Answer:
![W=\{\left[\begin{array}{ccc}a+2b\\b\\-3a\end{array}\right]: a,b\in\mathbb{R} \}](https://tex.z-dn.net/?f=W%3D%5C%7B%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Da%2B2b%5C%5Cb%5C%5C-3a%5Cend%7Barray%7D%5Cright%5D%3A%20a%2Cb%5Cin%5Cmathbb%7BR%7D%20%5C%7D)
Observe that if the vector ![x=\left[\begin{array}{ccc}x\\y\\z\end{array}\right]](https://tex.z-dn.net/?f=x%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%5Cright%5D)
is in W then it satisfies:
![\left[\begin{array}{ccc}x\\y\\z\end{array}\right]=\left[\begin{array}{c}a+2b\\b\\-3a\end{array}\right]=a\left[\begin{array}{c}1\\0\\-3\end{array}\right]+b\left[\begin{array}{c}2\\1\\0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Da%2B2b%5C%5Cb%5C%5C-3a%5Cend%7Barray%7D%5Cright%5D%3Da%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D1%5C%5C0%5C%5C-3%5Cend%7Barray%7D%5Cright%5D%2Bb%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D2%5C%5C1%5C%5C0%5Cend%7Barray%7D%5Cright%5D)
This means that each vector in W can be expressed as a linear combination of the vectors ![\left[\begin{array}{c}1\\0\\-3\end{array}\right], \left[\begin{array}{c}2\\1\\0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D1%5C%5C0%5C%5C-3%5Cend%7Barray%7D%5Cright%5D%2C%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D2%5C%5C1%5C%5C0%5Cend%7Barray%7D%5Cright%5D)
Also we can see that those vectors are linear independent. Then the set
![\{\left[\begin{array}{c}1\\0\\-3\end{array}\right], \left[\begin{array}{c}2\\1\\0\end{array}\right]\}](https://tex.z-dn.net/?f=%5C%7B%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D1%5C%5C0%5C%5C-3%5Cend%7Barray%7D%5Cright%5D%2C%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D2%5C%5C1%5C%5C0%5Cend%7Barray%7D%5Cright%5D%5C%7D)
is a basis for W and the dimension of W is 2.
-2x + 8 = -4x + 1
- 1 -2x
7 = -6x
—
-6x
7/-6 or -7/6
Multiply the number in the tens place of the bottom
number by the number in ones place of the top number. So
multiply the 1 by the 5, which makes 5. Multiply the number in
the tens place of the bottom number by the number in tens
place of the top number. Multiply 1 by 2, which equals 2.
Yes 12/15 can be simplified.
To simplify 12/15 find the greatest common multiple. The <span> greatest common multiple of 12/15 is 3. Now divide 12/15 by 3.
</span>
12 ÷ 3 = 4
15 ÷ 3 = 5
4/5 is the simplified version of 12/15
Answer:
C) y = 60x
Step-by-step explanation:
