Download Summary 1 Solving systems by elimination and substitution. 1. Solve

Summary 1
Solving systems by elimination and substitution.
1. Solve the following system of equation:
7 x  4y  27

5x  2y  29
7x  4y  27
 7x  4y  27
 equate the y coefficients 

2 10x  4y  58
5x  2y  29
 17x  85  x  5
5x  2y  29 and x  5  25  2y  29  2y  4  y  2
The solution of the system is given by x = 5 and y = 2, that is, the point (5, 2)
2. Find the intersection of the lines 3x – 5y = 21 and 4x + 3y = -1
Solution:
 3x  5y  21
The intersection is given by the solution of the system 
 4x  3y  1
3  3x  5y  21
 9x  15y  63

 29x  58  x  2

5  4x  3y  1
 20x  15y  5
3x  5y  21 and x  2  6  5y  21  5y  15  y  3
The intersection is (2, -3)
 y  3x  11
3. Use the substitution technique to solve the system 
 4x  3y  6
Replace y in the second equation by 3x – 11.
4x + 3(3x – 11) = 6
4x + 9x – 33 = 6
13x=39
x = 3 and y = 3(3) – 11 = -2
The solution of the system is x = 3 and y = -2
4. Use the substitution principle to find the points where the line y = 3x - 13 intersects the
parabola y  x 2  3x  5 .
The points of intersection are given by the solution of the system
 y  x 2  3x  5
 replacing y by 3x  13 in the first equation 3x  13  x 2  3x  5

 y  3x  13
 0  x 2  6x  8 
0  ( x  4)( x  2)  x  2 or x  4
If x  2, y  3( 2)  13  7
If x  4, y  3( 4)  13  1
The po int s of intersecti on are ( 2,  7 ) and ( 4,  1)
5. Find the points of intersection of the line x + 2y = 6 and the circle x 2  y 2  4x  6y  12  0
The points of intersection is given by the solution of the system
 x 2  y 2  4x  6y  12  0

 x  2y  6
Solve the second equation for x and substitute in the first equation
x  2y  6  ( 2y  6) 2  y 2  4( 2y  6)  6y  12  0
 4y 2  24y  36  y 2  8y  24  6y  12  0
 5y 2  10y  0  5y( y  2)  0  y  0 or y  2
If y  0, x  2(0)  6  6
If y  2, x  2( 2)  6  2
The line intersects the circle at the point (6, 0) and (2, 2)
6) Use a system of linear equations to find the equation of the line that contains the points (-3, 4)
and (5, -6). Give answer in the form y = mx + b
If (-3, 4) belongs to the line, replace x = -3 and y = 4 to obtain 4 = -3m + b
If (5, -6) belongs to the line, replace x = 5 and y = -6 to obtain -6 = 5m +b
 1   3m  b  4
  3m  b  4
3m  b  4
Solve the system 



 5m  b  6
 5m  b  6
5m  b  6
10
5
 8m  10  m    
8
4
15 16 1
 5
 3m  b  4  b  3m  4  3    4   

4
4 4
 4
5
1
The equation of the line is y   x 
4
4
7. Solve the following system by the elimination method:
 y  3z  1
 2x

 3x  2 y  z  11
 2 x  3 y  5 z  9

 y  3z  1
3  2x
 2x


 3x  2y  z  11  2  3x
  2x  3y  5z  9
3   2x

 5  y
  6y  6z  30



  5y  17z  49
  5y
y
 3z
1
y  ____
z  ____
 6x  3y  9z  3

 2y  z  11   6x  4y  2z  22
  6x  9y  15z  27
 3y  5z  9

z
5
 5z  25
 5y
 
 17z  49
  5y  17z  49
 12z  24  z  2
 y  z  5 and z  2   y  2  5  y  3
2x  y  3z  1, z  2, y  3  2x  3  6  1  x  5
Answer : x  5, y  3, z  2
x  ____
8. Solve the following system by the elimination method:
5
 3x  2y  5z

 4y
1
 x
  2x  6y  3z  12

5
3  3x  2y  5z
5
 3x  2y  5z


 4y
1 
 4y
1
 x
 x
  2x  6y  3z  12

 5   2x  6y  3z  12

 9x  6y  15z  15
1
9  x  4y
1
 x  4y

  x
 4y
1  
 
19  36y  75
19  36y  75
10x  30y  15z  60

 9x  36y  9
 
 28x  84  x  3
19x  36z  75
x  4y  1 and x  3
 3  4y  1  y  
3x  2y  5z  5, x  3, y  
1
2
1
2
 9  1  5z  5
 5z   5  z   1
1
Answer : x  3, y   , z  1
2
9) Find the equation of the circle that contains the points (-3, 6), (-10, -1) and (14, -11)
Replacing the x values and y values in the general form of a circle, x 2  y 2  Dx  Ey  F  0
results in the following system:
 3D  6E  F  45
Divide first equation by 17
17 D  17E  272


 10D  E  F  101  
Divide second equation by 2
 24D  10E  216
14D  11E  F  317

 5 D
 E  16
  5D  5E  80

 7 D  28  D  4

12D  5E  108
 12D  5E  108
D  E  16 , D  4   4  E  16  E  12

3D  6E  F  45   12  72  F  45   F  129  F  129
The equation of the circle is x 2  y 2  4x  12y  129  0
c) Find the center and radius of the circle.
x 2  y 2  4x  12y  129  0  x 2  4x  y 2  12y  129
  

 x 2  4x  4   y 2  12y  36   129  4  36
 x 2  4x  y 2  12y  129
 x  2 2  ( y  6) 2  169
Center is (2, -6) and r  169  13
10. Find the equation of the parabola of the form y  ax 2  bx  c which contains the points
(5, -2), (6, 1) and (3, 10)
The point (5, -2) belongs to the parabola y  ax 2  bx  c implies that -2 = 25a +5b + c
The point (6, 1) belongs to the parabola y  ax 2  bx  c implies that 1 = 36a + 6b + c
The point (3, 10) belongs to the parabola y  ax 2  bx  c implies that 10 = 9a + 3b + c
 25a  5b  c  2
3 16a  2b  12
16a  2b  12

1  

 36a  6b  c

 2  27a  3b  9
 27a  3b  9
  9a  3b  c  10

 48a  6b  36

 6a  18  a  3
  54a  6b  18
 48  2b  12  2b  60  b  30
 27  90  c  10  63  c  10  c  73
The equation of the parabola is y  3x 2  30x  73
11. Find the equation of the parabola of the form x  ay 2  by  c which contains the points
(5, -2), (6, 1) and (3, 10)
The point (5, -2) belongs to the parabola x  ay 2  by  c implies that 5 = 4a -2b + c
The point (6, 1) belongs to the parabola x  ay 2  by  c implies that 6 = a + b + c
The point (3, 10) belongs to the parabola x  ay 2  by  c implies that 3 = 100a + 10b + c
 2b  c  5
 4a
 3a  3b  1
 3a  3b  1

 b  c  6  

 a
99a  9b  3
33a  3b  1
100a  10b  c  3

1
18
11
5
5
 1
 33a  3b  1  33    3b  1   3b  1  3b   b 
6
6
18
 18 
1
5
1
5
4
14 104
 abc  6    c  6  c  6 
or c  6 
5 
18 18
18 18
18
18 18
1
5
104
1 2
The equation of the parabola is x   y 2  y 
or x  
y  5y  104
18
18
18
18
 36a  2  a  

