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, 06:55 PM #1

- Oct 27th 2009, 07:03 PM #2
You can rewrite it using matrices:

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

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

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

Does this help?

- Oct 27th 2009, 07:05 PM #3

- Oct 27th 2009, 07:16 PM #4

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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, 07:18 PM #5
The other way then would be to use substitutions.

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

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

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

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

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

Simplifying, we get

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

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

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

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

- Oct 27th 2009, 07:22 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, 07:24 PM #7

- Oct 27th 2009, 08:56 PM #8