# [SOLVED]307.1.1 Use row reduction on the appropriate augmented

#### karush

##### Well-known member
$\tiny{307.1.1}$
Use row reduction on the appropriate augmented matrix to solve the following system of equations:
$\begin{array}{ll}3x+2y&=2\\x-y&=1\end{array} \sim\left[\begin{array}{rr|r}3&2&2\\ \:1&-1&1\end{array}\right]\sim \begin{bmatrix}1&0&\frac{4}{5}\\ 0&1&-\frac{1}{5}\end{bmatrix}$

$x=\dfrac{4}{5}\quad y= -\dfrac{1}{5}$

hopefully no typos!!
suggestions??

Last edited:

#### Country Boy

##### Well-known member
MHB Math Helper
Well, 3(4/5)+ 2(-1/5)= 12/5- 2/5= 10/5= 2
and 4/5- (-1/5)= 4/5+ 1/5= 5/5= 1

So you certainly got the right answers!

#### karush

##### Well-known member
as simple as it looks the steps were very confusing
spent over an hour on it

#### skeeter

##### Well-known member
MHB Math Helper
as simple as it looks the steps were very confusing
spent over an hour on it
reduction of matrices is arithmetic torture ...that's why calculators were invented

#### Klaas van Aarsen

##### MHB Seeker
Staff member
I don't consider it that bad. It's mostly just a matter of practice like arithmetic.
$\begin{array}{c}-3\\*\end{array}\left[\begin{array}{cc|c}3&2&2\\1&-1&1\end{array}\right] \to \updownarrow\left[\begin{array}{cc|c}0&5&-1\\1&-1&1\end{array}\right] \to \begin{array}{c}\\*\frac 15\end{array}\left[\begin{array}{cc|c}1&-1&1\\0&5&-1\end{array}\right] \to \begin{array}{c}+1\\*\end{array}\left[\begin{array}{cc|c}1&-1&1\\0&1&-\frac 15\end{array}\right] \to \left[\begin{array}{cc|c}1&0&\frac 45\\0&1&-\frac 15\end{array}\right]$