Can anyone explain how to solve x, y, and z in the following systems of three equations?

Question

a. X + 2Y + Z = 6
X + Y = 4
3X + Y + Z = 8

b. 10X + Y + Z = 12
8X + 2Y +Z = 11
20X - 10Y - 2Z = 8

c. 22X + 5Y + 7Z = 12
10X + 3Y + 2Z = 5
9X + 2Y + 12Z = 14

Answer 1

You have to first eliminate one variable from two of the equations which would leave two equations with two variables.
Then eliminate one of those two variables to get one value. Plug this in to get a second value; plug these in to get the third.

For example in your problem b, if you multiply the second row by -1 and add it to the first row, that would eliminate z

(2x - 1y = 1)

Then divide the third row by 2 and add it to the (original) first row, which also eliminates z

(20x - 4y = 16)

Now take those 2 rows, eliminate either x or y, and start plugging in

Answer 2

Elimination method

Problem b.

10x + y + z = 12- - - - - -Original Equation 1
8x + 2y + z = 11- - - - - -Original Equation 2
20x - 10y - 2z = 8- - - - Original Equation 3
- - - - - - - - - - - -

Multiply equation 1 by - 1

- 1(10x) + (- 1)(y) + (- 1)(z) = - 1(12)

- 10zx - y - z = - 12.. . .New equation 1
- - - - - - - - - - - - - - -

Elimination of z new equation 1 and equation 2

- 10x - y - z = - 12
8x + 2y + z = 11
- - - - - - - - - - - - -
- 2x + y = - 1. . . Equation 4

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

Multiply original equation 1 by 2

10x + y + z = 12

2(10x) +2(y) + 2(z) = 2(12)

20x + 2y + 2z = 24 . . second new equation 1

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Elimination of z Second new equation 1 and equation 3

20 x + 2y + 2z = 24
20x - 10y -- 2z = 8
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40 x - 8y = 32 . . .Equation 5

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Multiply equation 4 by 8

- 2x + y = - 1
8(- 2x) + 8(y) = 8(- 1)

-16x + 8y = - 8. . . .New equation 4

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Elimination of y New equation 4 and equation 5

- 16x + 8y = -8
40x - 8y = 32
- - - - - - - - - -

24x = 24

24x/24 = 24/24

x = 1

The answer is x = 1

Insert the x value into equation 4

- 2x + y = - 1

- 2(1) + y = - 1

- 2 + y = - 1

- 2 + y + 2 = - 1 + 2

y = 1

The answer is y = 1

Insert the y value into the original equation 1

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

Solving for z

10x + y + z = 12

10(1) + 1 + z = 12

10 + 1 + z = 12

11 + z = 12

11 + z - 11 = 12 - 11

z = 1

The answer is z = 1

Insert the z value into original equation 1

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Check original equation 1

10x + y + z = 12

10(1) + 1 + 1 = 12

12 = 12

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Check for original equation 2

8z + 2y + 2z = 11

8(1) + 2(1) + 1) = 11

8 + 2 + 1 = 11

11 = 11

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

Check for original equation 3

20x - 10y - 2x = 8

20(1) - 10(1) - 2(1) = 8

20 - 10 - 2 = 8

20 - 12 = 8

8 = 8

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The solution set { 1, 1, 1 }

- - - - - - s-

Answer 3

For each of these, you first need to get to a point where you get a two-variable equation, and then express one of the variables in terms of the other. In a, this is easy because you're given a two-variable equation:
x + y = 4
x = 4 - y

Plug the "new" value of x into the other two equations to get:
(4 - y) + 2y + z = 6
4 + y + z = 6
y + z = 2
and
3(4 - y) + y + z = 8
12 - 3y + y + z = 8
-2y + z = -4

Now subtract the second equation from the first to get:
y + z = 2
-(-2y + z = -4)
------------------
3y = 6
y = 2

Using this, you can then figure out that x = 2 and z = 0.

For the other two problems, you don't have a convenient two-variable equation given to you, so you need to make your own by adding equations to eliminate variables. It's not as hard as it sounds, though. In b, combine the first equation with each of the other two to make two two-variable equations:
10x + y + z =12
-(8x + 2y + z = 11)
------------------------
2x -y = 1

2(10x + y + z = 12) = 20x + 2y + 2z = 24
20x + 2y + 2z =24
+20x -10y -2z = 8
-----------------------
40x - 8y = 32 (simplify to 5x - y = 4)

Combine these two equations to get:
2x -y = 1
-(5x - y = 4)
---------------
-3x = -3
x = 1

Plug this back in to get y = 1 and z = 1

Following the same set of steps, you should be able to figure problem c out on your own.

Answer 4

There are many paths you can take. For instance, you could cam take the first one to mean x = 5y - 2z and substitute this into the other two, which will eliminate x from those equations Then you have two equations with two variables so 5y - 2z + 4y - z = 12 ==> 9y -3z = 12 ==> 3y - z = 4 ==> z = 4 - 3y and 10y - 4z - y + 3z = 10 ==>two equations, so eliminate z and solve for y and then replace y in the others to get z and finally use the y and z values to find x As long as you remember what you are doing, it isn't really confusing - it's just a series of simple steps you already know how to do.

Answer 5

Solving systems of three equations is essentially done the same way as with two... you have to eliminate one variable at a time.

For your first system, you could subtract the first equation from the third:
-x - 2y - z = -6
3x + y + z = 8 gives you
2x - y = 2. Now combine this with the second equation and you've a system of two equations with two variables.

2x - y = 2
x + y = 4. Add these.

3x = 6, therefore x = 2. Now substitute the value for x in either of your x and y equations to solve for y.

x + y = 4
(2) + y = 4
y = 2. Now substitute your values for x and y into any of the orignal x, y, and z equations to solve for z.

x + 2y + z = 6
(2) + 2(2) + z = 6
2 + 4 + z = 6
6 + z = 6
z = 0.

Your solution is x = 2, y = 2, and z = 0.

Answer 6

a.
1) X + 2Y + Z = 6
2) X + Y = 4
3) 3X + Y + Z = 8
You have 3 equations, and 3 variables. Let's try to get 2 equations with the same 2 variables in them. equation 2 has only X and Y, so let's manipulate equations 1 and 3 to get rid of the Z.
subtract equation 1 from equation 3 to get:
4) 2X - Y = 2
Now you have 2 equations w/ just X and Y
2) X + Y = 4
add equation 4 to equation 2 to get:
3x = 6
solve for X
substitute back into either equation 2 or equation 4 to find Y
substitute these values of X and Y into either equation 1 or equation 3 to find Z

Hopefully this will help you understand the process/logic and you can apply it yourself to parts b and c

Answer 7

a. X + 2Y + Z = 6.....(1)
X + Y = 4 (2)
3X + Y + Z = 8 (3)
(3)-(1)
2x-y=2
x+y=4
adding
3x=6
x=2
substituting
y=2
from(1)
z=0

Answer 8

In A. x and y must be 2 and z 0 cuz when u put it in it works. Don't know B. B would have to be negative numbers and same for C

Answer 9

First try to eliminate Z by taking two eq at a time,
For eg. In (b) part take first two eq and subtract,
Then take second and third, multiply (2) by two and subtract from (3)
In this way u get two eq in X & Y
Now solve them for X&Y
And put value of X&Y in any given eq to get Z