Help on this question

**Rheanna** · October 27th 2009, 06:55 PM

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.

**Chris L T521** · October 27th 2009, 07:03 PM

Originally Posted by Rheanna

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.

You can rewrite it using matrices:

$\begin{bmatrix}7 & 7 & 1 & 1\\ 1 & 8 & 8 & 8\\ 9 & 1 & 9 & 9\end{bmatrix}$

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

Where $\left(x_0,y_0,z_0\right)$ is the solution to the system.

Does this help?

**Rheanna** · October 27th 2009, 07:05 PM

I don't get it. I got the matrix part, but what you did after that you lost me.

**Open that Hampster!** · October 27th 2009, 07:16 PM

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.

**Chris L T521** · October 27th 2009, 07:18 PM

Originally Posted by Rheanna

I don't get it.

The other way then would be to use substitutions.

Note that in the first equation $7x+7y+z=1\implies z=1-7x-7y$

If we now substitute this into the second equation, we have $x+8y+8(1-7x-7y)=8\implies x+8y+8-56x-56y=8\implies -55x-48y=0$ $\implies y=-\frac{55x}{48}$ .

Now if we substitute this into the first equation, we get $z=1-7x-7\left(-\frac{55x}{48}\right)\implies z=1-7x+\frac{7\cdot 55x}{48}\implies z=1-\frac{49x}{48}$

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

$9x+\left(-\frac{55x}{48}\right)+9\left(1-\frac{49x}{48}\right)=9$

Simplifying, we get

$9x-\frac{55x}{48}+9-\frac{9\cdot49x}{48}=9\implies \frac{9(48-49)x}{48}-\frac{55x}{48}=0\implies -\frac{64x}{48}=0\implies x=0$

Since $x=0$ , then $y=-\frac{55(0)}{48}=0$ and $z=1-\frac{49(0)}{48}=1$ .

Thus, the solution to this system of equations is $\left(0,0,1\right)$

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

**Rheanna** · October 27th 2009, 07:22 PM

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}

**Rheanna** · October 27th 2009, 07:24 PM

Originally Posted by Chris L T521

You can rewrite it using matrices:

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

Where $\left(x_0,y_0,z_0\right)$ is the solution to the system.

Does this help?

This is where you lost me. How you managed to get the 0, and the 1 , Did you subtract it? Duh, nevermind.. lol reduce=subtract haha lol

**Rheanna** · October 27th 2009, 08:56 PM

Does it matter what order it comes in?

Like the Answer is 0,0,1

but if you put 1,0,0 would that be wrong?

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