I need help solving these simultaneous linear equations, not just the answers but show me how to work them?

Question

2x-y+z=15
x-y+2z=12
x+2y-z=-3

x-2y+z=-5
2x-y+z=1
2x+y+2z=10

x+2y-z=-3
2x+y+z=6
-2x+2y+3z=3

Answer 1

I will show one example on how to do it and then you take that example and apply it to the other two problems:

On the first problem, start with these two equations:
2x-y+z=15
x-y+2z=12

Take the second equation and multiply it by -2:
2x-y+z=15
-2x+2y-4z=-24
y-3z=-9

Let's focus on the other part:
x-y+2z=12
x+2y-z=-3

Multiply the second equation by -1:
x-y+2z=12
-x-2y+z=3
-3y+3z=15

Let's put those two equations together:
y-3z=-9
-3y+3z=15
-2y=6
y=-3

Plug it into a y-value:
-3-3z=-9
-3z=-6
z=2

Put the y and z value into this equation:
x-(-3)+2(2)=12
x+3+4=12
x+7=12
x=5

(5,-3,2)

Check:
5-(-3)+2(2)=12
5+3+4=12
8+4=12
12=12

I hope that helps!

P.S. I checked with the original equations and they equaled out also. You can check the answers with a graphing calculator. Whoever is giving the people for thumbs down for their answers, whether it is right or wrong, is being very unkind and rude for doing so. I am sure that I am not only person who does not appreciate the thumbs down given too.

Answer 2

2x - y + z = 15 - - - - -Equation 1
x - y + 2z = 12- - - - - Equation 1
x + 2y - z = - 3- - - - -Equation 3
- - - - - - - - - - - - - -

Multiply equation 2 by - 1
x - y + 2z = 12

-1(x) -1(-y) +(-1)(2z) = -1(12)

-x + y - 2z = - 12. . . New equation 2
- - - - - - - - - - - - - - -

elimionation of y Equation 1 and new equation 2

2x - y + z = 15
-x + y - 2z = - 12
- - - -- - - - - - -
x - z = 3 . . . .Equation 4

- - - - - - - - - - -
Multiply Equation 1 by 2

2x - y + z = 15

2(2x) - 2(Y) + 2(z) = 2(15)

4x - 2y + 2x = 30 . . .New equation 1
- - - - - - - - - - - -

Elimination of y new equation 1 and equation 3

4x - 2y + 2z = 30
x + 2y - z = - 3
- - - - - - - - - - - -

5x + z = 27. . . . .Equation 5

- - - - - - - - - -

Elimination of z equation 4 and equation 5

5x + z = 27
x - z = 3
- - - - - - - -
6x = 30

6x/6 = 60/6

x = 5

Insert the x value into equation 4

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

x - z = 3

5 - z = 3

5 - z _ 5 = 3 - 5

- z = - 2

-1(- z) = -1(- 2)

z = 2

Insert the z value into equation 1

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

Solving for y

2x - y + z = 15

2(5) - y + 2 = 15

10 - y + 2 = 15

12 - y = 15

12 - y - 12 = 15 - 12

- y = 3

-1(- y) = -1(3)

y = - 3

Insert the y value into equation 1

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

Check for equation 1

2x - y + z = 15

2(5) - (-3) + 2 = 15

10 + 3 + 2 = 15

15 = 15

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

Check for equation 2

x - y + 2z = 12

5 - (- 3) + 2 (2) = 12

5 + 3 + 4 = 12

12 + 12

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

Check for equation 3

x + 2y - z = - 3

5 + 2(- 3) - 3

5 +( -6) - 2 = - 3

5 - 6 - 2 = - 3

-3 = -3

- - - - - - - - - -

The solution set { 5, - 3, 2 }

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

Answer 3

Perhaps the other Answers are sufficient for your requirements, but I'll tackle the first system as an example for you:

(1) 2x-y+z=15
(2) x-y+2z=12
(3) x+2y-z=-3

I could choose any one of the 3 variables to eliminate, but I will elect 'x' ... and I will begin with equations (1) and (2).

[2x-y+z=15]
-2 * [x-y+2z=12]
=============
y - 3z = -9

Let's eliminate the same variable, using equations (2) and (3):

[x-y+2z=12]
-1 * [x+2y-z=-3]
============
-3y + 3z = 15

We can now use the two resulting equations to eliminate another variable (y or z).

I leave the remainder of this exercise for you.

Hope this helps!

Answer 4

question 1
2x-y+z=15 equation 1
x-y+2z=12 equation 2
x+2y-z=-3 equation 3

equation 3 tells
z = x + 2y + 3

equation 1 can now become
2x - y + (x+2y+3) = 15
that is
3x + y = 12 (new equation 1)

equation 2 (replacing z with x+2y+3) becomes
3x+3y = 6 (new equation 2)

9x + 3y = 36 new equation 1. multiplied by 3
subtract terms in new equation 2
6x = 30
x = 5

3x + y = 12 so y = -3
z = x+2y+3 so z = 2

main thing now is to check that the answer works

lemme say that these might seem very difficult.
please don't be discouraged.

question 2 now :
x-2y+z=-5 eq 1
2x-y+z=1 eq 2
2x+y+2z=10 eq 3
eq 1 says z = -x+2y-5
eq 2 becomes 2x -y +(-x+2y-5) = 1
that is x +y = 6
eq 3 becomes 2x+y+2(-x+2y-5) = 10
that is
5y = 20, so y = 4, so x = 2, so z = 1

honest, the concepts are really really interesting when used in
cartesian geometry. i know it must seem terribly terribly boring.
(it nearly killed me just writing out this answer)
any questions about maths write me anytime
[e-mail]

Answer 5

In series of equations, the goal is to eliminate variables until only one is left, solve for it, then substitute the result into an original equation, and continue until all variables are solved.

Let's start with your first set...

2x - 1y + 1z = 15
1x - 1y + 2z = 12
1x + 2y - 1z = -3

I like to write them this way so you can add down in columns. This series is easier than normal, because we don't need to multiply any of the equations to eliminate a variable... the y variables already add to 0!

(2x + 1x + 1x) + (-1y + -1y + 2y) + (1z + 2z + -1z) = (15 + 12 + -3)
4x + 0y + 2z = 24
4x + 2z = 24

Now we can go a few different ways here, but the easiest for me is to use the original equations and eliminate another variable:

2x - 1y + 1z = 15
1x - 1y + 2z = 12
1x + 2y - 1z = -3

this time I'll eliminate x by multiplying the first equation by -1:

-2x + 1y - 1z = -15
1x - 1y + 2z = 12
1x + 2y - 1z = -3

which gives us:

2y = -6
y = -3

We lucked out because we eliminated both x and z at the same time in this one, but the procedure is still the same. Now we substitue y = -3 in any of the original equations:

x + 2(-3) -z = -3
x -6 - z = -3
x - z = 3

So now we have 2 equations with 2 variables:

4x + 2z = 24 and
x - z = 3

so we can solve by eliminating a variable... let's multiply the second equation by 2:

4x + 2z = 24
2x - 2z = 6

now we add them:

6x = 30
x = 5

so y = -3 and x = 5

to find z, we just substitute these into any of the equations:

2x - y + z = 15
2(5) -(-3) + z = 15
10 + 3 + z = 15
13 + z = 15
z = 2

Hope this helps!

Answer 6

Start with two equations and cancel out a variable.
Then choose two other equations and cancel out the same variable.
solve those two like you would a normal system.
Then substitute your answer in for whatever variable you found,
continue solving like a normal system until you've solved for all three variables.
You can do it on a calc with matrices but that's hard to explain.
If you have a graphing calculator most of them show you how.

Answer 7

To the asker:
It's good that the people are showing you at least one example on how to do at least one problem. It's not right for us to show you how to do all of them. That would be too much work! You need to only learn one example and the rest of the problems you need to do on your own. Understood!

Answer 8

You can write them in matrix form and then either use row reduction operations to solve or find the inverse of the matrix. Substitution would be very messy.

Answer 9

i will start with the third one
x+2y-z=-3 *2(eqn.1 *3)
2x+4y-2z=-6
2x+y+z=6 (eqn.2)
subtracting
3y-3z=-12.........(4)
2x+4y-2z=-6 (eqn 1*3)
-2x+2y+3z=3(eqn 3)
adding
6y+z=-3
y-z=-4 (eqn.(1)/3)
adding
7y=-7
so y=-1
sub in (4)
-3-3z=-12
-3z=-9
z=3
substituting
x-2-3=-3
x=2
the solution set is{2,-1,3}