1. ## Matrices!

I'm not familiar with LaTex, so bear with my notation (I apologize in advance)...
Find the values of x and y for which...
[2y+5 = [x-1
y-2] = 3x]
is true.
My work thus far
2y+5=x-1; (0,-3)
y-2=x-1; (-2/3, 0)
...I just don't know what steps to take particularly.

2. Hello, puzzledwithpolynomials!

Solve for $\displaystyle x\text{ and }y\!:\;\;\begin{bmatrix}2y+5 \\ y-2\end{bmatrix} \:=\:\begin{bmatrix}x-1 \\ 3x \end{bmatrix}$

This can be written: .$\displaystyle \begin{bmatrix}x-2y \\ 3x-y\end{bmatrix} \:=\:\begin{bmatrix}6\\\text{-}2\end{bmatrix}$

We have: .$\displaystyle \left[\begin{array}{cc|c}1 & \text{-}2 & 6 \\ 3 & \text{-}1 & \text{-}2 \end{array}\right]$

. $\displaystyle \begin{array}{c} \\ R_3-3R_1 \end{array} \left[\begin{array}{cc|c}1 & \text{-}2 & 6 \\ 0 & 5 & \text{-}20 \end{array}\right]$

. . . . $\displaystyle \begin{array}{c} \\ \frac{1}{5}R_2 \end{array} \left[\begin{array}{cc|c}1 & \text{-}2 & 6 \\ 0 & 1 & \text{-}4 \end{array}\right]$

.$\displaystyle \begin{array}{c}R_1+2R_2 \\ \\ \end{array} \left[\begin{array}{cc|c}1 & 0 & \text{-}2 \\ 0 & 1 & \text{-}4 \end{array}\right]$

Therefore: .$\displaystyle \begin{Bmatrix}x &=& \text{-}2 \\ y &=& \text{-}4 \end{Bmatrix}$

3. Originally Posted by puzzledwithpolynomials
I'm not familiar with LaTex, so bear with my notation (I apologize in advance)...
Find the values of x and y for which...
[2y+5 = [x-1
y-2] = 3x]
is true.
If the previous reply's interpretation is correct, you have the following matrix equality:

. . . . .$\displaystyle \left[\begin{array}{c}2y\, +\, 5\\y\, -\, 2\end{array}\right]\, =\, \left[\begin{array}{c}x\, -\, 1\\3x\end{array}\right]$

It may be simpler, in this case, to use what you know about matrix equality to create regular equations:

. . . . .$\displaystyle 2y\, +\, 5\, =\, x\, -\, 1$
. . . . .$\displaystyle y\, -\, 2\, =\, 3x$

Rearrange to get the variables on one side of the "equals" signs, and the constant terms on the other:

. . . . .$\displaystyle -x\, +\, 2y\, =\, -1\, -\, 5$
. . . . .$\displaystyle -3x\, +\, y\, =\, +2$

. . . . .$\displaystyle -1x\, +\, 2y\, =\, -6$
. . . . .$\displaystyle -3x\, +\, 1y\, =\, 2$

Multiply the first row by -3 and add down to get:

. . . . . ..$\displaystyle 3x\, -\, 6y\, =\, 18$
. . . . .$\displaystyle \underline{-3x\, +\, 1y\, =\, \mbox{ }\, 2}$

. . . . . . . . ..$\displaystyle -5y\, =\, 20$

Etc, etc.