After putting the value of y from the second equation to the first equation, the resultant equation is 
.
GIven:
The equations are:
It is required to put the value of y from second equation to the first equation.
<h3>How to solve equations?</h3>
The value of y from the second equation is,
Now, put this value of y in the first equation as,
Therefore, after putting the value of y from the second equation to the first equation, the resultant equation is .
For more details about equations, refer to the link:
brainly.com/question/2263981
Answer:
(I rotated the trapezoid on the origin)
T' (-2, 2)
R' (-2, 5)
A' (-6, 2)
P' (-7, 5)
Step-by-step explanation:
The original points of the trapezoid were (2, -2), (2, -5), (6, -2) and (7, -5). Flipping trapezoid TRAP on the origin has the x and y coordinates showing their opposites from the original. So, find the opposite of each x and y coordinate to get the coordinates of the rotated trapezoid T'R'A'P'.
<h2>
The required solution is x = 6 and y = 11 </h2>
Step-by-step explanation:
Given system of equations are
x+5y = 11 and x-y =5
![A= \left[\begin{array}{cc}1&5\\1&-1\end{array}\right]](https://tex.z-dn.net/?f=A%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%265%5C%5C1%26-1%5Cend%7Barray%7D%5Cright%5D)
![X=\left[\begin{array}{c}x\\y\end{array}\right]](https://tex.z-dn.net/?f=X%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%5C%5Cy%5Cend%7Barray%7D%5Cright%5D)
and ![B= \left[\begin{array}{c}11\\5\end{array}\right]](https://tex.z-dn.net/?f=B%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D11%5C%5C5%5Cend%7Barray%7D%5Cright%5D)
∴AX=B
![adj A = \left[\begin{array}{cc}{-1}&{-5}\\{-1}&1\end{array}\right]](https://tex.z-dn.net/?f=adj%20A%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D%7B-1%7D%26%7B-5%7D%5C%5C%7B-1%7D%261%5Cend%7Barray%7D%5Cright%5D)
∴
So,![A^{-1} =\frac{ \left[\begin{array}{cc}{-1}&{-5}\\{-1}&1\end{array}\right]}{-6}](https://tex.z-dn.net/?f=A%5E%7B-1%7D%20%3D%5Cfrac%7B%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D%7B-1%7D%26%7B-5%7D%5C%5C%7B-1%7D%261%5Cend%7Barray%7D%5Cright%5D%7D%7B-6%7D)
![A^{-1} ={ \left[\begin{array}{c \c} {{\frac{1}{6} }}&{\frac{5}{6}}\ \\ {{\frac{1}{6} }}&{\frac{-1}{6}} \end{array}\right]}](https://tex.z-dn.net/?f=A%5E%7B-1%7D%20%3D%7B%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%20%5Cc%7D%20%20%7B%7B%5Cfrac%7B1%7D%7B6%7D%20%7D%7D%26%7B%5Cfrac%7B5%7D%7B6%7D%7D%5C%20%5C%5C%20%20%7B%7B%5Cfrac%7B1%7D%7B6%7D%20%7D%7D%26%7B%5Cfrac%7B-1%7D%7B6%7D%7D%20%5Cend%7Barray%7D%5Cright%5D%7D)
⇒![\left[\begin{array}{c}x\\y\end{array}\right] ={ \left[\begin{array}{c \c} {{\frac{1}{6} }}&{\frac{5}{6}}\ \\ {{\frac{1}{6} }}&{\frac{-1}{6}} \end{array}\right]} \times \left[\begin{array}{c}11\\5\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%5C%5Cy%5Cend%7Barray%7D%5Cright%5D%20%3D%7B%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%20%5Cc%7D%20%20%7B%7B%5Cfrac%7B1%7D%7B6%7D%20%7D%7D%26%7B%5Cfrac%7B5%7D%7B6%7D%7D%5C%20%5C%5C%20%20%7B%7B%5Cfrac%7B1%7D%7B6%7D%20%7D%7D%26%7B%5Cfrac%7B-1%7D%7B6%7D%7D%20%5Cend%7Barray%7D%5Cright%5D%7D%20%5Ctimes%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D11%5C%5C5%5Cend%7Barray%7D%5Cright%5D)
⇒![\left[\begin{array}{c}x\\y\end{array}\right] ={ \left[\begin{array}{c} {6}\\ {11} \end{array}\right]}](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%5C%5Cy%5Cend%7Barray%7D%5Cright%5D%20%3D%7B%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D%20%20%7B6%7D%5C%5C%20%20%7B11%7D%20%5Cend%7Barray%7D%5Cright%5D%7D)
∴ x= 6 and y = 11
The required solution is x = 6 and y = 11
Answer:
Two 50 bottles
Step-by-step explanation:
because the 50 bottles is more cheaper than buying a n 80 ml bottle,
it's just a waste of money.
