~ Chapter 7 ~ Systems of Equations & Inequalities Algebra I Lesson 7-1 Solving Systems by Graphing Lesson 7-2 Solving Systems Using Substitution Lesson

~ Chapter 7 ~Systems of Equations &

Inequalities

Algebra I

Lesson 7-1 Solving Systems by Graphing

Lesson 7-2 Solving Systems Using Substitution

Lesson 7-3 Solving Systems Using Elimination

Lesson 7-4 Applications of Linear Systems

Lesson 7-5 Linear Inequalities

Lesson 7-6 Systems of Linear Inequalities

Chapter Review

Algebra I

Solving Systems by

Graphing Cumulative Review Chap 1-6

Lesson 7-1

Solving Systems by

Graphing Cumulative Review Chap 1-10

Lesson 7-1

Solving Systems by

Graphing NotesLesson 7-1

System of linear equations – Two or more linear equations together… One way to solve a system of linear equations is by… Graphing.

Solving a System of Equations

Step 1: Graph both equations on the same plane.

(Hint: Use the slope and the y-intercept or x- & y-intercepts to graph.)

Step 2: Find the point of intersection

Step 3: Check to see if the point of intersection makes both equations true.

Solve by graphing. Check your solution.

y = x + 5

y = -4x

Your turn…

y = -1/2 x + 2

y = -3x - 3

~ Try another one ~

x + y = 4

x = -1

Solving Systems by

Graphing NotesLesson 7-1

Systems with No Solution

When two lines are parallel, there are no points of intersection; therefore, the system has NO SOLUTION!

y = -2x + 1

y = -2x – 1

Systems with Infinitely Many Solutions

y = 1/5x + 9

5y = x + 45

Since they are graphs of the same line… There are an infinite number of solutions.

Solving Systems by Graphing

HomeworkLesson 7-1

Homework – Practice 7-1

#1-28 odd

Solving Systems by Substitution

Practice 7-1Lesson 7-2

Solving Systems Using Substitution

NotesLesson 7-2

Using Substitution

Step 1: Start with one equation.

Step 2: Substitute for y using the other equation.

Step 3: Solve the equation for x.

Step 4: Substitute solution for x and solve for y

Step 5: Your x & y values make the intersection point (x, y).

Step 6: Check your solution.

y = 2x

7x – y = 15

Your turn…

y = 4x – 8

y = 2x + 10

~ Another example~

c = 3d – 27

4d + 10c = 120


NotesLesson 7-2

Using Substitution & the Distributive Property

3y + 2x = 4

-6x + y = -7

Step 1: Solve the equation in which y has a coefficient of 1…

-6x + y = -7

+6x +6x

y = 6x -7

Step 2: Use the other equation (substitute using the equation from Step 1.)

3y + 2x = 4

3(6x – 7) + 2x = 4

18x – 21 + 2x = 4

20x = 25

x = 1 1/4

Step 3: Solve for the other variableSubstitute 1 ¼ or 1.25 for xy = 6(1.25) – 7y = 7.5 -7y = 0.5Solution is (1.25, 0.5)


NotesLesson 7-2

Your turn… 6y + 8x = 28

3 = 2x – y

Solution is (2.3, 1.6) or (2 3/10, 1 3/5)

A rectangle is 4 times longer than it is wide. The perimeter of the rectangle is 30 cm. Find the dimensions of the rectangle.

Let w = width

Let l = length

l = 4w

2l + 2w = 30

Solve for l…

l = 4(3)

l = 12

Use substitution to solve.2(4w) + 2w = 30 8w + 2w = 3010w = 30w = 3


HomeworkLesson 7-2

Homework ~ Practice 7-2 even

Solving Systems Using Elimination Practice 7-2

Lesson 7-3

Solving Systems Using Elimination

NotesLesson 7-3

Adding Equations

Step 1: Eliminate the variable which has a coefficient sum of 0 and solve.

Step 2: Solve for the eliminated variable.

Step 3: Check the solution.

5x – 6y = -32

3x + 6y = 48

8x + 0 = 16

x = 2

Solution is (2, 7)

Check

3(2) + 6(7) = 48

6 + 42 = 48

48 = 48

Your turn… 6x – 3y = 3 & -6x + 5y = 3

5x – 6y = - 325(2) – 6y = - 3210 – 6y = -32-6y = -42y = 7


NotesLesson 7-3

Multiplying One Equation

Step 1: Eliminate one variable.

-2x + 15y = -32

7x – 5y = 17

Step 2: Multiply one equation by a number that will eliminate a variable.

-2x + 15y = -32

3(7x – 5y = 17)

Step 3: Solve for the variable

19x = 19

x = 1

Step 4: Solve for the eliminated variable using either original equation.

-2(1) + 15y = -32

Solution (1, -2)

-2x + 15y = -32

21x - 15y = 51

19x + 0 = 19

-2 + 15y = -32 15y = -30 y = -2


NotesLesson 7-3

Your turn… 3x – 10y = -25

4x + 40y = 20

Solution (-5, 1)

Multiply Both Equations

Step 1: Eliminate one variable.

4x + 2y = 14

7x – 3y = -8

Step 2: Solve for the variable

26x = 26

x = 1

Try this one… 15x + 3y = 9

10x + 7y = -4

3(4x + 2y = 14)

2(7x – 3y = -8)

12x + 6y = 42

14x – 6y = -16

26x + 0 = 26Step 3: Solve for the eliminated variable

4(1) + 2y = 14

2y = 10

y = 5 Solution (1, 5)


HomeworkLesson 7-3

Homework – Practice 7-3 odd

Applications of Linear Systems Practice 7-3Lesson 7-4

Applications of Linear Systems NotesLesson 7-4

Applications of Linear Systems

HomeworkLesson 7-4

Homework – Practice 7-4 #6-10

Linear InequalitiesPractice 7-4Lesson 7-5

Linear InequalitiesNotesLesson 7-5

Using inequalities to describe regions of a coordinate plane:

x < 1

y > x + 1

y ≤ - 2x + 4

Steps for graphing inequalities…

(1) First graph the boundary line.

(2) Determine if the boundary line is a dashed or solid line.

(3)Shade above or below the boundary line… (< below or > above)

Graph y ≥ 3x - 1

Rewriting to Graph an Inequality

Graph 3x – 5y ≤ 10

Solve for y… (remember if you divide by a negative, the inequality sign changes direction) then apply the steps for graphing an inequality.

Graph 6x + 8y ≥ 12

Linear InequalitiesHomework

Lesson 7-5

Homework ~ Practice 7-5 odd

Systems of Linear Inequalities Practice 7-5Lesson 7-6



Systems of Linear Inequalities

NotesLesson 7-6

Solve by graphing…

x ≥ 3 & y < -2

You can describe each quadrant using inequalities…

Quadrant I?

Quadrant II?

Quadrant III?

Quadrant IV?

Graph a system of Inequalities…

(1) Solve each equation for y…

(2) Graph one inequality and shade.

(3) Graph the second inequality and shade.

(4) The solutions of the system are where the shading overlaps.

(5) Choose a point in the overlapping region and check in each inequality.

Systems of Linear InequalitiesNotesLesson 7-6

Graph to find the solution…

y ≥ -x + 2 & 2x + 4y < 4

Writing a System of Inequalities from a Graph

Determine the boundary line for the pink region…

y = x – 2

The region shaded is above the dashed line… so

y > x – 2

Determine the boundary line for the blue region…

y = -1/3x + 3

The region shaded is below the solid line… so

y ≤ -1/3x + 3

Your turn…

Systems of Linear InequalitiesPractice 7-6

Lesson 7-6

Homework 7-6 odd


Lesson 7-6


Lesson 7-6


Lesson 7-6

~ Chapter 7 ~Chapter Review

Algebra I Algebra I

~ Chapter 7 ~Chapter Review

Algebra I Algebra I

Documents

~ Chapter 7 ~ Systems of Equations & Inequalities Algebra I Lesson 7-1 Solving Systems by Graphing Lesson 7-2 Solving Systems Using Substitution Lesson