Answer with explanation:
1. The given equations are
3x -5 y=2
-x+2 y= 0
⇒The matrix in the form of , AX=B, is
![A=\left[\begin{array}{cc}3&-5\\-1&2\end{array}\right] ,\\\\ X=\left[\begin{array}{c}x&y\end{array}\right],\\\\B=\left[\begin{array}{c}2&0\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D3%26-5%5C%5C-1%262%5Cend%7Barray%7D%5Cright%5D%20%2C%5C%5C%5C%5C%20X%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%26y%5Cend%7Barray%7D%5Cright%5D%2C%5C%5C%5C%5CB%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D2%260%5Cend%7Barray%7D%5Cright%5D)
Adj.A=Transpose of cofactor of Matrix A
![Adj.A=\left[\begin{array}{cc}2&1\\5&3\end{array}\right] ,\\\\ |A|=6-5\\\\|A|=1\\\\\left[\begin{array}{c}x&y\end{array}\right]=\left[\begin{array}{cc}2&5\\1&3\end{array}\right] \times \left[\begin{array}{c}2&0\end{array}\right]\\\\x=4, y=2](https://tex.z-dn.net/?f=Adj.A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D2%261%5C%5C5%263%5Cend%7Barray%7D%5Cright%5D%20%2C%5C%5C%5C%5C%20%7CA%7C%3D6-5%5C%5C%5C%5C%7CA%7C%3D1%5C%5C%5C%5C%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%26y%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D2%265%5C%5C1%263%5Cend%7Barray%7D%5Cright%5D%20%5Ctimes%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D2%260%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5Cx%3D4%2C%20y%3D2)
2.
The given equations are
x+y-z=2
x+z=7
2 x +y+z=13
⇒The matrix in the form of , AX=B, is
![A=\left[\begin{array}{ccc}1&1&-1\\1&0&1\\2&1&1\end{array}\right]\\\\ X=\left[\begin{array}{ccc}x\\y\\z\end{array}\right]\\\\B= \left[\begin{array}{ccc}2\\7\\13\end{array}\right]\\\\\rightarrow X=A^{-1}B\\\\\rightarrow X=\frac{Adj.A}{|A|}\times B\\\\a_{11}=-1,a_{12}=1,a_{13}=1,a_{21}=-2,a_{22}=3,a_{23}=1,a_{31}=1,a_{32}=-2,a_{33}=-1\\\\|A|=1\times(0-1)-1\times(1-2)-1\times(1-0)\\\\=-1+1-1\\\\|A|=-1\\\\Adj.A=\left[\begin{array}{ccc}-1&-2&1\\1&3&-2\\1&1&-1\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%26-1%5C%5C1%260%261%5C%5C2%261%261%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5C%20X%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5CB%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%5C%5C7%5C%5C13%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5C%5Crightarrow%20X%3DA%5E%7B-1%7DB%5C%5C%5C%5C%5Crightarrow%20X%3D%5Cfrac%7BAdj.A%7D%7B%7CA%7C%7D%5Ctimes%20B%5C%5C%5C%5Ca_%7B11%7D%3D-1%2Ca_%7B12%7D%3D1%2Ca_%7B13%7D%3D1%2Ca_%7B21%7D%3D-2%2Ca_%7B22%7D%3D3%2Ca_%7B23%7D%3D1%2Ca_%7B31%7D%3D1%2Ca_%7B32%7D%3D-2%2Ca_%7B33%7D%3D-1%5C%5C%5C%5C%7CA%7C%3D1%5Ctimes%280-1%29-1%5Ctimes%281-2%29-1%5Ctimes%281-0%29%5C%5C%5C%5C%3D-1%2B1-1%5C%5C%5C%5C%7CA%7C%3D-1%5C%5C%5C%5CAdj.A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-1%26-2%261%5C%5C1%263%26-2%5C%5C1%261%26-1%5Cend%7Barray%7D%5Cright%5D)
![\frac{Adj.A}{|A|}=\left[\begin{array}{ccc}1&2&-1\\-1&-3&2\\-1&-1&1\end{array}\right]\\\\X=A^{-1}B\\\\\left[\begin{array}{ccc}x\\y\\z\end{array}\right]=\left[\begin{array}{ccc}1&2&-1\\-1&-3&2\\-1&-1&1\end{array}\right]\times\left[\begin{array}{ccc}2\\7\\13\end{array}\right]\\\\x=3,y=3,z=4](https://tex.z-dn.net/?f=%5Cfrac%7BAdj.A%7D%7B%7CA%7C%7D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%262%26-1%5C%5C-1%26-3%262%5C%5C-1%26-1%261%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5CX%3DA%5E%7B-1%7DB%5C%5C%5C%5C%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%262%26-1%5C%5C-1%26-3%262%5C%5C-1%26-1%261%5Cend%7Barray%7D%5Cright%5D%5Ctimes%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%5C%5C7%5C%5C13%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5Cx%3D3%2Cy%3D3%2Cz%3D4)
Answer:9000
Step-by-step explanation:
Answer:
54
Step-by-step explanation:
6x + 5x = 99
11x = 99
x=99/11
x=9
6x = 6x9 = 54
5x = 5x9 = 45
<em>ans: </em><u><em>54</em></u>
Answer: One POSSIBLE answer is (0,9)
Step-by-step explanation: This may look hard but it's actually super easy. All you have to do is substitute a value for x and see what you get as the y value. The first thing I would do is rearrange the equation so that the equation is in the format y = something. So the first thing you should do is subtract 6x from both sides of the equation and you should get -y=-6x+9. Right now, the y is being multiplied by -1, and to undo multiplication, you have to divide. So divide both sides of the equation by -1 (remember, what you do on one side of the equation, YOU HAVE TO DO THE SAME THING ON THE OTHER SIDE). So your new equation would be y=6x-9. Then, all you have to do is substitute a value for x and see what you get for y (x is basically the input and whatever you get for y is the output). S for example, let's say that x=0. So substitute a 0 in the equation wherever you see an x. So in this case, you would get y=(6x0)+9. Then you just simplify. So 6x0 is 0 and 0+9=9 So y=9
TO RECAP: Rearrange the equation so that the equation is in the format y = something. Then, input any number for x, and whatever you get for y is what you put in the brackets as your y value for your final answer.
Hope this helps!
