Answer:
2/5, 1/3 4/15
Step-by-step explanation:
3x2/5=6/15
5x1/3=5/15
4/15
<span>2n-n-4+7n in simplest form is found by combining like terms:
(2-1+7)n - 4 = 8n-4, or 4(2n-1)</span>
Let n = cost of 1 notebook
Let p = cost of 1 pencil
Then,
3n + 4p = 8.5
5n + 8p = 14.5
You can solve for one variable in terms of the other and then substitute into the remaining equation.
3n + 4p = 8.5
+ 5n + 8p = 14.5
Multiply the top equation by -2 so that the p-containing terms cancel each other out:
-2(3n + 4p = 8.5)
+ 5n + 8p = 14.5
-n + 0 = -2.5
So after dividing both sides by -1, we see that n = $2.5. Plugging into the first equation gives p = $0.25.
3n + 4p = 8.5
5n + 8p = 14.5
Answer:
We have the equation
![c_1\left[\begin{array}{c}0\\0\\0\\1\end{array}\right] +c_2\left[\begin{array}{c}0\\0\\3\\1\end{array}\right] +c_3\left[\begin{array}{c}0\\4\\3\\1\end{array}\right] +c_4\left[\begin{array}{c}8\\4\\3\\1\end{array}\right] =\left[\begin{array}{c}0\\0\\0\\0\end{array}\right]](https://tex.z-dn.net/?f=c_1%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D0%5C%5C0%5C%5C0%5C%5C1%5Cend%7Barray%7D%5Cright%5D%20%2Bc_2%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D0%5C%5C0%5C%5C3%5C%5C1%5Cend%7Barray%7D%5Cright%5D%20%2Bc_3%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D0%5C%5C4%5C%5C3%5C%5C1%5Cend%7Barray%7D%5Cright%5D%20%2Bc_4%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D8%5C%5C4%5C%5C3%5C%5C1%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D0%5C%5C0%5C%5C0%5C%5C0%5Cend%7Barray%7D%5Cright%5D)
Then, the augmented matrix of the system is
![\left[\begin{array}{cccc}0&0&0&8\\0&0&4&4\\0&3&3&3\\1&1&1&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D0%260%260%268%5C%5C0%260%264%264%5C%5C0%263%263%263%5C%5C1%261%261%261%5Cend%7Barray%7D%5Cright%5D)
We exchange rows 1 and 4 and rows 2 and 3 and obtain the matrix:
![\left[\begin{array}{cccc}1&1&1&1\\0&3&3&3\\0&0&4&4\\0&0&0&8\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%261%261%261%5C%5C0%263%263%263%5C%5C0%260%264%264%5C%5C0%260%260%268%5Cend%7Barray%7D%5Cright%5D)
This matrix is in echelon form. Then, now we apply backward substitution:
1.

2.

3.

4.

Then the system has unique solution that is
and this imply that the vectors
are linear independent.