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
Answer:
Step-by-step explanation:
Divide the tens column dividend by the divisor.
Multiply the divisor by the quotient in the tens place column.
Subtract the product from the divisor.
Bring down the dividend in the ones column and repeat.
okay so look at the image down below
2x - 3y = 3.....multiply by 4
5x - 4y = 4....multiply by -3
--------------
8x - 12y = 12 (result of multiplying by 4)
-15x + 12y = -12 (result of multiplying by -3)
-------------
- 7x = 0
x = 0 <===
Answer:
f(x) = -2x² - 8x - 2
General Formulas and Concepts:
- Order of Operations: BPEMDAS
- Expand by FOIL (First Outside Inside Last)
- Standard Form: f(x) = ax² + bx + c
- Vertex Form: f(x) = a(bx + c)² + d
Step-by-step explanation:
<u>Step 1: Define function</u>
Vertex Form: f(x) = -2(x + 2)² + 6
<u>Step 2: Find Standard Form</u>
- Expand by FOILing: f(x) = -2(x² + 4x + 4) + 6
- Distribute -2: f(x) = -2x² - 8x - 8 + 6
- Combine like terms (constants): f(x) = -2x² - 8x - 2
