Write the equations in matrix,
![\left[\begin{array}{ccc}5&-1&1\\1&2&-1\\2&3&-3\end{array}\right] \left[\begin{array}{ccc}x\\y\\z\end{array}\right] = \left[\begin{array}{ccc}4\\5\\5\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%26-1%261%5C%5C1%262%26-1%5C%5C2%263%26-3%5Cend%7Barray%7D%5Cright%5D%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%5Cright%5D%20%3D%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4%5C%5C5%5C%5C5%5Cend%7Barray%7D%5Cright%5D%20)
Using row transformation,
R₂ <---> R₃
![\left[\begin{array}{ccc}5&-1&1\\2&3&-3\\1&2&-1\end{array}\right] \left[\begin{array}{ccc}x\\y\\z\end{array}\right] = \left[\begin{array}{ccc}4\\5\\5\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%26-1%261%5C%5C2%263%26-3%5C%5C1%262%26-1%5Cend%7Barray%7D%5Cright%5D%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%5Cright%5D%20%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4%5C%5C5%5C%5C5%5Cend%7Barray%7D%5Cright%5D%20)
Using,
R₂ ---> R₂ - 2R₃
![\left[\begin{array}{ccc}5&-1&1\\0&-1&-1\\1&2&-1\end{array}\right] \left[\begin{array}{ccc}x\\y\\z\end{array}\right] = \left[\begin{array}{ccc}4\\-5\\5\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%26-1%261%5C%5C0%26-1%26-1%5C%5C1%262%26-1%5Cend%7Barray%7D%5Cright%5D%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%5Cright%5D%20%3D%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4%5C%5C-5%5C%5C5%5Cend%7Barray%7D%5Cright%5D%20)
Using,
R₂ --- > (-1)R₂
![\left[\begin{array}{ccc}5&-1&1\\0&1&1\\1&2&-1\end{array}\right] \left[\begin{array}{ccc}x\\y\\z\end{array}\right] = \left[\begin{array}{ccc}4\\5\\5\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%26-1%261%5C%5C0%261%261%5C%5C1%262%26-1%5Cend%7Barray%7D%5Cright%5D%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%5Cright%5D%20%3D%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4%5C%5C5%5C%5C5%5Cend%7Barray%7D%5Cright%5D%20)
Using row transformation,
R₂ <----> R₃
![\left[\begin{array}{ccc}5&-1&1\\1&2&-1\\0&1&1\end{array}\right] \left[\begin{array}{ccc}x\\y\\z\end{array}\right] = \left[\begin{array}{ccc}4\\5\\5\end{array}\right]](https://tex.z-dn.net/?f=%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%26-1%261%5C%5C1%262%26-1%5C%5C0%261%261%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4%5C%5C5%5C%5C5%5Cend%7Barray%7D%5Cright%5D%20)
Using,
R₂ ---> R₂ - R₁/5
![\left[\begin{array}{ccc}5&-1&1\\0&11/5&-6/5\\0&1&1\end{array}\right] \left[\begin{array}{ccc}x\\y\\z\end{array}\right] = \left[\begin{array}{ccc}4\\21/5\\5\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%26-1%261%5C%5C0%2611%2F5%26-6%2F5%5C%5C0%261%261%5Cend%7Barray%7D%5Cright%5D%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%5Cright%5D%20%3D%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4%5C%5C21%2F5%5C%5C5%5Cend%7Barray%7D%5Cright%5D%20)
Using,
R₃ ---> R₃ - 5R₂/11
![\left[\begin{array}{ccc}5&-1&1\\0&11/5&-6/5\\0&0&17/11\end{array}\right] \left[\begin{array}{ccc}x\\y\\z\end{array}\right] = \left[\begin{array}{ccc}4\\21/5\\34/11\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%26-1%261%5C%5C0%2611%2F5%26-6%2F5%5C%5C0%260%2617%2F11%5Cend%7Barray%7D%5Cright%5D%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%5Cright%5D%20%3D%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4%5C%5C21%2F5%5C%5C34%2F11%5Cend%7Barray%7D%5Cright%5D%20)
∴ 5x-y+z = 4 ====(i)
11y-6z = 21 === (ii)
17z=34 === (iii)
from iii,
z=2.
Plug z=2 in ii to get y,
∴y=3.
Plug y and z values in i to get x,
∴x=1
Therefore the solution to the system of equations is (1,3,2)
Answer:
1)-6a+-48
2)4+36x
3)-30n+18
4)18m+20
5)32-24n
6)-8b-32
7)5-35n
8)-6x+-24
9)15m-30
10)24p+-28
11)5b-5
12)5x+45
Step-by-step explanation:
Distributive property, Hope this helps!
Answer:
D. 1/3^3
Step-by-step explanation:
Divide 3/3 as usual and get 1.
When dividing exponents, you are actually subtracting them. 5-8 = -3.
D is the answer because positive exponents in the denominator of the fraction is the same as a negative exponent in the numerator.
X+ 3y = 1 -----(first equation)
-2x+7 = -23 ---(second equation)
-2x = -23-7
x = 30/2
x = 15
Now, Substitute the value of x in first equation,
15 + 3y = 1
3y = 1 - 15
y = -14/3
In short, Final Answers are: x = 15 & y = -14/3
Hope this helps!