my neice lives several states away, it's been many years since i've done algebra. can someone help to simplif

Question

2x-y+z=4
x+3y-z=11
4x+y-z=14

Answer 1

I think you are asking to find the values of x,y, and z that satisfy all of the three below equations:
2x-y+z=4 <--- Equation 1
x+3y-z=11 <--- Equation 2
4x+y-z=14 <--- Equation 3
First add Equation 1 to Equation 2 which gives:
3x +2y = 15 <--- Equation 4
Now add Equation 1 to Equation 3 getting:
6x= 18
x=3
Put x=3 into Equation 4 getting:
3*3 +2y = 15
2y =15 -9 =6
y = 3
Now put x= 3 and y =3 into Equation 1 getting
2*3 -3 +z=4
6-3 +z = 4
3 +z = 4
z = 4-3 = 1

Hope this helps.

Answer 2

1) 2x - y + z = 4
2) x + 3y - z = 11
3) 4x + y - z = 14

I guess you wnat to solve a system of three equations in three unknowns. There are various ways to do this. One approach follows:

Manipulate the equations to get two equations in two unknowns, and then solve that system.

Add equation 1 to equation 2

1) 2x - y + z = 4
2) x + 3y - z = 11
- - - - - - - - - - - -
4) 3x + 2y = 15

Subtract equation 3 from equation 2

2) x + 3y - z = 11
3) 4x + y - z = 14
- - - - - - - - - - - -
5) -3x + 2y = - 3

Now add equations 4 and 5

4) 3x + 2y = 15
5) -3x + 2y = - 3
- - - - - - - - - - - -
4y = 12
y = 3

Substitute this result into either equation 4 or 5 to find x

5) -3x + 2y = - 3
- 3x + 2(3) = - 3
- 3x + 6 = - 3
- 3x = - 9
x = 3

Substitute for x and y in one of the original equations to get z

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

x = 3, y = 3, z = 1

Check

2) x + 3y - z = 11
3 + 3(3) - 1 = 11
3 + 9 - 1 = 11
11 = 11

3) 4x + y - z = 14
4(3) + 3 - 1 = 14
12 + 3 - 1 = 14
14 = 14

Answer 3

I first rearranged two of the problems and set them equal to each other. I isolated the z variable first.

z = 4 + y - 2x (first problem)
z = 4x + y - 14 (third problem)

I set them equal to each other...

4 + y - 2x = 4x + y - 14 (then simplify... the ys cancel out)
18 = 6x
**x = 3**

Then I plugged the x=3 into 2 of the problems. First I had...
3 + 3y - z = 11
z = 3 + 3y - 11

So I took that and plugged the value of x (3) and that value of z (above) into the first problem.

2 (3) - y + 3 + 3y - 11 = 4 Simplify...
-2 + 2y = 4
2y = 6
**y = 3**

So now that we know that x = 3, and y = 3, we can plug those numbers into any problem to get the value of z. (I plugged them into the first problem.)

6 - 3 + z = 4
3 + z = 4
z =1

Plug all 3 numbers into all of the equations to see if it works, and it does.

Answer 4

If you add row 1 and row 3, then you cancel out 2 of the variables, leaving only the x so you can solve for x:

2x -y z = 4
4x y -z = 14

6x +0 +0 = 18
Solve for x:
6x = 18
x = 3

Then continue from there.

Answer 5

x=3 y=3 z=1

I used a graphing calculator to solve it:
1) Enter the coefficients of the left side into Matrix [A] like this:
2 -1 1
1 3 -1
4 1 -1
2) Enter the numbers on the right side into Matrix [B] like this:
4
11
14
3) Solve the system by pressing [A][X^-1][B][ENTER]

Answer 6

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

3y= 11 - x + z
y= (11-x+z)/3

-z= 14 - 4x - y
z= -14+4x+y

You want to isolate each variable (x,y,z) to move things to the opposite side of the = you need to do to one side as you do to the other.

Answer 7

Here's how you do it:
http://fp.academic.venturacollege.edu/rbrunner/3e_3s.htm