learn how to model a real-life situation using a system of linear inequalities

When Graphing Inequalities . . .

• Greater Than (>) and Less Than (>) symbols get dashed ( - - - - - ) lines

• Greater Than or Equal To (≥) and Less Than or Equal To (≤) symbols get solid ( ) lines

• Pick a point and shade true side Graph the following system of Inequalities:

y < 2

x ≥ – 1

y > x – 2

Hint: Use colored pencils when shading inequalities Graph y < 2 graph x ≥ – 1 graph y > x – 2 dotted line where y = 2, shade down

solid line at x = –1, shade right

Dashed line with a y-intercept at –2, and slope 11. Shade up and left

1 2 3 4 5–1–2–3–4–5 x

12345

–1–2–3–4–5

y

1 2 3 4 5–1–2–3–4–5 x

12345

–1–2–3–4–5

y

1 2 3 4 5–1–2–3–4–5 x

12345

–1–2–3–4–5

y


7.6 Solving Systems of Linear Inequalities Combining all three graphs, we find an area where all three regions intersect. This area is the solution to the system of inequalities.

2 4–2–4 x

2

4

–2

–4

y

Solution is the area where the inequalities intersect Graph and Shade: (you may want to use colored pencils)

2 4–2–4 x

2

4

–2

–4

y

Find the solution for this system:

y ≤ 4 y ≥ 0 y ≤ – 2x + 12 x ≥ 0


7.6 Solving Systems of Linear Inequalities Solve:

y ≥ 12 x + 5

y ≤ x + 2

y ≥ 12 x + 5

y ≤ x + 2

2 4 6 8 10 12 14 16–2–4–6–8–10–12–14–16 x

2468

10121416

–2–4–6–8

–10–12–14–16

y


7.7 Exploring Data: Linear Programming


Objectives:

• Learn how to solve a linear programming problem

• Learn how to model a real-life situation using linear programming

The process of Optimization means finding the minimum and/or maximum value of some quantity . . . one method for doing this is called Linear Programming. A Linear Programming problem consists of two parts:

1. An Objective Quantity - the numeric value of a quantity (expressed in terms of an

equation) we want to minimize and/or maximize

2. Constraints – limiting factors expressed in terms of a system of linear inequalities

To find the minimum or maximum value of the objective quantity, find all the vertices of

the graph of the constraint inequalities and then evaluate the objective quantity at each

vertex. The smallest value is the minimum, and the largest value is the maximum.

Example 1: Find the minimum value and maximum value of

C = 3x + 2y Objective Quantity

Subject to the following constraints:

x ≥ 0

y ≥ 0 Constraints

x + y ≤ 4



Solution: The three vertices are (0, 0), (4, 0), and (0, 4). To find

the minimum and maximum values of C, evaluate

C = 3x + 2y at each of the three vertices:

At (0, 0): C = 3(0) + 2(0) = 0 (Minimum Value of C)

At (4, 0): C = 3(4) + 2(0) = 12 (Maximum Value of C)

At (0, 4): C = 3(0) + 2(4) = 8

The minimum value of C is 0 and it occurs when x = 0

and y = 0. The maximum value is 12, when x = 4 and y = 0. When evaluating C at

other points in the graph, the value of C will always fall between 0 and and 12.

Example 2: Finding the Maximum Profit Suppose you own a skateboard manufacturing company that makes a premium

skateboard by using two very different processes. The hours of skilled labor and

machine time per skateboard can be found in the matrix below. You can use up to

1796 hours of skilled labor and up to 3500 hours of machine time. How many

skateboards should be made by each process to maximize your profits?

ProcessAssembly hours A B

Skilled labor 3 1

Machine Time 1 3

ProcessAssembly hours A B

Skilled labor 3 1

Machine Time 1 3

Skateboards made by Process A

cost $35 each, and skateboards

made by process B cost $15 each.



Solution:

Let a and b represent the number of skateboards assembled with each process.

Because you want to maximize profit, P, the objective quantity is:

P = 35a + 15b Profit factors for each process

The constraints are as follows:

3a + b ≤ 1796 Skilled labor: Up to 1796 hours

a + 3b ≤ 3500 Machine time: Up to 3500 hours

a ≥ 0, b ≥ 0 Cannot produce negative amounts

At (0, 1166 ⅔): P = 35(0) + 15(1166 ⅔) = $17,500

At (236, 1088): P = 35(236) + 15(1088) = $24,580 Maximum Profit

At (598 ⅔, 0): P = 35(598 ⅔) + 15(0) = $20,953

At (0, 0): P = 35(0) + 15(0) = $0

The maximum profit is obtained by making 236 skateboards with Process A and

1088 skateboards with Process B.

7.1 solving linear systems by graphing · 7.1 solving linear systems by graphing objectives: •...

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