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:
42
Step-by-step explanation:
83+q=125
q=125-83
q=42
In point for (2,7) and in Equation form x=2 y=7
2x - 3y = 7 and -3x + y = 7..multiply Equation 2 by THREE and add to Equation 1
-9x + 3y = 21...........................watch the y's disappear
-7x........ = 28
x = -4
substitute -4 instead of x in either of the ORIGINAL equations
2x - 3y = 7
2(-4) - 3y = 7
-8 -3y = 7..........add 8 to both sides
-3y = 15
y = -5
im not sure
Isolate one variable in the system of equations. Use substitution to create a one-variable equation. Then, set the quadratic equation equal to zero and find the discriminant. If the discriminant is negative, then there are no real number solutions. If the discriminant is zero, then there is one real number solution. If the discriminant is positive, then there are two real number solutions.
