Multiply 2x-5y= -21 by 3 to make it 6x-15y= -63
Multiply 3x-3y= -18 by -5 to make it -15x+15y=90
This cancels the y’s out which leaves us with
6x=-63
&
-15x=90
x for 6x=-63 equals - 10.5 so x is - 10.5 and for -15x=90, x= -6
Then you plug in x into any equation you’d like to find y.
Let’s plug in - 10.5 into 6x... equation.
6(- 10.5)-15y=-63
63-15y= -63
-63 -63
-15y=0
y=0 and x= - 10.5. When you plug in this values it makes the equation true!
But the correct answer is the first one north. Sorry if I’m doing too much hahah
If I’m confusing here’s the right answer...
6x-15y= -63
-15x+15y=90
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.
Answer:
-3
Step-by-step explanation:
ur welcome :)))))
- Multiply x-1 on both sides
Now
ATQ to the equation 
x should be greater than 1 .Because if it becomes 1 then the denominator will be 0 which is impossible.
Hence x>1
Thus
Now
Answer:
b = a + 2 or b = 2 - a
Step-by-step explanation:
Solve for b:
-4 + a^2 + 4 b - b^2 = 0
The left hand side factors into a product with two terms:
(2 + a - b) (-2 + a + b) = 0
Split into two equations:
2 + a - b = 0 or -2 + a + b = 0
Subtract a + 2 from both sides:
-b = -a - 2 or -2 + a + b = 0
Multiply both sides by -1:
b = a + 2 or -2 + a + b = 0
Subtract a - 2 from both sides:
Answer:
b = a + 2 or b = 2 - a
