solving systems of linear equations in three variables?

Question

Hey there, I am stuck on my Algebra 2 homework. I am so confused and lost on this problem. Can anyone please help me solve this problem and show me step by step what they did to solve it? Thanks very much in advance!

Solve the system using any algebraic method (either substitution or multiplication method):

x + 2y-z=3
-x + y + 3z=-5
3x + y + 2z=4

Please show the steps!

Answer 1

Add #1 and #2 to eliminate X

(X + -X) + (2Y+Y) + (-Z+3Z) = (3+ -5)

3Y+2Z =-2

Now add 3 times #2 to #3 to eliminate X

(-3X + 3X) + (3Y + Y) + (9Z + 2Z) = (-15+4)

4Y + 11Z = -11

Now you have two equations with two unknowns

3Y+2Z=-2
4Y+11Z=-11

You can eliminate Y by multiplying by -4 for the first and multiplying by 3 to the second and then adding them together

(-12Y+12Y) + (-8Z+33Z) = (8+-33)
25Z = -25
Z = -1

Now, plug Z back into the above equation

3Y+2Z=-2
3Y+2(-1) = -2
3Y-2=-2
3Y=0

Y=0

Now plug Y and Z back into #1

X+2Y-Z=3
X+0+1=3

X=2

Answer 2

x + 2y - z = 3- - - - - -Equation 1
-x + y + 3z = - 5- - - -Equation 2
3x + y + 2z = 4- - - - Equation 3
- - - - - - - - - - - --

Elimination of x equation 1 and equation 2

x + 2y - z = 3
-x + y + 3z = - 5
- - - - - - - - - - -
3y + 2z = - 2. . .Equation 4
- - - - - - - -
Elimination of x equation 1 and equation 3

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

Multiply equation 1 by - 3

-3(x) + (- 3)(2y) - (- 3)(z) = -3(3)

- 3x + (- 6y) - (- 3z) = - 9

- 3x - 6y + 3z = - 9
- - - - - - - - - - - - - - -

-3x - 6y + 3z = - 9
3x + y + 2z = 4
- - - - - - - - - - - -
- 5y + 5x = - 5. . .Equation 5
- - - - - - - - - - -

Elimination of y equation 4 and equation 5

3y + 2z = - 2
5y + 5z = - 5
- - - - - - - -
Multiply equation 4 by - 5 and equation 5 by 3

-5(3y) + (- 5)(2z) = -5(-2

- 15y + (- 10z) = 10

- 15y - 10z = 10
- - - - - - - - - -

5y + 5z = - 5

3(5y) +3(5z) = - 3(5)

15y + 15z = - 15

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

- 15y - 10z = 10
15y + 15z = - 15
- - - - - - - - - - - -

5z = - 5

5z/5 = - 5/5

z = - 1

The answer is z = - 1

Insert the z value into equation 4
- - - - - - - - - - - - - - - - - - - - - -
3y + 2z = - 2

3y + 2(- 1) = - 2

3y + ( - 2) = - 2

3y - 2 = - 2

3y - 2 + 2 = - 2 + 2

3y = 0

3y/3 = 0/3

y = 0/3

y = 0

The answer is y = 0

Insert the y value into equation 1

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

x + 2y - z = 3

x + 2(0) - (- 1) = 3

x + 0 + 1 = 3

x + 1 - 1 = 3 - 1

x = 2

The answer is x = 2

Insert the x value into equation 1

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

Check for equation 1

x + 2y - z = 3

2 + 2(0) - (- 1) = 3

2 + 0 + 1 = 3

3 = 3

- - - - - - - -

Check for equation 2

-x + y + 3z = - 5

- (2) + 0 + 3(- 1) = - 5

- 2 + (- 3) = - 5

- 2 - 3 = - 5

- 5 = - 5

- - - - - - - - - -

Check for equation 3

3x + y + 2z = 4

3(2) + 0 + 2(- 1) = 4

6 + (- 2) = 4

6 - 2 = 4

4 = 4

- - - - - - - - -

The solutiojn set is { 2, 0, - 1 }

- - - - - - - - - - s-

Answer 3

I will solve this system by substitution.

x + 2y - z = 3
-x + y + 3z = -5
3x + y + 2z = 4

Isolating z in the first equation:
x + 2y - z = 3
-z = 3 - x - 2y
z = x + 2y - 3

Substituting this value for z in the other two equations:
-x + y + 3z = -5
3x + y + 2z = 4

-x + y + 3(x + 2y - 3) = -5
3x + y + 2(x + 2y - 3) = 4

-x + y + 3x + 6y - 9 = -5
3x + y + 2x + 4y - 6 = 4

2x + 7y = 4
5x + 5y = 10

2x + 7y = 4
x + y = 2

The system reduced itself to 2 equations on 2 variables, which you probably know how to solve. I will pace myself quicker in this system of equations.

Again, by substitution: y = 2 - x (2nd equation), so 2x + 7(2 - x) = 4.

2x + 14 - 7x = 4
-5x = -10
x = 2

Substituting back in the equations:

y = 2 - x = 2 - 2 = 0

x + 2y - z = 3
2 + 2*0 - z = 3
2 - z = 3
z = -1

Final result: x = 2, y = 0, z = -1. Please check back the values in the original equations, to see if they match.

Answer 4

let me see
1. x + 2y-z=3
2. -x + y + 3z=-5
3. 3x + y + 2z=4
* for #1
x = 3-2y+z

Answer 5

Let these be equ.[1],[2],[3]

add [1]+[2]: 3y + 2z= -2 -[4])
add [1]*(3) - [3]: 5y -5z= 5 -[5] >*3 means multiplied with 3

[4]*(5) - [5]*3 > 10z + 15z = -10 -15 = -25 >>
z= -1 put in [5]

>> 5y - (-5*1) = 5 or 5y = 0
y = 0

put y, z in [1] >> x = 3 - (0) + (-1)
= 3 - 1
=2

(x,y,z) = (2, 0 , -1)

Answer 6

1. If the system in three variables has one solution, it is an ordered triple (x, y, z) that is a solution to ALL THREE equations. In other words, when you plug in the values of the ordered triple, it makes ALL THREE equations TRUE. If you do get one solution for your final answer, is this system consistent or inconsistent? If you said consistent, give yourself a pat on the back! If you do get one solution for your final answer, would the equations be dependent or independent? If you said independent, you are correct! 2. If the three planes are parallel to each other, they will never intersect. This means they do not have any points in common. In this situation, you would have no solution. If you get no solution for your final answer, is this system consistent or inconsistent? If you said inconsistent, you are right! If you get no solution for your final answer, would the equations be dependent or independent? If you said independent, you are correct! 3. If the three planes end up lying on top of each other, then there is an infinite number of solutions. In this situation, they would end up being the same plane, so any solution that would work in one equation is going to work in the other. If you get an infinite number of solutions for your final answer, is this system consistent or inconsistent? If you said consistent you are right! If you get an infinite number of solutions for your final answer, would the equations be dependent or independent? If you said dependent you are correct!