Answer:
(0, 1).
Method 1 (Substitution):
Substituting our two y's, we get the following:

Thus, the only set of solutions is (0, 1). A quick sketch (either by hand or on Desmos) can confirm this.
Method 2 (Elimination):
We have two equations. We'll let the top one be equation 1 and the bottom one be equation 2. Eliminating as many variables as we can, we subtract (2) from (1) to get:
0 = 3x => x = 0.
So the only set of solutions is (0, 1).
Method 3 (Gaussian elimination):
We can place this in an augmented matrix and row reduce.
![\left[\begin{array}{cccc}1&5&1 & 1\\1&2&1 & 1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%265%261%20%26%201%5C%5C1%262%261%20%26%201%5Cend%7Barray%7D%5Cright%5D)
Row reducing this gives us:
![\left[\begin{array}{cccc}1&5&1 & 1\\0&3&0 & 0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%265%261%20%26%201%5C%5C0%263%260%20%26%200%5Cend%7Barray%7D%5Cright%5D)
This tells us that the only solution for x is x = 0 (since we read this as "3x = 0") and thus, the only solution we get is (0, 1).
Answer:
f(x) = 2x – 3
f(0) = 2(0) – 3 = -3
f(1) = 2(1) – 3 = 2 – 3 = -1
(f(0) + f(1))/2 = (-3 - 1)/2 = -4/2 (f(0) + f(1))/2 = - 2
In vertex form the value of the vertex is easily identified in the equation. I’m standard form the value of the x intercepts are evident.

Since it can be expressed as fraction, it's a rational number.
“A” because none of the coordinates repeat.