Help with SYSTEMS OF LINEAR EQUATIONS IN THREE VARIABLES?

Question

So im solving linear equations in 3 variables and got a huge fraction trying to decipher this problem out. Here's the problem. The answer in the back of the book says it is (1,4,-3) but i am not getting that. Here is the problem. Help please!

3x + 2y + z = 8
2x -3y + 2z = -16
x + 4y -z = 20

Answer 1

There are several ways to do this problem. One way is to solve this problem algebraically, which is a hassle. Another way is to solve this using a calculator which involves using linear algebra and matrices. Here's how you do it:

Algebraically, here are the steps. First let me define (1) as 3x+2y+x=8, (2) as 2x-3y+2=-16, (3) as x+4y-z=20

Now, you need to simplify two equations to two unknowns then solve for the unknown variable. Looking at the equations, you can add (1) and (3) to eliminate 'z'.

3x+2y+z=8
x+4y-z=20

Adding these two equations, you get 4x+6y=28 Let's label this as 'A'

Now eliminate 'z' using another combination of equations. One such is adding (2) with twice (3). The result is:

2x-3y+2z=-16
2x+8y-2z=40

Adding these two equations, you get 4x+5y=24. Let's call this 'B'

Putting 'A' and 'B' together, you have two equations and two unknowns.

4x+6y=28
4x+5y=24, now you can solve for x and y

Subtracting these two equations you are left with y = 4

Plugging this value of 'y' into either equation and you will get 'x'.

Plugging this into 'A', you get 4x+6(4)=28 -----> 4x+24=28,
thus 4x=4, which makes x = 1

Now that you have 'x' and 'y', plug these values into (1), (2) or (3) and solve for 'z'.

Plugging into (1), you get 3(1)+2(4)+z=8 --->3+8+z=8
Thus, z = 8-8-3 = -3

Thus, you solution is (1,4,-3)

A faster and MUCH EASIER method is using a graphing calculator, like any TI-84, 83, 86, or 89 version.

If you have one of these calculations, look under the MATH function and there should be a button called 'rref'. This function is called the 'reduced row echleon form', which is commonly used in linear algebra. Basically, it simplifies any matrix into its simpliest form. Using this method, we can solve for this system of linear equations.

Now to use 'rref', you must create a matrix to use. Go under your MATRIX button and create a 3x4 matrix. For the values of this matrix, it's fairly simple. Each row has 4 components, the first 3 being the coefficients of 'x', 'y', and 'z' and the last component as the answer.

Using the first equation, you would get a first row matrix of 3 2 1 8

Doing this for the other two equations and combining, you get the following 4x3 matrix:

3 2 1 8
2 -3 2 -16
1 4 -1 20

Label this matrix as 'A'

Now, using the calculator, perform the following operation:

rref(A). This simplifies your matrix into its simpliest form. After doing this, I got the resulting matrix:

1 0 0 1
0 1 0 4
0 0 1 -3

Since the first column represents x, the 2nd representing y and the 3rd representing 'z', you can conclude that this matrix concludes that:

x=1, y=4, z=-3, which is the solution of the equations. Use this trick over and over again.

Hope this helps

Answer 2

multiply the last equation by 3 and the second by 3/2.
Adding the first and the third will eliminate z.

you will notice that several of the coefficients are now equal.
subtract one equation from another to eliminate a variable until you have one equation with 1 unknown. Continue this and eliminate another variable. Now you've solved for 2 of the 3. Plug in and solve for the remaining variable.

Answer 3

Add the first and third equations together:

4x + 6y = 28

Now double the third and add the second:

4x + 5y = 24

I think you can solve these two equations to get y = 4, and then
take it from there to find the back of the book is right. If not, email me at
[e-mail]

Answer 4

Let x = number of $23 subscriptions let y = ----------------$17 ------------------- let z = ----------------$21 ------------------ x+y+z=32 y=x-2 23x+17y+21z=646 You can either solve the above three equations or you can rewrite equation (2) as -x+y+0z= -2, so you'll have a 3 x 3 sets of equations. (whichever is easier for you)

Answer 5

i have never solved a linear function with 3 given points. sorry