# Thread: Help on this question

1. ## Help on this question

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.

2. 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?

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

4. 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.

5. 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.

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}

7. 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

8. ## The Matrice?

Does it matter what order it comes in?