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Linear Systems of Equations
CHAPTER 6 Linear Systems of Equations
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Slope of a Line and Slope-Intercept Form
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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is an equation whose graph is a straight line.
LINEAR EQUATION is 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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FORMULA FOR SLOPE Or m = rise run m = change in y-coordinate
change in x-coordinate Or m = rise run
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SLOPE OF A LINE m = 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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is the point where the line intersects the y -axis.
Y-Intercept is the point where the line intersects the y -axis.
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is the point where the line intersects the
X-Intercept is the point where the line intersects the x -axis.
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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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where m is the slope and b is the y -intercept
SLOPE-INTERCEPT FORM y = mx + b where 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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Parallel and Perpendicular Lines
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.
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Find the slope of a line parallel to the line containing points M and N.
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Find the slope of a line parallel to the line containing points M and N.
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Two lines are perpendicular if the product of their slopes is -1
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.
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Find the slope of a line perpendicular to the line containing points M and N.
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Find the slope of a line perpendicular to the line containing points M and N.
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Determine whether each pair of lines is parallel, perpendicular, or neither
7x + 2y = 14 7y = 2x - 5
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Determine whether each pair of lines is parallel, perpendicular, or neither
-5x + 3y = 2 3x – 5y = 15
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Determine whether each pair of lines is parallel, perpendicular, or neither
2x – 3y = 6 8x – 4y = 4
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Write Equations for Lines
SECTION 6-3 Write Equations for Lines
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where m is the slope and (x1 ,y1) is a point on the line.
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=-2 P(-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=-1/3 P(1, 1)
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Write an equation of a line perpendicular to y=2x+1 containing the point (2,1)
M=2 P(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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The ordered pair that makes both equations true.
SOLUTION The ordered pair that makes both equations true.
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The point of intersection of the two lines.
SOLUTION The point of intersection of the two lines.
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The graph of each equation intersects in one point.
INDEPENDENT SYSTEM The graph of each equation intersects in one point.
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The graphs of each equation do not intersect.
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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Solve Systems by Substitution
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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The ordered pair that makes both equations true.
SOLUTION The ordered pair that makes both equations true.
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The point of intersection of the two lines.
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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3x – y = 6 x + 2y = 2 Solve for y in terms of x. 3x = 6 + y
SOLVE BY SUBSTITUTION 3x – y = 6 x + 2y = 2 Solve for y in terms of x. 3x = 6 + y 3x – 6 = y then
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Substitute the value of y into the second equation
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
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Substitute the value of x into the first equation
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
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2x + y = 0 x – 5y = -11 Solve for y in terms of x. y = -2x then
SOLVE BY SUBSTITUTION 2x + y = 0 x – 5y = -11 Solve for y in terms of x. y = -2x then
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Substitute the value of y into the second equation
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
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Substitute the value of x into the first equation
SOLVE BY SUBSTITUTION Substitute the value of x into the first equation 2x + y = 0 y = -2x y = -2(-1) y = 2
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Solve Systems by Adding and Multiplying
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 = 10 3y = 2x – 7
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3x – 4y = 10 -2x +3y = -7 Multiply equation 1 by 2
SOLUTION 3x – 4y = 10 -2x +3y = -7 Multiply equation 1 by 2 Multiply 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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Substitute the value of y into either equation and solve for
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
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Determinants & Matrices
SECTION 6-7 Determinants & Matrices
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An array of numbers arranged in rows and columns.
MATRIX An array of numbers arranged in rows and columns.
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An array with the same number of rows and columns.
SQUARE MATRIX An array with the same number of rows and columns.
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Another method of solving a system of equations.
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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Is the difference of the product of the diagonals (ad – bc).
DETERMINANT VALUE Is the difference of the product of the diagonals (ad – bc). a b c 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 y = det Ay /det A
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SOLVE USING DETERMINANTS
det Ay = -1(1) – (4)(-2) = =7 thus y = 7/7 = 1
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Systems of Inequalities
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 < 5 2x – 3y ≤ 1
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SOLVE BY GRAPHING THE INEQUALITIES
4x - y 5 8x + 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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Limits to available resources
CONSTRAINTS Limits to available resources
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The intersection of the graphs of a system of constraints.
FEASIBLE REGION The intersection of the graphs of a system of constraints.
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Used to determine how to maximize profit while minimizing cost
OBJECTIVE FUNCTION Used to determine how to maximize profit while minimizing cost
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END
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