Answer:
First Box: Factor by difference of squares
Second Box: It has all perfect squares
Third Box: see my work
Step-by-step explanation:
We have the equation
Before we do anything, let us bring all of the terms to the left side
Now that we have moved the terms to one side, it is clear that we have a difference of squares.
-Work-
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Answer:
The ordered pair is indeed a solution to the system:
.
Step-by-step explanation:
Consider a system of equations about variables and
. An ordered pair
(where
and
are constant) is a solution to that system if and only if all equations in that system hold after substituting in
and
.
For the system in this question, would be a solution only if both equations in the system hold after replacing all
in equations of the system with
and all
with
.
The of the equation
would become
. The
of that equation would become
. The two sides are indeed equal.
Similarly, the of the equation
would become
. The
of that equation would become
. The two sides are indeed equal.
Thus, and
simultaneously satisfy both equations of the given system. Therefore, the ordered pair
would indeed be a solution to that system.