# Thread: i am testing the code

1. ## i am testing the code

$\vec{r}(t)= (-t+ 11/10)\vec{i}- 3t/vec{j}$

2. Hello, quah135791

I assume you're hoping for:

$\vec{r}(t) \;=\;\left(-t + \tfrac{11}{10}\right)\vec{i} - 3t\vec{j}$

Click on QUOTE to see how it's done.

3. Originally Posted by Soroban
Hello, quah135791

I assume you're hoping for:

$\vec{r}(t) \;=\;\left(-t + \tfrac{11}{10}\right)\vec{i} - 3t\vec{j}$

Click on QUOTE to see how it's done.
thanks!

one more question, what is the code for the following those three equations
x_1−x_2+x_3 = 2
3x_1+x_2− x_3 = 2
2x_1+x_2− x_3 = k

4. Originally Posted by quah13579
thanks!

one more question, what is the code for the following those three equations
x_1−x_2+x_3 = 2
3x_1+x_2− x_3 = 2
2x_1+x_2− x_3 = k
There are lots of ways of doing this, the one I like (because it involves remembering fewer commands that some of the others) is:

$\begin{array}{rcrcrcr}
x_1&-&x_2&+&x_3&=&2\\
3x_1&+&x_2&-&x_3&=&2\\
2x_1&+&x_2&-&x_3&=&k
\end{array}$

Which is produced by:

\begin{array}{rcrcrcr}
x_1&-&x_2&+&x_3&=&2\\
3x_1&+&x_2&-&x_3&=&2\\
2x_1&+&x_2&-&x_3&=&k
\end{array}

CB