Answer:
Graphs A and B
Step-by-step explanation:
Graphs A & B are using x-values 1-6 and stop there, which is what the domain is trying to find. Though, for the other graphs, these graphs are rather starting at 1 via the domain, while not going up to 6 on the domain.
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).
Let's call the width
and the length
based on this, we can make two equations based on the first and second sentences in the problem.


If we substitute
for
into the second equation, we can make the following:


- Use the Distributive Property


- Subtract 4 from both sides

- Divide both sides by 4 and simplify
Now, let's find
:

The length of the rectangle is
and the width of the rectangle is
.