2.5y + 3x = 27
5x - 2.5y = 5
Align the variables to make solving this easier.
2.5y + 3x = 27
-2.5y + 5x = 5
You can see that the 2.5y and -2.5y cancel each other out. Then add the 3x and 5x, and 27 + 5.
3x = 27
5x = 5
8x = 32
Divide both sides by 8.
x = 4
Now input that x into one of the equations to get y.
2.5y + 3(4) = 27
2.5y + 12 = 27
2.5y = 15
y = 6
The answer to this system of equations is x = 4, y = 6. (4, 6)
Answer:
A) JK and AB =
![[tex] \boxed{( \theta) = \f{12}{( \theta)} }](https://tex.z-dn.net/?f=%5Btex%5D%20%5Cboxed%7B%28%20%5Ctheta%29%20%3D%20%5Cf%7B12%7D%7B%28%20%5Ctheta%29%7D%20%7D)
So, =》 
=》 
[/tex]
B) AABC
![[tex] \boxed{( \theta) = {1}{( \theta)} }](https://tex.z-dn.net/?f=%5Btex%5D%20%5Cboxed%7B%28%20%5Ctheta%29%20%3D%20%7B1%7D%7B%28%20%5Ctheta%29%7D%20%7D)
So, 
C) AABM:
![[tex] e \fbox{: }](https://tex.z-dn.net/?f=%5Btex%5D%20e%20%5Cfbox%7B%3A%20%7D)
The value of k for which 
is 
ans[/tex]
----------
