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