Download - CHAPTER 6 Linear Systems of Equations. SECTION 6-1 Slope of a Line and Slope-Intercept Form
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CHAPTER 6
Linear Systems of Equations
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SECTION 6-1
Slope of a Line and Slope-Intercept Form
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COORDINATE PLANE consists of two
perpendicular number lines, dividing the plane into four regions called
quadrants
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X-AXIS - the horizontal number line
Y-AXIS - the vertical number line
ORIGIN - the point where the
x-axis and y-axis cross
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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)
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COORDINATE PLANE
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LINEAR EQUATIONis an equation whose
graph is a straight line.
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SLOPE
is the ratio of vertical change to the
horizontal change. The variable m is used to
represent slope.
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m = change in y-coordinate change in x-coordinate
Or m = rise run
FORMULA FOR SLOPE
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SLOPE OF A LINEm = y2 – y1
x2 – x1
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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)
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Y-Intercept
is the point where the line intersects the y -
axis.
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X-Intercept
is the point where the line intersects the
x -axis.
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HORIZONTAL LINE
a horizontal line containing the point
(a, b) is described by the equation y = b
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VERTICAL LINE
a vertical line containing the point (c, d) is described by the equation x = c
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SLOPE-INTERCEPT FORM
y = mx + bwhere m is the slope and b
is the y -intercept
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Find the Slope and Intercept
y = 2x - 7
2y = 4x – 8
2x + 2y = 4
-4x + 7y = 28
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SECTION 6-2
Parallel and Perpendicular Lines
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SLOPE of PARALLEL LINES
Two lines are parallel if their slopes are equal
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Find the slope of a line parallel to the line containing points M and N.
M(-2, 5) and N(0, -1)
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Find the slope of a line parallel to the line containing points M and N.
M(3, 5) and N(0, 6)
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Find the slope of a line parallel to the line containing points M and N.
M(-2, -6) and N(2, 1)
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SLOPE of PERPENDICULAR LINES
Two lines are perpendicular if the
product of their slopes is -1
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Find the slope of a line perpendicular to the line containing points M and N.
M(4, -1) and N(-5, -2)
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Find the slope of a line perpendicular to the line containing points M and N.
M(3, 5) and N(0, 6)
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Find the slope of a line perpendicular to the line containing points M and N.
M(-2, -6) and N(2, 1)
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Determine whether each pair of lines is parallel, perpendicular, or neither
7x + 2y = 147y = 2x - 5
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Determine whether each pair of lines is parallel, perpendicular, or neither
-5x + 3y = 23x – 5y = 15
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Determine whether each pair of lines is parallel, perpendicular, or neither
2x – 3y = 68x – 4y = 4
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SECTION 6-3
Write Equations for Lines
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POINT-SLOPE FORM
y – y1 = m (x – x1)where m is the slope and
(x1 ,y1) is a point on the line.
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Write an equation of a line with the given slope and through a given point
m=-2P(-1, 3)
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Write an equation of a line through the given points
A(1, -3) B(3,2)
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Write an equation of a line with the given point and y-intercept
b=3 P(2, -1)
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Write an equation of a line parallel to y=-1/3x+1 containing the point (1,1)
m=-1/3P(1, 1)
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Write an equation of a line perpendicular to y=2x+1 containing the point (2,1)
M=2P(2, 1)
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SECTION 6-4
Systems of Equations
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SYSTEM OF EQUATIONS
Two linear equations with the same two variable
form a system of equations.
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SOLUTION
The ordered pair that makes both equations
true.
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SOLUTION
The point of intersection of the two lines.
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INDEPENDENT SYSTEM
The graph of each equation intersects in
one point.
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INCONSISTENT SYSTEM
The graphs of each equation do not
intersect.
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DEPENDENT SYSTEM
The graph of each equation is the same. The lines coincide and
any point on the line is a solution.
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SOLVE BY GRAPHING
4x + 2y = 8
3y = -6x + 12
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SOLVE BY GRAPHING
y = 1/2x + 3
2y = x - 2
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SOLVE BY GRAPHING
x + y =8
x-y = 4
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SECTION 6-5
Solve Systems by Substitution
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SYSTEM OF EQUATIONS
Two linear equations with the same two variable
form a system of equations.
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SOLUTION
The ordered pair that makes both equations
true.
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SOLUTION
The point of intersection of the two lines.
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PRACTICE USING DISTRIBUTIVE LAW
x + 2(3x - 6) = 2
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PRACTICE USING DISTRIBUTIVE LAW
-(4x – 2) = 2(x + 7)
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SUBSTITUTION
A method for solving a system of equations by solving for one variable
in terms of the other variable.
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SOLVE BY SUBSTITUTION
3x – y = 6x + 2y = 2
Solve for y in terms of x.3x – y = 63x = 6 + y
3x – 6 = y then
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SOLVE BY SUBSTITUTION
Substitute the value of y into the second equation
x + 2y = 2x + 2(3x – 6) = 2x + 6x – 12 = 2
7x = 14x = 2 now
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SOLVE BY SUBSTITUTION
Substitute the value of x into the first equation
3x – y = 6y = 3x – 6
y = 3(2 – 6)y = 3(-4)y = -12
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SOLVE BY SUBSTITUTION
2x + y = 0x – 5y = -11
Solve for y in terms of x.2x + y = 0
y = -2xthen
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SOLVE BY SUBSTITUTION
Substitute the value of y into the second equation
x – 5y = -11x – 5(-2x) = -11x+ 10x = -11
11x = -11x = -1
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SOLVE BY SUBSTITUTION
Substitute the value of x into the first equation
2x + y = 0y = -2x
y = -2(-1)y = 2
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SECTION 6-6
Solve Systems by Adding and Multiplying
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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.
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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.
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STEPS FOR ADDITION OR SUBTRACTION METHOD
Solve the resulting equation for the remaining variable.
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STEPS FOR ADDITION OR SUBTRACTION METHOD
Substitute the value for the variable in one of the original equations and solve for the unknown variable.
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STEPS FOR ADDITION OR SUBTRACTION METHOD
Check the solution in both of the original equations.
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MULTIPLICATION AND ADDITION METHOD
This method combines the multiplication property of
equations with the addition/subtraction
method.
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SOLVE BY ADDING AND MULTIPLYING
3x – 4y = 103y = 2x – 7
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SOLUTION
3x – 4y = 10-2x +3y = -7
Multiply equation 1 by 2Multiply equation 2 by 3
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SOLUTION
6x – 8y = 20-6x +9y = -21
Add the two equations.
y = -1
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SOLUTION
Substitute the value of y into either equation and solve for
3x – 4y = 103x – 4(-1) = 10
3x + 4 = 103x = 6x = 2
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SECTION 6-7
Determinants & Matrices
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MATRIX
An array of numbers arranged in rows and
columns.
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SQUARE MATRIX
An array with the same number of rows and
columns.
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DETERMINANT
Another method of solving a system of
equations.
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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.
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DETERMINANT VALUE
Is the difference of the product of the
diagonals (ad – bc).a bc d
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SOLVE USING DETERMINANTS
x + 3y = 4-2x + y = -1
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SOLVE USING DETERMINANTS
x + 3y = 4-2x + y = -1
Matrix A = 1 3 -2 1
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SOLVE USING DETERMINANTS
Matrix Ax = 4 3 -1 1
x = det Ax /det A
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SOLVE USING DETERMINANTS
det Ax = 4(1) – (3)(-1)
= 4 + 3=7
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SOLVE USING DETERMINANTS
Det A = 1(1) – (3)(-2)
= 1 + 6=7 thus
x = 7/7 = 1
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SOLVE USING DETERMINANTS
Matrix Ay = 1 4 -2 -1
y = det Ay /det A
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SOLVE USING DETERMINANTS
det Ay = -1(1) – (4)(-2)
= -1 + 8=7 thus
y = 7/7 = 1
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SECTION 6-8
Systems of Inequalities
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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.
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SYSTEM OF LINEAR INEQUALITIES
The intersection of the graphs of the inequalities
is the solution set.
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SOLVE BY GRAPHING THE INEQUALITIES
x + 2y < 52x – 3y ≤ 1
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SOLVE BY GRAPHING THE INEQUALITIES
4x - y 58x + 5y ≤ 3
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SECTION 6-9
Linear Programming
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LINEAR PROGRAMMING
A method used by business and government to help manage resources and
time.
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CONSTRAINTS
Limits to available resources
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FEASIBLE REGION
The intersection of the graphs of a system of
constraints.
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OBJECTIVE FUNCTION
Used to determine how to maximize profit while
minimizing cost
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ENDEND