Answer:
The reduced row-echelon form of the linear system is ![\left[\begin{array}{cccc}1&0&-5&0\\0&1&3&0\\0&0&0&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%260%26-5%260%5C%5C0%261%263%260%5C%5C0%260%260%261%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
We will solve the original system of linear equations by performing a sequence of the following elementary row operations on the augmented matrix:
- Interchange two rows
- Multiply one row by a nonzero number
- Add a multiple of one row to a different row
To find the reduced row-echelon form of this augmented matrix
![\left[\begin{array}{cccc}2&3&-1&14\\1&2&1&4\\5&9&2&7\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D2%263%26-1%2614%5C%5C1%262%261%264%5C%5C5%269%262%267%5Cend%7Barray%7D%5Cright%5D)
You need to follow these steps:
- Divide row 1 by 2

![\left[\begin{array}{cccc}1&3/2&-1/2&7\\1&2&1&4\\5&9&2&7\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%263%2F2%26-1%2F2%267%5C%5C1%262%261%264%5C%5C5%269%262%267%5Cend%7Barray%7D%5Cright%5D)
- Subtract row 1 from row 2

![\left[\begin{array}{cccc}1&3/2&-1/2&7\\0&1/2&3/2&-3\\5&9&2&7\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%263%2F2%26-1%2F2%267%5C%5C0%261%2F2%263%2F2%26-3%5C%5C5%269%262%267%5Cend%7Barray%7D%5Cright%5D)
- Subtract row 1 multiplied by 5 from row 3

![\left[\begin{array}{cccc}1&3/2&-1/2&7\\0&1/2&3/2&-3\\0&3/9&9/2&-28\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%263%2F2%26-1%2F2%267%5C%5C0%261%2F2%263%2F2%26-3%5C%5C0%263%2F9%269%2F2%26-28%5Cend%7Barray%7D%5Cright%5D)
- Subtract row 2 multiplied by 3 from row 1

![\left[\begin{array}{cccc}1&0&-5&16\\0&1/2&3/2&-3\\0&3/9&9/2&-28\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%260%26-5%2616%5C%5C0%261%2F2%263%2F2%26-3%5C%5C0%263%2F9%269%2F2%26-28%5Cend%7Barray%7D%5Cright%5D)
- Subtract row 2 multiplied by 3 from row 3

![\left[\begin{array}{cccc}1&0&-5&16\\0&1/2&3/2&-3\\0&0&0&-19\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%260%26-5%2616%5C%5C0%261%2F2%263%2F2%26-3%5C%5C0%260%260%26-19%5Cend%7Barray%7D%5Cright%5D)
- Multiply row 2 by 2

![\left[\begin{array}{cccc}1&0&-5&16\\0&2&3&-6\\0&0&0&-19\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%260%26-5%2616%5C%5C0%262%263%26-6%5C%5C0%260%260%26-19%5Cend%7Barray%7D%5Cright%5D)
- Divide row 3 by −19

![\left[\begin{array}{cccc}1&0&-5&16\\0&2&3&-6\\0&0&0&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%260%26-5%2616%5C%5C0%262%263%26-6%5C%5C0%260%260%261%5Cend%7Barray%7D%5Cright%5D)
- Subtract row 3 multiplied by 16 from row 1

![\left[\begin{array}{cccc}1&0&-5&0\\0&1&3&-6\\0&0&0&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%260%26-5%260%5C%5C0%261%263%26-6%5C%5C0%260%260%261%5Cend%7Barray%7D%5Cright%5D)
- Add row 3 multiplied by 6 to row 2

![\left[\begin{array}{cccc}1&0&-5&0\\0&1&3&0\\0&0&0&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%260%26-5%260%5C%5C0%261%263%260%5C%5C0%260%260%261%5Cend%7Barray%7D%5Cright%5D)
Answer:
75
Step-by-step explanation:
Giving Q was not necessary because you aren't doing P(Q) or anything like that....
Anyways,

Since you're doing p(8), you would replace the x's with 8s.

So your final answer would be 75.
Answer:
120 cm^2.
Step-by-step explanation:
The area of one triangle = 1/2 * base * height
= 1/2 * 8 * 6
= 1/2 * 48
= 24 cm^2
So area of the pentagon = 5 * 24
= 120 cm^2.
Answer:
blue = 110
red = 83
Step-by-step explanation:
110 + 83 = 193
and the blue jar has 27 more coins than the red jar , so this is correct!