Showing posts with label 3x3 Linear Systems of Equations. Show all posts
Showing posts with label 3x3 Linear Systems of Equations. Show all posts

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

Using x we can find y:
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 



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)

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 

:)