# Linear Algebra 1 equation with 3 variables

The question:

solve the equation

$8x + 3y - 7z = 12 \\ \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} -\dfrac 3 8 \\ -\dfrac 8 3 \\ \dfrac 8 7 \end{bmatrix} + s \begin{bmatrix} \dfrac 7 8 \\ \dots \\ \dots \end{bmatrix} + t \begin{bmatrix} \dfrac 3 2 \\ \dots \\ \dots \end{bmatrix}$

I do not understand what to put into the remaining values. I tried to solve for the $y$ and $z$ like I did for the $x$, but the system is telling me that is incorrect. Some help would be appreciated.

• How did you "solve for the $x$"? Look at the first vector you filled in, which is the final result of your formula when $s=t=0$. Do those values satisfy the given equation? It seems to me they are not even close. This question will likely be much more instructive for you if you show us all your work, otherwise we cannot say what misconceptions you have. – David K Feb 15 '16 at 18:43
• I had the correct values, the mistake was that for some reason I thought the first and second column were the s and t values instead of the second and third. Thank you – Allan Feb 15 '16 at 19:02
• The vectors to multiply by $s$ and $t$ should solve the equation $8x+3y-7z=0$. Again, not even close. Wouldn't you like to know what wrong step you took so you can avoid taking it again on some other problem? That would be a useful question rather than just another instance of "what's the answer to my homework?" – David K Feb 15 '16 at 19:09
• Dear @Lovsovs, if you really want to edit, then do it properly: you should have replaced the image with equivalent MathJax code. – Alex M. Feb 15 '16 at 21:30
• @AlexM. You do have a point, but since my edits must go through peer-review, I see my edit as being accepted by the community. I respect that you'd want to use the extra time to code up (i.e. other than the basics) a downvoted answer regarding homework, but I must admit that I probably wouldn't. – Bobson Dugnutt Feb 15 '16 at 21:39

Your equation is $8x+3y-7z=12$, hence $\displaystyle x=\frac{12-3y+7z}{8}=\frac{3}{2}-\frac{3}{8}y+\frac{7}{8}z$, thus the set of solutions is $\displaystyle S=\left\{\begin{pmatrix}\frac{3}{2}-\frac{3}{8}t+\frac{7}{8}s \\t\\s\end{pmatrix}\right\}$ or equivalently $\displaystyle \begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}\frac{3}{2}\\0\\0\end{pmatrix}+t\begin{pmatrix}-\frac{3}{8}\\1\\0\end{pmatrix}+s\begin{pmatrix}\frac{7}{8}\\0\\1\end{pmatrix}$.