Answer:
x1=-4/5, x2=18/5 and x3=7/5
Step-by-step explanation:
![\left[\begin{array}{ccc|c}1&0&-3&-5\\3&1&2&4\\2&2&1&7\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%260%26-3%26-5%5C%5C3%261%262%264%5C%5C2%262%261%267%5Cend%7Barray%7D%5Cright%5D)
you can do linear combination between the rows:
2nd row=R2-3R1 and 3th row=R3-2R1
![\left[\begin{array}{ccc|c}1&0&-3&-5\\0&1&11&19\\0&2&7&17\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%260%26-3%26-5%5C%5C0%261%2611%2619%5C%5C0%262%267%2617%5Cend%7Barray%7D%5Cright%5D)
3th row=(3R2-R3)/15
![\left[\begin{array}{ccc|c}1&0&-3&-5\\0&1&11&19\\0&0&1&\frac{7}{5} \end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%260%26-3%26-5%5C%5C0%261%2611%2619%5C%5C0%260%261%26%5Cfrac%7B7%7D%7B5%7D%20%5Cend%7Barray%7D%5Cright%5D)
1st row=R1+3R3 and R2-11R3
![\left[\begin{array}{ccc|c}1&0&0&-4/5 \\0&1&0&18/5 \\0&0&1&7/5\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%260%260%26-4%2F5%20%5C%5C0%261%260%2618%2F5%20%5C%5C0%260%261%267%2F5%5Cend%7Barray%7D%5Cright%5D)
x1=-4/5, x2=18/5 and x3=7/5
Answer:
2/8 is simplest form is 1/4 and 5/8 in simplest form is just 5/8
Step-by-step explanation:
Answer:
i think the answer is 99 im not completely sure though
but i hope this helps
This is something you would do through trial and error. At least, that's the approach I took. I'm not sure if there is any algorithm to solve. The solution I got is shown in the attached image below. There are probably other solutions possible. The trick is to keep each number separate but not too far away so that the other numbers to be filled in later don't get too crowded to their neighbor.
Side note: any mirror copy of what I posted would work as well since you can flip the page around and it's effectively the same solution.
<span>https://www.youtube.com/watch?v=GjYL0-BlejI So, to answer your question yes.</span>