So we are given the system:

Written in matrix form we get:
![\left[\begin{array}{cc}2&4\\6&3\end{array}\right] \left[\begin{array}{c}x\\y\end{array}\right] = \left[\begin{array}{c}8\\-3\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D2%264%5C%5C6%263%5Cend%7Barray%7D%5Cright%5D%20%0A%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%5C%5Cy%5Cend%7Barray%7D%5Cright%5D%20%3D%0A%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D8%5C%5C-3%5Cend%7Barray%7D%5Cright%5D%20)
We compute the solution like this:
![ \left[\begin{array}{c}x\\y\end{array}\right] = \left[\begin{array}{cc}2&4\\6&3\end{array}\right] ^{-1} \left[\begin{array}{c}8\\-3\end{array}\right] \\= \left[\begin{array}{cc}-3&4\\6&-2\end{array}\right] \left[\begin{array}{c}8\\-3\end{array}\right] \dfrac{1}{18}\\= \left[\begin{array}{c}2\\-3\end{array}\right]](https://tex.z-dn.net/?f=%20%0A%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%5C%5Cy%5Cend%7Barray%7D%5Cright%5D%20%3D%0A%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D2%264%5C%5C6%263%5Cend%7Barray%7D%5Cright%5D%20%5E%7B-1%7D%0A%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D8%5C%5C-3%5Cend%7Barray%7D%5Cright%5D%20%20%5C%5C%3D%0A%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-3%264%5C%5C6%26-2%5Cend%7Barray%7D%5Cright%5D%0A%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D8%5C%5C-3%5Cend%7Barray%7D%5Cright%5D%20%5Cdfrac%7B1%7D%7B18%7D%5C%5C%3D%0A%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D2%5C%5C-3%5Cend%7Barray%7D%5Cright%5D)
The solution is :
There is one solution, (0,4) is your answer
Note that x = 0, y = 4
(4) = 1/3(0) + 4
4 = 0 + 4
4 = 4 (true)
3(4) - (0) = 12
12 - 0 = 12
12 = 12 (true)
There is one solution, (0,4) is your answer
hope this helps
Hello There!
The "Inter-Quartile Range" is 9
The best thing to do is to divide 78 by 100, and then multiply this by 9.5
78/100= 0.78. 0.78x9.5=7.41
You've then got to add 7.41 to 78, and this gives you 85.41
Hope this helps :)