chapter 6 –5y = -11 solve for y in terms of x. 2x + y = 0 y = -2x then solve by substitution...

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 EQUATION

is 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 LINE

m = y2– y

1

x2– x

1

Find the slope of the line that

contains the given points.

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

m = 3 –(-6)

-2 – 4

m = 3 + 6

-6

m = - 9/6 or -3/2

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 + b

where m is the slope and

b is the y -intercept

SECTION 6-2

Parallel and

Perpendicular Lines

SLOPE of PERPENDICULAR

LINES

Two lines are

perpendicular if the

product of their

slopes is -1

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 = -1 -5

0 – (-2)

m = -6

2

m = - 3

Find the slope of a line perpendicular to

the line containing points M and N.

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

m = -2 – (-1)

-5 - 4

m = -1

-9

m = -9

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=-2

P(-1, 3)

y-3=-2[x-(-1)]

y-3= -2(x+1)

y-3= -2x-2

y=-2x+1

Write an equation of a line

through the given points

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

y-(-3)=[2-(-3)](x-1)

3-1

y+3=5(x-1)

2

y+3=5x-5

2 2

y=5x-11

2 2

Write an equation of a line with

the given point and y-intercept

y=3 P(2, -1)

y-(-1)=3(x-2)

y+1=3x-6

y+1-1=3x-6-1

y=3x -7

Write an equation of a line parallel to

y=-1/3x+1 containing the point (1,1)

m=-1/3

P(1, 1)

y-1=-1/3(x-1)

y-1= -1/3x+1/3

y= -1/3x+4/3

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

SECTION 6-5

Solve Systems by

Substitution

PRACTICE USING

DISTRIBUTIVE LAW

x + 2(3x - 6) = 2

x + 6x – 12 = 2

7x -12 = 2

7x =14

x = 2

PRACTICE USING

DISTRIBUTIVE LAW

-(4x – 2) = 2(x + 7)

-4x + 2 = 2x +14

2 = 6x + 14

-12 = 6x

-2 = x

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 = 6

x + 2y = 2

Solve for y in terms of x.

3x – y = 6

3x = 6 + y

3x – 6 = y then

SOLVE BY SUBSTITUTION

Substitute the value of y

into the second equation

x + 2y = 2

x + 2(3x – 6) = 2

x + 6x – 12 = 2

7x = 14

x = 2 now

SOLVE BY SUBSTITUTION

Substitute the value of x

into the first equation

3x – y = 6

y = 3x – 6

y = 3(2 – 6)

y = 3(-4)

y = -12

SOLVE BY SUBSTITUTION

2x + y = 0

x – 5y = -11

Solve for y in terms of x.

2x + y = 0

y = -2x

then

SOLVE BY SUBSTITUTION

Substitute the value of y

into the second equation

x – 5y = -11

x – 5(-2x) = -11

x+ 10x = -11

11x = -11

x = -1

SOLVE BY SUBSTITUTION

Substitute the value of x

into the first equation

2x + y = 0

y = -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.

STEPS FOR ADDITION OR

SUBTRACTION METHOD

Solve the resulting

equation for the

remaining variable.

STEPS FOR ADDITION OR

SUBTRACTION METHOD

Substitute the value for

the variable in one of the

original equations and

solve for the unknown

variable.

STEPS FOR ADDITION OR

SUBTRACTION METHOD

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 = 10

3y = 2x – 7

SOLUTION

3x – 4y = 10

-2x +3y = -7

Multiply equation 1 by 2

Multiply 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 = 10

3x – 4(-1) = 10

3x + 4 = 10

3x = 6

x = 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 b

c d

SOLVE USING

DETERMINANTS

x + 3y = 4

-2x + y = -1

SOLVE USING

DETERMINANTS

x + 3y = 4

-2x + y = -1

Matrix A = 1 3

-2 1

SOLVE USING

DETERMINANTS

Matrix Ax

= 4 3

-1 1

x = det Ax

/det A

SOLVE USING

DETERMINANTS

det Ax

= 4(1) – (3)(-1)

= 4 + 3

=7

SOLVE USING

DETERMINANTS

Det A = 1(1) – (3)(-2)

= 1 + 6

=7 thus

x = 7/7 = 1

SOLVE USING

DETERMINANTS

Matrix Ay

= 1 4

-2 -1

y = det Ay /det A

SOLVE USING

DETERMINANTS

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 < 5

2x – 3y ≤ 1

SOLVE BY GRAPHING THE

INEQUALITIES

4x - y 5

8x + 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

END