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Transcript of 1 Standards 2, 5, 25 SLOPE INTERCEPT FORM AND POINT-SLOPE FORM PROBLEM 1 PROBLEM 3 PROBLEM 4 PROBLEM...
![Page 1: 1 Standards 2, 5, 25 SLOPE INTERCEPT FORM AND POINT-SLOPE FORM PROBLEM 1 PROBLEM 3 PROBLEM 4 PROBLEM 2 PROBLEM 12 SOLVING SYSTEMS OF TWO VARIABLES LINEAR.](https://reader035.fdocuments.in/reader035/viewer/2022062715/56649db45503460f94aa507c/html5/thumbnails/1.jpg)
1
Standards 2, 5, 25
SLOPE INTERCEPT FORM AND POINT-SLOPE FORM
PROBLEM 1
PROBLEM 3 PROBLEM 4
PROBLEM 2
PROBLEM 12
SOLVING SYSTEMS OF TWO VARIABLES LINEAR EQUATIONS
BY GRAPHING, ELIMINATION AND SUBSTITUTION.
PROBLEM 13
PROBLEM 15 PROBLEM 16
PROBLEM 14
END SHOWPRESENTATION CREATED BY SIMON PEREZ. All rights reserved
PROBLEM 5 PROBLEM 6
PROBLEM 7 PROBLEM 8
PROBLEM 9 PROBLEM 10
PROBLEM 11
SPECIAL FUNCTIONS
![Page 2: 1 Standards 2, 5, 25 SLOPE INTERCEPT FORM AND POINT-SLOPE FORM PROBLEM 1 PROBLEM 3 PROBLEM 4 PROBLEM 2 PROBLEM 12 SOLVING SYSTEMS OF TWO VARIABLES LINEAR.](https://reader035.fdocuments.in/reader035/viewer/2022062715/56649db45503460f94aa507c/html5/thumbnails/2.jpg)
2
Standard 2:
Students solve systems of linear equations and inequalities (in two or three variables) by substitution, with graphs, or with matrices.
Estándar 2:
Los estudiantes resuelven sistemas de ecuaciones lineares y desigualdades (en 2 o tres variables) por substitución, con gráficas o con matrices.
Standard 5:
Students demonstrate knowledge of how real and complex numbers are related both arithmetically and graphically. In particular, they can plot complex numbers as points in the plane.
Estándar 5:
Los estudiantes demuestran conocimiento de como los números reales y complejos se relacionan tanto aritméticamente como gráficamente. En particular, ellos grafican números como puntos en el plano.
Standard 25:
Students use properties from number systems to justify steps in combining and simplifying functions.
Estándar 25:
Los estudiantes usan propiedades de sistemas numéricos para justificar pasos en combinar y simplificar funciones.
ALGEBRA II STANDARDS THIS LESSON AIMS:
PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
![Page 3: 1 Standards 2, 5, 25 SLOPE INTERCEPT FORM AND POINT-SLOPE FORM PROBLEM 1 PROBLEM 3 PROBLEM 4 PROBLEM 2 PROBLEM 12 SOLVING SYSTEMS OF TWO VARIABLES LINEAR.](https://reader035.fdocuments.in/reader035/viewer/2022062715/56649db45503460f94aa507c/html5/thumbnails/3.jpg)
3
Standard 5
Write the slope-intercept form for this equation and graph it.
4x + 2y = 10-4x -4x
2y = -4x + 102 2
y = - x +42
10 2
y = -2x + 5
m= -2
y- intercept = (0,5)
42 6-2-4-6
2
4
6
-2
-4
-6
8 10-8-10
8
-8
10
x
y
+1
21+
-=
-
2
y= mx + b
Slope Intercept
Form
PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
![Page 4: 1 Standards 2, 5, 25 SLOPE INTERCEPT FORM AND POINT-SLOPE FORM PROBLEM 1 PROBLEM 3 PROBLEM 4 PROBLEM 2 PROBLEM 12 SOLVING SYSTEMS OF TWO VARIABLES LINEAR.](https://reader035.fdocuments.in/reader035/viewer/2022062715/56649db45503460f94aa507c/html5/thumbnails/4.jpg)
4
Standard 5
Write the slope-intercept form for this equation and graph it.
-9x + 3y = 3+9x +9x
3y = 9x + 33 3
y = x +93
3 3
y = 3x + 1
m= 3
y- intercept = (0, 1)
42 6-2-4-6
2
4
6
-2
-4
-6
8 10-8-10
8
-8
10
x
y
31+
+=
y= mx + b
3++1
Slope Intercept
Form
PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
![Page 5: 1 Standards 2, 5, 25 SLOPE INTERCEPT FORM AND POINT-SLOPE FORM PROBLEM 1 PROBLEM 3 PROBLEM 4 PROBLEM 2 PROBLEM 12 SOLVING SYSTEMS OF TWO VARIABLES LINEAR.](https://reader035.fdocuments.in/reader035/viewer/2022062715/56649db45503460f94aa507c/html5/thumbnails/5.jpg)
5
Standard 5
42 6-2-4-6
2
4
6
-2
-4
-6
8 10-8-10
8
-8
10
x
y
(y – ) = (x – )54
y - 2 = (x – 4)54
y - 2 = x - 20 4
54
y = x -3 54
(y – ) = m(x – )x1y1
Point = (4, 2)x1
4
y1
2
Write the slope-intercept form of the equation that passes through (4,2) and has a slope of , then graph it.5
4m = 5
4
y - 2 = x - 554
+2 +2
y- intercept = (0,-3)
m 54+
+=
y= mx + b
+4
5+
Point-Slope Form
Slope Intercept
Form
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![Page 6: 1 Standards 2, 5, 25 SLOPE INTERCEPT FORM AND POINT-SLOPE FORM PROBLEM 1 PROBLEM 3 PROBLEM 4 PROBLEM 2 PROBLEM 12 SOLVING SYSTEMS OF TWO VARIABLES LINEAR.](https://reader035.fdocuments.in/reader035/viewer/2022062715/56649db45503460f94aa507c/html5/thumbnails/6.jpg)
6
Standard 5
(y – ) = (x – )37
y - 1 = (x – 7)37
y - 1 = x - 21 7
37
y = x -2 37
(y – ) = m(x – )x1y1
Point = (7, 1)x1
7
y1
1
Write the slope-intercept form of the equation that passes through (7,1) and has a slope of , then graph it.3
7m = 3
7
y - 1 = x - 337
+1 +1
y- intercept = (0,-2)
m 37+
+=
y= mx + b
Point-Slope Form
Slope Intercept
Form
42 6-2-4-6
2
4
6
-2
-4
-6
8 10-8-10
8
-8
10
x
y
+7
3+
PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
![Page 7: 1 Standards 2, 5, 25 SLOPE INTERCEPT FORM AND POINT-SLOPE FORM PROBLEM 1 PROBLEM 3 PROBLEM 4 PROBLEM 2 PROBLEM 12 SOLVING SYSTEMS OF TWO VARIABLES LINEAR.](https://reader035.fdocuments.in/reader035/viewer/2022062715/56649db45503460f94aa507c/html5/thumbnails/7.jpg)
7
Standard 5
(y – ) = (x – )
y - 1 = 1(x – 2)
y - 1 = x - 2
y = x -1 (y – ) = m(x – )x1y1
point = (2, 1)x1
2
y1
1
Write the slope-intercept form of the equation that passes through points (2,1) and (-3,-4), then graph it.
m = 1 +1 +1
y- intercept = (0,-1)
m 11+
+=
y= mx + b
First we find the slope:
= 1
(2,1) (-3,-4)x2
2
x1
-3
y2
1
y1
-4=
5 5
m =--
= 1+4
2+3
Then using
1
y - 1 = x - 2
21 3-1-2-3
1
2
3
-1
-2
-3
4 5-4-5
4
-4
5
x
y
+11+
PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
![Page 8: 1 Standards 2, 5, 25 SLOPE INTERCEPT FORM AND POINT-SLOPE FORM PROBLEM 1 PROBLEM 3 PROBLEM 4 PROBLEM 2 PROBLEM 12 SOLVING SYSTEMS OF TWO VARIABLES LINEAR.](https://reader035.fdocuments.in/reader035/viewer/2022062715/56649db45503460f94aa507c/html5/thumbnails/8.jpg)
8
Standard 5
42 6-2-4-6
2
4
6
-2
-4
-6
8 10-8-10
8
-8
10
x
y
43
Point = (6, 9)
Write the slope-intercept form of the equation that passes through (6,9) and has a slope of , then graph it.4
3m = 4
3
y- intercept = (0,1)
m 43+
+=
( ) = ( ) +b
y= mx + b
+3
4+
Slope Intercept
Form
PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
(x,y)
69
24 3
9 = +b
9 = 8 + b -8 -8
1 = b
b = 1 y= x 43
+ 1
![Page 9: 1 Standards 2, 5, 25 SLOPE INTERCEPT FORM AND POINT-SLOPE FORM PROBLEM 1 PROBLEM 3 PROBLEM 4 PROBLEM 2 PROBLEM 12 SOLVING SYSTEMS OF TWO VARIABLES LINEAR.](https://reader035.fdocuments.in/reader035/viewer/2022062715/56649db45503460f94aa507c/html5/thumbnails/9.jpg)
9
Standard 5
42 6-2-4-6
2
4
6
-2
-4
-6
8 10-8-10
8
-8
10
x
y
52
Point = (4, 12)
Write the slope-intercept form of the equation that passes through (4,12) and has a slope of , then graph it.5
2m = 5
2
y- intercept = (0,2)
m 52+
+=
( ) = ( )+b
y= mx + b
+2
5+Slope Intercept
Form
PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
(x,y)
412
20 2
12 = + b
12 = 10 + b -10 -10
b = 2y= x
52
+ 2
![Page 10: 1 Standards 2, 5, 25 SLOPE INTERCEPT FORM AND POINT-SLOPE FORM PROBLEM 1 PROBLEM 3 PROBLEM 4 PROBLEM 2 PROBLEM 12 SOLVING SYSTEMS OF TWO VARIABLES LINEAR.](https://reader035.fdocuments.in/reader035/viewer/2022062715/56649db45503460f94aa507c/html5/thumbnails/10.jpg)
10
Standard 5
72
Point = (2, 9)
Write the standard form of the equation that passes through (2,9) and has a slope of .7
2m = 7
2
( ) = ( ) +b
y= mx + b Slope Intercept
Form
PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
(x,y)
29
9 = 7 + b -7 -7
2 = b
b =2
y= x 72
+ 2
y= x 72
+ 2 (2)
2y = 7x + 4
-7x -7x
-7x + 2y = 4
or
7x – 2y = -4
Standard Form
![Page 11: 1 Standards 2, 5, 25 SLOPE INTERCEPT FORM AND POINT-SLOPE FORM PROBLEM 1 PROBLEM 3 PROBLEM 4 PROBLEM 2 PROBLEM 12 SOLVING SYSTEMS OF TWO VARIABLES LINEAR.](https://reader035.fdocuments.in/reader035/viewer/2022062715/56649db45503460f94aa507c/html5/thumbnails/11.jpg)
11
Standard 5
Find the slope-intercept form of the equation that passes through (3,1) and it is parallel to the equation
m = - 92
( ) = ( ) + b
y= mx + b Slope Intercept
Form
PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
(x,y)
3 1
y = - x + 1. 92
Two lines are parallel when they have the same slope, therefore the equation must have slope: point = (3, 1)and
92
-
( ) = + b 1 27
2-
27 2
+ 27 2
+
b = 1 27 2
+
b = 27 2
+22
22
b = 29 2
y= x 92
- + 29 2
![Page 12: 1 Standards 2, 5, 25 SLOPE INTERCEPT FORM AND POINT-SLOPE FORM PROBLEM 1 PROBLEM 3 PROBLEM 4 PROBLEM 2 PROBLEM 12 SOLVING SYSTEMS OF TWO VARIABLES LINEAR.](https://reader035.fdocuments.in/reader035/viewer/2022062715/56649db45503460f94aa507c/html5/thumbnails/12.jpg)
12
Standard 5
Find the slope-intercept form of the equation that passes through (1,2) and it is parallel to the equation
m = - 43
( ) = ( ) + b
y= mx + b Slope Intercept
Form
PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
(x,y)
1 2
y = - x + 3. 43
Two lines are parallel when they have the same slope, therefore the equation must have slope: point = (1, 2)and
43
-
( ) = + b 2 4
3-
43
+43
+
b = 2 43
+
b = 43
+63
33
b = 10 3
y= x 43
- + 10 3
![Page 13: 1 Standards 2, 5, 25 SLOPE INTERCEPT FORM AND POINT-SLOPE FORM PROBLEM 1 PROBLEM 3 PROBLEM 4 PROBLEM 2 PROBLEM 12 SOLVING SYSTEMS OF TWO VARIABLES LINEAR.](https://reader035.fdocuments.in/reader035/viewer/2022062715/56649db45503460f94aa507c/html5/thumbnails/13.jpg)
13
Standard 5
Find the slope-intercept form of the equation that passes through (2,3) and it is perpendicular to the equation
( ) = ( ) + b
y= mx + b Slope Intercept
Form
PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
(x,y)
23
y = - x + 1. 23
Two lines that are perpendicular have slopes whose product is -1. One is the reciprocal of the other:
point = (2, 3)and
32
m =2
m1
-1
- 23
m =2
-132
- 32
-
32
m =2
- 23
m =1
and
3 = 3 + b -3 -3
0 = b
b = 0 y= x 32
+ 0
y= x
32
![Page 14: 1 Standards 2, 5, 25 SLOPE INTERCEPT FORM AND POINT-SLOPE FORM PROBLEM 1 PROBLEM 3 PROBLEM 4 PROBLEM 2 PROBLEM 12 SOLVING SYSTEMS OF TWO VARIABLES LINEAR.](https://reader035.fdocuments.in/reader035/viewer/2022062715/56649db45503460f94aa507c/html5/thumbnails/14.jpg)
14
Standard 5
PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
SPECIAL FUNCTIONS
Constant Function:
f(x)= 3 Means that the function has the same value for all the domain.
.5-.5 1-1 1.5-1.5 2-2 2.5-2.5 3-3 3.5-3.5
1
-1
2
-2
3
-3
4
-4
5
-5
x
y
f(x)=3
![Page 15: 1 Standards 2, 5, 25 SLOPE INTERCEPT FORM AND POINT-SLOPE FORM PROBLEM 1 PROBLEM 3 PROBLEM 4 PROBLEM 2 PROBLEM 12 SOLVING SYSTEMS OF TWO VARIABLES LINEAR.](https://reader035.fdocuments.in/reader035/viewer/2022062715/56649db45503460f94aa507c/html5/thumbnails/15.jpg)
15
Standard 5
42 6-2-4-6
2
4
6
-2
-4
-6
8 10-8-10
8
-8
10
x
y
Graph the following absolute value equation:
y = |x|
For x < 0
y = -x y = x
For x 0>
Now let’s shift it two units above:
y = |x| + 2
PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
SPECIAL FUNCTIONS
![Page 16: 1 Standards 2, 5, 25 SLOPE INTERCEPT FORM AND POINT-SLOPE FORM PROBLEM 1 PROBLEM 3 PROBLEM 4 PROBLEM 2 PROBLEM 12 SOLVING SYSTEMS OF TWO VARIABLES LINEAR.](https://reader035.fdocuments.in/reader035/viewer/2022062715/56649db45503460f94aa507c/html5/thumbnails/16.jpg)
16
Standard 5
42 6-2-4-6
2
4
6
-2
-4
-6
8 10-8-10
8
-8
10
x
y
y = |x|
For x < 0
y = -x y = x
For x 0>
Now let’s shift it two units above:
y = |x| + 2
Now let’s shift it three units to the right:
y = |x-3| + 2
PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
Graph the following absolute value equation:SPECIAL FUNCTIONS
![Page 17: 1 Standards 2, 5, 25 SLOPE INTERCEPT FORM AND POINT-SLOPE FORM PROBLEM 1 PROBLEM 3 PROBLEM 4 PROBLEM 2 PROBLEM 12 SOLVING SYSTEMS OF TWO VARIABLES LINEAR.](https://reader035.fdocuments.in/reader035/viewer/2022062715/56649db45503460f94aa507c/html5/thumbnails/17.jpg)
17
Standard 5
42 6-2-4-6
2
4
6
-2
-4
-6
8 10-8-10
8
-8
10
x
y
y = |x|
For x < 0
y = -x y = x
For x 0>
Now let’s shift it two units above:
y = |x| + 2
Now let’s shift it three units to the right:
y = |x-3| + 2
Now let’s graph it upside down
y = – |x-3| + 2
PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
Graph the following absolute value equation:SPECIAL FUNCTIONS
![Page 18: 1 Standards 2, 5, 25 SLOPE INTERCEPT FORM AND POINT-SLOPE FORM PROBLEM 1 PROBLEM 3 PROBLEM 4 PROBLEM 2 PROBLEM 12 SOLVING SYSTEMS OF TWO VARIABLES LINEAR.](https://reader035.fdocuments.in/reader035/viewer/2022062715/56649db45503460f94aa507c/html5/thumbnails/18.jpg)
18
Standard 5
42 6-2-4-6
2
4
6
-2
-4
-6
8 10-8-10
8
-8
10
x
y
y = |x|
For x < 0
y = -x y = x
For x 0>
Now let’s shift it two units above:
y = |x| + 2
Now let’s shift it three units to the right:
y = |x-3| + 2
Now let’s graph it upside down
y = – |x-3| + 2
Now let’s make it skinner
y = – 6|x-3| + 2PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
Graph the following absolute value equation:SPECIAL FUNCTIONS
![Page 19: 1 Standards 2, 5, 25 SLOPE INTERCEPT FORM AND POINT-SLOPE FORM PROBLEM 1 PROBLEM 3 PROBLEM 4 PROBLEM 2 PROBLEM 12 SOLVING SYSTEMS OF TWO VARIABLES LINEAR.](https://reader035.fdocuments.in/reader035/viewer/2022062715/56649db45503460f94aa507c/html5/thumbnails/19.jpg)
19
Standard 5
42 6-2-4-6
2
4
6
-2
-4
-6
8 10-8-10
8
-8
10
x
y
y = |x|
For x < 0
y = -x y = x
For x 0>
Now let’s shift it two units above:
y = |x| + 2
Now let’s shift it three units to the right:
y = |x-3| + 2
Now let’s graph it upside down
y = – |x-3| + 2
Now let’s make it skinner
y = – 2|x-3| + 2
So, we have learn how the different parameters in an absolute value equation affect our graph.
PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
Graph the following absolute value equation:SPECIAL FUNCTIONS
![Page 20: 1 Standards 2, 5, 25 SLOPE INTERCEPT FORM AND POINT-SLOPE FORM PROBLEM 1 PROBLEM 3 PROBLEM 4 PROBLEM 2 PROBLEM 12 SOLVING SYSTEMS OF TWO VARIABLES LINEAR.](https://reader035.fdocuments.in/reader035/viewer/2022062715/56649db45503460f94aa507c/html5/thumbnails/20.jpg)
20
42 6-2-4-6
2
4
6
-2
-4
-6
8 10-8-10
8
-8
10
x
y
Standard 5
PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
SPECIAL FUNCTIONS
f(x)=x Identity function
(1,1)
(6,6)
(-6,-6)
At all points in the line the x-coordinate and the y-coordinate are the same.
![Page 21: 1 Standards 2, 5, 25 SLOPE INTERCEPT FORM AND POINT-SLOPE FORM PROBLEM 1 PROBLEM 3 PROBLEM 4 PROBLEM 2 PROBLEM 12 SOLVING SYSTEMS OF TWO VARIABLES LINEAR.](https://reader035.fdocuments.in/reader035/viewer/2022062715/56649db45503460f94aa507c/html5/thumbnails/21.jpg)
21
Standard 5
PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
SPECIAL FUNCTIONS
GREATEST INTEGER FUNCTION:
f(x)=[x] Step function were [x] means the greatest integer not greater than x.
.5-.5 1-1 1.5-1.5 2-2 2.5-2.5 3-3 3.5-3.5
1
-1
2
-2
3
-3
4
-4
5
-5
x
y
2
1
1
1
1
1
1
1
1
1
1
Means:
y
2
1.9
1.8
1.7
1.6
1.5
1.4
1.3
1.2
1.1
1
x
![Page 22: 1 Standards 2, 5, 25 SLOPE INTERCEPT FORM AND POINT-SLOPE FORM PROBLEM 1 PROBLEM 3 PROBLEM 4 PROBLEM 2 PROBLEM 12 SOLVING SYSTEMS OF TWO VARIABLES LINEAR.](https://reader035.fdocuments.in/reader035/viewer/2022062715/56649db45503460f94aa507c/html5/thumbnails/22.jpg)
22
Standard 5
PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
SPECIAL FUNCTIONS
GREATEST INTEGER FUNCTION:
f(x)=[x] Step function were [x] means the greatest integer not greater than x.
.5-.5 1-1 1.5-1.5 2-2 2.5-2.5 3-3 3.5-3.5
1
-1
2
-2
3
-3
4
-4
5
-5
x
y
3
2
2
2
2
2
2
2
2
2
2
Means:
y
3
2.9
2.8
2.7
2.6
2.5
2.4
2.3
2.2
2.1
2
x
![Page 23: 1 Standards 2, 5, 25 SLOPE INTERCEPT FORM AND POINT-SLOPE FORM PROBLEM 1 PROBLEM 3 PROBLEM 4 PROBLEM 2 PROBLEM 12 SOLVING SYSTEMS OF TWO VARIABLES LINEAR.](https://reader035.fdocuments.in/reader035/viewer/2022062715/56649db45503460f94aa507c/html5/thumbnails/23.jpg)
23
Standard 2
Graph the following system of equations, find the solution if it exist, and state if it is consistent and independent, consistent and dependent or inconsistent:
6x –2y = 02x – 2y = -4
6x – 2y = 0
-6x -6x
-2y = -6x-2 -2
y = 3x
2x – 2y = -4
-2x -2x
-2y = -2x -4-2 - 2
y = x + 2 2
4 2
y = x + 2
y- intercept = (0,2)
m 11+
+=
m 31+
+=
y- intercept = (0,0)
The system is consistent and independent, with a unique solution at (1,3)
21 3-1-2-3
1
2
3
-1
-2
-3
4 5-4-5
4
-4
5
x
y
+11+
3+
+1
Solution(1,3)
PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
![Page 24: 1 Standards 2, 5, 25 SLOPE INTERCEPT FORM AND POINT-SLOPE FORM PROBLEM 1 PROBLEM 3 PROBLEM 4 PROBLEM 2 PROBLEM 12 SOLVING SYSTEMS OF TWO VARIABLES LINEAR.](https://reader035.fdocuments.in/reader035/viewer/2022062715/56649db45503460f94aa507c/html5/thumbnails/24.jpg)
24
Standard 2Solve the following system of equations by elimination and verify by graphing:
2x + y = 45x + y = 7
2x + y = 45x + y = 7
Multiplying one of the equations by -1 to eliminate y:
-1 -1
2x + y = 4-5x - y = -7
-3x = -3-3 -3
x=1
2x + y = 4
2( ) + y = 41
2 + y = 4
-2 -2
y = 2
Changing both equations to Slope Intercept Form:
2x + y = 4 5x + y = 7
-2x -2x
y = -2x + 4
-5x -5x
y = -5x + 7
+1 +1
42 6-2-4-6
2
4
6
-2
-4
-6
8 10-8-10
8
-8
10
x
y
+1
2-3-
+1Solution
(1,2)
The system is consistent and independent, with a unique solution at (1,2)
PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
![Page 25: 1 Standards 2, 5, 25 SLOPE INTERCEPT FORM AND POINT-SLOPE FORM PROBLEM 1 PROBLEM 3 PROBLEM 4 PROBLEM 2 PROBLEM 12 SOLVING SYSTEMS OF TWO VARIABLES LINEAR.](https://reader035.fdocuments.in/reader035/viewer/2022062715/56649db45503460f94aa507c/html5/thumbnails/25.jpg)
25
Standard 2Solve the following system of equations by elimination and verify by graphing:
4x + 2y = 105x + y = 17
4x + 2y = 105x + y = 17
Multiplying one of the equations by -2 to eliminate y:
-2 -2
4x + 2y = 10-10x - 2y = -34
- 6x = -24-6 -6
x=4
4x + 2y = 10
4( ) + 2y = 104
16 + 2y = 10
-16 -16
2y =-6
Changing both equations to Slope Intercept Form:
4x + 2y = 10 5x + y = 17
-4x -4x
2y = -4x + 10
-5x -5x
y = -5x + 17
+1
+1The system is consistent and independent, with a unique solution at (4,-3)
2 2
y= -3
y
84 12-4-8-12
4
8
12
-4
-8
-12
16 20-16-20
16
-16
20
xSolution(4,-3)
2 2y = -2x + 5
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X – 5 = 2Y + 12+5 +5
X= 2Y + 17 (2)(2)
4Y – 30 = X
2Y +17 = 4Y - 30
- 2Y -2Y
17 = 2Y - 30
+30 +30
47 = 2Y 2 2
Y = 23.5X= 2Y + 17= 2( ) + 1723.5= 47 + 17
X = 64
Standard 2
2Y –15 = X
Solve the following two equations by substitution:
The system is consistent and independent, with a unique solution at (64,23.5)
12
2Y –15 = X12
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y + 2 = 4x
2y = 7x
-2 -2
y = 4x -2
2( ) = 7x4x-2
8x – 4 = 7x
-8x -8x
- 4 = -x (-1)(-1)
x = 4
y = 4x -2
= 4( ) -24
= 16 -2
y = 14
Standard 2
y + 2 = 4x
2y = 7x
Solve the following equations by substitution:
The system is consistent and independent, with a unique solution at (4,14)
PRESENTATION CREATED BY SIMON PEREZ. All rights reserved