Answer:
i think the second option
Step-by-step explanation:
follow me on my insta a.cc.o.u.n.t below
<h2>
<em><u>"</u></em><em><u>A</u></em><em><u>T</u></em><em><u>T</u></em><em><u>I</u></em><em><u>C</u></em><em><u>U</u></em><em><u>S</u></em><em><u>_</u></em><em><u>S</u></em><em><u>W</u></em><em><u>E</u></em><em><u>E</u></em><em><u>T</u></em><em><u>L</u></em><em><u>O</u></em><em><u>V</u></em><em><u>E</u></em><em><u>R</u></em><em><u>"</u></em></h2>
Answer:
-51/7
Step-by-step explanation:
-51/7 cannot be simplified
Answer: D
<u>Step-by-step explanation:</u>
The first matrix contains the coefficients of the x- and y- values for both equations (top row is the top equation and the bottom row is the bottom equation. The second matrix contains what each equation is equal to.
![\begin{array}{c}2x-y\\x-6y\end{array}\qquad \rightarrow \qquad \left[\begin{array}{cc}2&-1\\1&-6\end{array}\right] \\\\\\\begin{array}{c}-6\\13\end{array}\qquad \rightarrow \qquad \left[\begin{array}{c}-6\\13\end{array}\right]](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Bc%7D2x-y%5C%5Cx-6y%5Cend%7Barray%7D%5Cqquad%20%5Crightarrow%20%5Cqquad%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D2%26-1%5C%5C1%26-6%5Cend%7Barray%7D%5Cright%5D%20%5C%5C%5C%5C%5C%5C%5Cbegin%7Barray%7D%7Bc%7D-6%5C%5C13%5Cend%7Barray%7D%5Cqquad%20%5Crightarrow%20%5Cqquad%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D-6%5C%5C13%5Cend%7Barray%7D%5Cright%5D)
The product will result in the solution for the x- and y-values of the system.