Showing posts with label 3x3 Linear Systems of Equations. Show all posts
Showing posts with label 3x3 Linear Systems of Equations. Show all posts
Tuesday, January 6, 2015
Friday, January 2, 2015
Solve the system of equations. If the system has no solution, say it is inconsistent. (1) x + y + z = 3,(2) x - z = 1,(3) y - z = -4
Adding (1) and (2) leads to
(x+y+z)+(x-z)=3+1
2x+y=4
(4) y=4-2x
Adding (1) and (3) leads to
(x+y+z)+(y-z)=3-4
x+2y=-1
Replacing y with (4) leads to the solution for x:
x+2(4-2x)=-1
x+8-4x=-1
-3x=-9
x=3
(x+y+z)+(x-z)=3+1
2x+y=4
(4) y=4-2x
Adding (1) and (3) leads to
(x+y+z)+(y-z)=3-4
x+2y=-1
Replacing y with (4) leads to the solution for x:
x+2(4-2x)=-1
x+8-4x=-1
-3x=-9
x=3
Using x we can find y:
y=4-2x=4-2(3)=4-6=-2
y=-2
y=4-2x=4-2(3)=4-6=-2
y=-2
Next we find z:
x+y+z=3
3-2+z=3
-2+z=0
z=2
x+y+z=3
3-2+z=3
-2+z=0
z=2
solution: (3, -2, 2)
Wednesday, August 21, 2013
Solve by back substitution. x = -2, x + y + z = -1, 2x + 2z = 0 Thank you
Solve by back substitution.
x = -2
x + y + z = -1
2x + 2z = 0
Thank you
Your Answer:
1---> x = -2
2---> x + y + z = -1
3---> 2x + 2z = 0
3---> 2(-2)+2z=0
2z=4
z=2
2---> (-2)+y+(2)=-1
y=-1
solution: (-2,-1,2)
2---> x + y + z = -1
3---> 2x + 2z = 0
3---> 2(-2)+2z=0
2z=4
z=2
2---> (-2)+y+(2)=-1
y=-1
solution: (-2,-1,2)
Tuesday, August 13, 2013
Tuesday, August 6, 2013
Saturday, August 3, 2013
Solve: x+y+z=6, 2x-y+z=3, 3x-z=0}
from 3:
z=3x
1 plus 2:
(2x-y+z)+(x+y+z)=3+6
3x+2z=9
3x+2(3x)=9
9x=9
x=1
so z=3
from 1:
x+y+z=1+y+3=6
y+4=6
y=2
Solution: (1,2,3)
:)
z=3x
1 plus 2:
(2x-y+z)+(x+y+z)=3+6
3x+2z=9
3x+2(3x)=9
9x=9
x=1
so z=3
from 1:
x+y+z=1+y+3=6
y+4=6
y=2
Solution: (1,2,3)
:)
Timmy enjoys playing old-fashioned video games. He found 3 game systems on eBay and wants to purchase a Nintendo game system, an Atari game system, and a Sega arcade game system. The Sega system costs $600 more than the sum of the Atari and the Nintendo systems. All together, the three game systems cost $11,000. The Atari system costs $1100 less than the Nintendo. Determine how much Timmy paid for each game system.
1 S=N+A+600
2 N+S+A=11000
3 A=N-1100
A=2050
N=3150
S=5800
2 N+S+A=11000
3 A=N-1100
A=2050
N=3150
S=5800
Sunday, July 21, 2013
Derive a quadratic equation using the following points: (1, -4), (2, -5), (3, -10)
y = ax^2+bx+c
(1) (1, -4) -> -4=a+b+c ,
(2) (2, -5) -> -5= 4a+2b+c,
(3) (3, -10) -> -10=9a+3b+c
(2)-(1):
-5+4=4a-a+2b-b+c-c
-1=3a+b (or 3a+b=-1)
(3)-(1):
-6=8a+2b (or 4a+b=-3)
taking the difference of these 2 we get
4a-3a+b-b=-3+1 (or a=-2)
and thus b=5
Now from (1):
-4=-2+5+c (so c=-7)
y=-2x^2+5x+-7
:)
(1) (1, -4) -> -4=a+b+c ,
(2) (2, -5) -> -5= 4a+2b+c,
(3) (3, -10) -> -10=9a+3b+c
(2)-(1):
-5+4=4a-a+2b-b+c-c
-1=3a+b (or 3a+b=-1)
(3)-(1):
-6=8a+2b (or 4a+b=-3)
taking the difference of these 2 we get
4a-3a+b-b=-3+1 (or a=-2)
and thus b=5
Now from (1):
-4=-2+5+c (so c=-7)
y=-2x^2+5x+-7
:)
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