CHAPTER 6 Linear Systems of Equations. SECTION 6-1 Slope of a Line and Slope-Intercept Form.

CHAPTER 6

Linear Systems of Equations

SECTION 6-1

Slope of a Line and Slope-Intercept Form

COORDINATE PLANE consists of two

perpendicular number lines, dividing the plane into four regions called

quadrants

X-AXIS - the horizontal number line

Y-AXIS - the vertical number line

ORIGIN - the point where the

x-axis and y-axis cross

ORDERED PAIR - a unique assignment of real

numbers to a point in the coordinate plane

consisting of one x-coordinate and one y-

coordinate

(-3, 5), (2,4), (6,0), (0,-3)

COORDINATE PLANE

LINEAR EQUATIONis an equation whose

graph is a straight line.

SLOPE

is the ratio of vertical change to the

horizontal change. The variable m is used to

represent slope.

m = change in y-coordinate change in x-coordinate

Or m = rise run

FORMULA FOR SLOPE

SLOPE OF A LINEm = y2 – y1

x2 – x1

Find the slope of the line that contains the given points.

M(4, -6) and N(-2, 3)

M(-2,1) and N(4, -2)

M(0, 0) and N(5, 5)

Y-Intercept

is the point where the line intersects the y -

axis.

X-Intercept

is the point where the line intersects the

x -axis.

HORIZONTAL LINE

a horizontal line containing the point

(a, b) is described by the equation y = b

VERTICAL LINE

a vertical line containing the point (c, d) is described by the equation x = c

SLOPE-INTERCEPT FORM

y = mx + bwhere m is the slope and b

is the y -intercept

Find the Slope and Intercept

y = 2x - 7

2y = 4x – 8

2x + 2y = 4

-4x + 7y = 28

SECTION 6-2

Parallel and Perpendicular Lines

SLOPE of PARALLEL LINES

Two lines are parallel if their slopes are equal

Find the slope of a line parallel to the line containing points M and N.

M(-2, 5) and N(0, -1)


M(3, 5) and N(0, 6)


M(-2, -6) and N(2, 1)

SLOPE of PERPENDICULAR LINES

Two lines are perpendicular if the

product of their slopes is -1

Find the slope of a line perpendicular to the line containing points M and N.

M(4, -1) and N(-5, -2)


M(3, 5) and N(0, 6)


M(-2, -6) and N(2, 1)

Determine whether each pair of lines is parallel, perpendicular, or neither

7x + 2y = 147y = 2x - 5


-5x + 3y = 23x – 5y = 15


2x – 3y = 68x – 4y = 4

SECTION 6-3

Write Equations for Lines

POINT-SLOPE FORM

y – y1 = m (x – x1)where m is the slope and

(x1 ,y1) is a point on the line.

Write an equation of a line with the given slope and through a given point

m=-2P(-1, 3)

Write an equation of a line through the given points

A(1, -3) B(3,2)

Write an equation of a line with the given point and y-intercept

b=3 P(2, -1)

Write an equation of a line parallel to y=-1/3x+1 containing the point (1,1)

m=-1/3P(1, 1)

Write an equation of a line perpendicular to y=2x+1 containing the point (2,1)

M=2P(2, 1)

SECTION 6-4

Systems of Equations

SYSTEM OF EQUATIONS

Two linear equations with the same two variable

form a system of equations.

SOLUTION

The ordered pair that makes both equations

true.

SOLUTION

The point of intersection of the two lines.

INDEPENDENT SYSTEM

The graph of each equation intersects in

one point.

INCONSISTENT SYSTEM

The graphs of each equation do not

intersect.

DEPENDENT SYSTEM

The graph of each equation is the same. The lines coincide and

any point on the line is a solution.

SOLVE BY GRAPHING

4x + 2y = 8

3y = -6x + 12

SOLVE BY GRAPHING

y = 1/2x + 3

2y = x - 2

SOLVE BY GRAPHING

x + y =8

x-y = 4

SECTION 6-5

Solve Systems by Substitution

SYSTEM OF EQUATIONS

Two linear equations with the same two variable

form a system of equations.

SOLUTION

The ordered pair that makes both equations

true.

SOLUTION

The point of intersection of the two lines.

PRACTICE USING DISTRIBUTIVE LAW

x + 2(3x - 6) = 2

PRACTICE USING DISTRIBUTIVE LAW

-(4x – 2) = 2(x + 7)

SUBSTITUTION

A method for solving a system of equations by solving for one variable

in terms of the other variable.

SOLVE BY SUBSTITUTION

3x – y = 6x + 2y = 2

Solve for y in terms of x.3x – y = 63x = 6 + y

3x – 6 = y then


Substitute the value of y into the second equation

x + 2y = 2x + 2(3x – 6) = 2x + 6x – 12 = 2

7x = 14x = 2 now


Substitute the value of x into the first equation

3x – y = 6y = 3x – 6

y = 3(2 – 6)y = 3(-4)y = -12


2x + y = 0x – 5y = -11

Solve for y in terms of x.2x + y = 0

y = -2xthen


Substitute the value of y into the second equation

x – 5y = -11x – 5(-2x) = -11x+ 10x = -11

11x = -11x = -1


Substitute the value of x into the first equation

2x + y = 0y = -2x

y = -2(-1)y = 2

SECTION 6-6

Solve Systems by Adding and Multiplying

ADDITION/SUBTRACTION METHOD

Another method for solving a system of equations where

one of the variables is eliminated by adding or

subtracting the two equations.

STEPS FOR ADDITION OR SUBTRACTION METHOD

If the coefficients of one of the variables are opposites, add the equations to eliminate one of the variables. If the coefficients of one of the variables are the same, subtract the equations to eliminate one of the variables.


Solve the resulting equation for the remaining variable.


Substitute the value for the variable in one of the original equations and solve for the unknown variable.


Check the solution in both of the original equations.

MULTIPLICATION AND ADDITION METHOD

This method combines the multiplication property of

equations with the addition/subtraction

method.

SOLVE BY ADDING AND MULTIPLYING

3x – 4y = 103y = 2x – 7

SOLUTION

3x – 4y = 10-2x +3y = -7

Multiply equation 1 by 2Multiply equation 2 by 3

SOLUTION

6x – 8y = 20-6x +9y = -21

Add the two equations.

y = -1

SOLUTION

Substitute the value of y into either equation and solve for

3x – 4y = 103x – 4(-1) = 10

3x + 4 = 103x = 6x = 2

SECTION 6-7

Determinants & Matrices

MATRIX

An array of numbers arranged in rows and

columns.

SQUARE MATRIX

An array with the same number of rows and

columns.

DETERMINANT

Another method of solving a system of

equations.

DETERMINANT OF A SYSTEM OF EQUATIONS

The determinant of a system of equations is

formed using the coefficient of the variables

when the equations are written in standard from.

DETERMINANT VALUE

Is the difference of the product of the

diagonals (ad – bc).a bc d

SOLVE USING DETERMINANTS

x + 3y = 4-2x + y = -1


x + 3y = 4-2x + y = -1

Matrix A = 1 3 -2 1


Matrix Ax = 4 3 -1 1

x = det Ax /det A


det Ax = 4(1) – (3)(-1)

= 4 + 3=7


Det A = 1(1) – (3)(-2)

= 1 + 6=7 thus

x = 7/7 = 1


Matrix Ay = 1 4 -2 -1

y = det Ay /det A


det Ay = -1(1) – (4)(-2)

= -1 + 8=7 thus

y = 7/7 = 1

SECTION 6-8

Systems of Inequalities

SYSTEM OF LINEAR INEQUALITIES

A system of linear inequalities can be solved by graphing

each equation and determining the region where

the inequality is true.

SYSTEM OF LINEAR INEQUALITIES

The intersection of the graphs of the inequalities

is the solution set.

SOLVE BY GRAPHING THE INEQUALITIES

x + 2y < 52x – 3y ≤ 1

SOLVE BY GRAPHING THE INEQUALITIES

4x - y 58x + 5y ≤ 3

SECTION 6-9

Linear Programming

LINEAR PROGRAMMING

A method used by business and government to help manage resources and

time.

CONSTRAINTS

Limits to available resources

FEASIBLE REGION

The intersection of the graphs of a system of

constraints.

OBJECTIVE FUNCTION

Used to determine how to maximize profit while

minimizing cost

ENDEND

CHAPTER 6 Linear Systems of Equations. SECTION 6-1 Slope of a Line and Slope-Intercept Form.

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