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2x+y-z=4
-x-3y+3z=3
3x+4y-4z=1
and
x+2y-2z=4
3x-y-z=6
-4x-y+3z=-10

2006-11-04 05:07:22 · 5 answers · asked by Anonymous in Science & Mathematics Mathematics

5 answers

2x + y + z = 4- -- - - - - Equation 1
-x - 3y + 3z = 3 - - - - - Equation 2
3x + 4y - 4z = 1- - - - - Equation 3
- - - - - - - - - - -
Multiply equation 2 by 2
2(-x) -2(3y) + 2(3z) = 2(3)
- 2x -6y + 6x = 5 New equation 2
- - - - - - - - - - -
Elimination of x
2x + y - z = 4
- 2x - 6y + 6z = 6
- - - - - - --
- 5y- 5z = 10- - - - -Equation 4

- - - - - - - - -
Multiply equation 1 by -3 and equation 3 by 2

2x + y - z = 4
-3(2x + (-3)(y) - (-3)(z)= -3(4)
-6z - 3y - 3z = -12 - - New equation 1
- - - - - - - - -

3x + 4y - 4z = 1
2(3x + 2(4y) - 2(4z) = 4(1)
6x + 8y - 8z = 4- - -New equation 3

- - - - - - - - - -
Elimination of x

- 6x - 3y - 3z = - 12
6x + 8y - 8z = 4
- - - - - - - - - -
5y - 6z = - 8 - - Equation 5

- - - - - - - - - - - - -
Elimination of y from equation 4 and 5.

Solving for z.

- 5y + 5z = 10
5y - 6z = -8
- - - - - - - -
-z = 2
-1(-z) = -1(2)

z = - 2

The answer is z = -2
Insert the z value into equation 4

- - - - - - - - - - -

Solving for y

- 5y + 5z = 10
- 5y + 5(-2) = 10
-5y + (- 10) = 10
- 5y - 10 = 10
- 5y - 10 + 10 = 10 + 10
- 5y = 20

-5y/- 5 = 20/-5

y = - 4

The answer is y = - 4

Insert thy y value into equation 1

- - - - - - - - -
Solving for x

2x + y - z = 4
2x + (- 4) - (-2) = 4
2x - 4 + 2 = 4
2x - 2 = 4
2x - 2 + 2 = 4 + 2
2x = 6

2x/2 = 6/2

x = 3

The answer is x = 3

Insert the x value into equation 1

- - - - - - - - - - - - -
Check For equation 1

2x + y - z = 4

2(3) + (- 4) -(-2) = 4

6 - 4 + 2 = 4

2 + 2 = 4

4 = 4
- - - - - - - - - - - -

Check for equation 2

-x - 3y + 3z = 3

-(3) - 3(- 4) + 3(-2) = 3

-3+ 12 + (- 6) = 3

-3 + 12 - 6 = 3

9 - 6 = 3

3 = 3

- - - - - - - - - - - - - - - - -

Check for equation 3

3x + 4yy - 4x = 1

3(3) + 4(- 4) - 4(- 2) = 1

9+ (-16) - (- 8) = 1

9 - 16 + 8 = 1

- 7 + 8 = 1

1 = 1

The solution set { 3, -4, - 2 }

- - - - - - - -s-

Follow this procedure of elimination


- - - - - - - -s-

2006-11-04 08:38:13 · answer #1 · answered by SAMUEL D 7 · 0 0

You are going to have to use either the substitution method of solving for one variable then substituting it into one of the equations OR you can use the elimination method, where you set your coefficient to a number that will cancel out the other cofficient in the other equation, when you add them together. Does this make sense? If not go to algebrahelp.com and they have an online 10 minute step by step process. or the matriz approach is easier and faster!

2006-11-04 05:15:42 · answer #2 · answered by Xochitl E 1 · 0 0

i was so determined to answer this.

x=3
y= -1
z=1

if you plug this back into the first equation, you get
2(3)+(-1)-(1)=4
6-1-1=4
5-1=4
4=4

so it works out

sorry i didn't show you all the work i did...but it took up a lot of space

2006-11-04 06:05:14 · answer #3 · answered by emphatical_me 1 · 0 0

Use the 3x3 Matrix approach.

2006-11-04 05:14:51 · answer #4 · answered by Halo 5 · 0 0

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