This isn't in our books. How would you solve this system of equation when it's set up like this?

7x+7y+z=1

x+8y+8z=8

9x+y+9z=9

??

17x+16y+18z=18?? or this a graphing linear equation? Thank you.

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- Oct 27th 2009, 07:55 PM #1

- Oct 27th 2009, 08:03 PM #2

- Oct 27th 2009, 08:05 PM #3

- Oct 27th 2009, 08:16 PM #4

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- Oct 2009
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If you don't want to do the matrix method (which is fairly easy), you can do it just like any other system of three equations:

(A) 7x+7y+z=1

(B) x+8y+8z=8

(C) 9x+y+9z=9

Let's solve for either y or z first, so we need to eliminate x. Multiple equation (B) by 7 and then do A-B. Then do a similar subtraction process with either (A) and (C) or (B) and (C).

After that, you'll have a system of equations with just y,z. So you can treat it like a two variable system of equations. Solve for y and z, then plug back in to get x.

- Oct 27th 2009, 08:18 PM #5
The other way then would be to use substitutions.

Note that in the first equation

If we now substitute this into the second equation, we have .

Now if we substitute this into the first equation, we get

Finally, we substitute both values into the final equation to get

Simplifying, we get

Since , then and .

Thus, the solution to this system of equations is

Does this make sense? Its a lot of messy work, but it shouldn't be too bad to follow.

- Oct 27th 2009, 08:22 PM #6
okie doke, thank you. I'll keep re writing this over and over until I understand it but I think I pretty much get what your saying.

On the matrix I was trying to follow you but lost me when you did this

Then reduce it to the form \begin{bmatrix}1 & 0 & 0 & x_0\\ 0 & 1 & 0 & y_0\\ 0 & 0 & 1 & z_0\end{bmatrix}

- Oct 27th 2009, 08:24 PM #7

- Oct 27th 2009, 09:56 PM #8