chapter 6 –5y = -11 solve for y in terms of x. 2x + y = 0 y = -2x then solve by substitution...
TRANSCRIPT
![Page 1: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/1.jpg)
CHAPTER 6
Linear Systems of
Equations
![Page 2: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/2.jpg)
SECTION 6-1
Slope of a Line and
Slope-Intercept Form
![Page 3: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/3.jpg)
COORDINATE PLANE
consists of two
perpendicular number
lines, dividing the
plane into four regions
called quadrants
![Page 4: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/4.jpg)
X-AXIS - the horizontal
number line
Y-AXIS - the vertical
number line
ORIGIN - the point where
the x-axis and y-axis
cross
![Page 5: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/5.jpg)
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)
![Page 6: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/6.jpg)
COORDINATE PLANE
![Page 7: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/7.jpg)
LINEAR EQUATION
is an equation whose
graph is a straight
line.
![Page 8: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/8.jpg)
SLOPE
is the ratio of vertical
change to the
horizontal change.
The variable m is used
to represent slope.
![Page 9: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/9.jpg)
m = change in y-coordinate
change in x-coordinate
Or
m = rise
run
FORMULA FOR SLOPE
![Page 10: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/10.jpg)
SLOPE OF A LINE
m = y2– y
1
x2– x
1
![Page 11: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/11.jpg)
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
![Page 12: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/12.jpg)
Y-Intercept
is the point where
the line intersects
the y -axis.
![Page 13: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/13.jpg)
X-Intercept
is the point where the
line intersects the
x -axis.
![Page 14: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/14.jpg)
HORIZONTAL LINE
a horizontal line
containing the point
(a, b) is described by
the equation y = b
![Page 15: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/15.jpg)
VERTICAL LINE
a vertical line
containing the point
(c, d) is described by
the equation x = c
![Page 16: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/16.jpg)
SLOPE-INTERCEPT FORM
y = mx + b
where m is the slope and
b is the y -intercept
![Page 17: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/17.jpg)
SECTION 6-2
Parallel and
Perpendicular Lines
![Page 18: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/18.jpg)
SLOPE of PERPENDICULAR
LINES
Two lines are
perpendicular if the
product of their
slopes is -1
![Page 19: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/19.jpg)
SLOPE of PARALLEL
LINES
Two lines are parallel if
their slopes are equal
![Page 20: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/20.jpg)
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
![Page 21: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/21.jpg)
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
![Page 22: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/22.jpg)
SECTION 6-3
Write Equations for
Lines
![Page 23: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/23.jpg)
POINT-SLOPE FORM
y – y1
= m (x – x1)
where m is the slope
and (x1 ,
y1) is a point
on the line.
![Page 24: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/24.jpg)
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
![Page 25: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/25.jpg)
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
![Page 26: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/26.jpg)
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
![Page 27: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/27.jpg)
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
![Page 28: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/28.jpg)
SECTION 6-4
Systems of Equations
![Page 29: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/29.jpg)
SYSTEM OF EQUATIONS
Two linear equations
with the same two
variable form a system
of equations.
![Page 30: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/30.jpg)
SOLUTION
The ordered pair that
makes both equations
true.
![Page 31: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/31.jpg)
SOLUTION
The point of
intersection of the two
lines.
![Page 32: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/32.jpg)
INDEPENDENT SYSTEM
The graph of each
equation intersects in
one point.
![Page 33: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/33.jpg)
INCONSISTENT SYSTEM
The graphs of each
equation do not
intersect.
![Page 34: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/34.jpg)
DEPENDENT SYSTEM
The graph of each
equation is the same.
The lines coincide and
any point on the line is
a solution.
![Page 35: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/35.jpg)
SOLVE BY GRAPHING
4x + 2y = 8
3y = -6x + 12
![Page 36: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/36.jpg)
SOLVE BY GRAPHING
y = 1/2x + 3
2y = x - 2
![Page 37: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/37.jpg)
SECTION 6-5
Solve Systems by
Substitution
![Page 38: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/38.jpg)
PRACTICE USING
DISTRIBUTIVE LAW
x + 2(3x - 6) = 2
x + 6x – 12 = 2
7x -12 = 2
7x =14
x = 2
![Page 39: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/39.jpg)
PRACTICE USING
DISTRIBUTIVE LAW
-(4x – 2) = 2(x + 7)
-4x + 2 = 2x +14
2 = 6x + 14
-12 = 6x
-2 = x
![Page 40: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/40.jpg)
SUBSTITUTION
A method for solving a
system of equations by
solving for one variable
in terms of the other
variable.
![Page 41: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/41.jpg)
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
![Page 42: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/42.jpg)
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
![Page 43: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/43.jpg)
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
![Page 44: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/44.jpg)
SOLVE BY SUBSTITUTION
2x + y = 0
x – 5y = -11
Solve for y in terms of x.
2x + y = 0
y = -2x
then
![Page 45: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/45.jpg)
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
![Page 46: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/46.jpg)
SOLVE BY SUBSTITUTION
Substitute the value of x
into the first equation
2x + y = 0
y = -2x
y = -2(-1)
y = 2
![Page 47: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/47.jpg)
SECTION 6-6
Solve Systems by
Adding and Multiplying
![Page 48: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/48.jpg)
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.
![Page 49: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/49.jpg)
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.
![Page 50: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/50.jpg)
STEPS FOR ADDITION OR
SUBTRACTION METHOD
Solve the resulting
equation for the
remaining variable.
![Page 51: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/51.jpg)
STEPS FOR ADDITION OR
SUBTRACTION METHOD
Substitute the value for
the variable in one of the
original equations and
solve for the unknown
variable.
![Page 52: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/52.jpg)
STEPS FOR ADDITION OR
SUBTRACTION METHOD
Check the solution in
both of the original
equations.
![Page 53: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/53.jpg)
MULTIPLICATION AND
ADDITION METHOD
This method combines
the multiplication
property of equations
with the
addition/subtraction
method.
![Page 54: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/54.jpg)
SOLVE BY ADDING AND
MULTIPLYING
3x – 4y = 10
3y = 2x – 7
![Page 55: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/55.jpg)
SOLUTION
3x – 4y = 10
-2x +3y = -7
Multiply equation 1 by 2
Multiply equation 2 by 3
![Page 56: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/56.jpg)
SOLUTION
6x – 8y = 20
-6x +9y = -21
Add the two equations.
y = -1
![Page 57: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/57.jpg)
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
![Page 58: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/58.jpg)
SECTION 6-7
Determinants &
Matrices
![Page 59: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/59.jpg)
MATRIX
An array of numbers
arranged in rows and
columns.
![Page 60: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/60.jpg)
SQUARE MATRIX
An array with the
same number of rows
and columns.
![Page 61: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/61.jpg)
DETERMINANT
Another method of
solving a system of
equations.
![Page 62: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/62.jpg)
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.
![Page 63: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/63.jpg)
DETERMINANT VALUE
Is the difference of the
product of the
diagonals (ad – bc).
a b
c d
![Page 64: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/64.jpg)
SOLVE USING
DETERMINANTS
x + 3y = 4
-2x + y = -1
![Page 65: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/65.jpg)
SOLVE USING
DETERMINANTS
x + 3y = 4
-2x + y = -1
Matrix A = 1 3
-2 1
![Page 66: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/66.jpg)
SOLVE USING
DETERMINANTS
Matrix Ax
= 4 3
-1 1
x = det Ax
/det A
![Page 67: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/67.jpg)
SOLVE USING
DETERMINANTS
det Ax
= 4(1) – (3)(-1)
= 4 + 3
=7
![Page 68: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/68.jpg)
SOLVE USING
DETERMINANTS
Det A = 1(1) – (3)(-2)
= 1 + 6
=7 thus
x = 7/7 = 1
![Page 69: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/69.jpg)
SOLVE USING
DETERMINANTS
Matrix Ay
= 1 4
-2 -1
y = det Ay /det A
![Page 70: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/70.jpg)
SOLVE USING
DETERMINANTS
det Ay = -1(1) – (4)(-2)
= -1 + 8
=7 thus
y = 7/7 = 1
![Page 71: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/71.jpg)
SECTION 6-8
Systems of
Inequalities
![Page 72: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/72.jpg)
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.
![Page 73: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/73.jpg)
SYSTEM OF LINEAR
INEQUALITIES
The intersection of the
graphs of the
inequalities is the
solution set.
![Page 74: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/74.jpg)
SOLVE BY GRAPHING THE
INEQUALITIES
x + 2y < 5
2x – 3y ≤ 1
![Page 75: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/75.jpg)
SOLVE BY GRAPHING THE
INEQUALITIES
4x - y 5
8x + 5y ≤ 3
![Page 76: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/76.jpg)
SECTION 6-9
Linear Programming
![Page 77: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/77.jpg)
LINEAR PROGRAMMING
A method used by
business and
government to help
manage resources and
time.
![Page 78: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/78.jpg)
CONSTRAINTS
Limits to available
resources
![Page 79: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/79.jpg)
FEASIBLE REGION
The intersection of the
graphs of a system of
constraints.
![Page 80: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/80.jpg)
OBJECTIVE FUNCTION
Used to determine how to
maximize profit while
minimizing cost
![Page 81: CHAPTER 6 –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+](https://reader031.vdocuments.net/reader031/viewer/2022013113/5ab775b07f8b9a684c8b85ba/html5/thumbnails/81.jpg)
END