We can see that, for this choice of values, the expressions do not give the same result.
c
For this part, we are asked to evaluate each expression for arbitrary values of x and y. We will use x=3 and y=2. Let's start with the first expression.
We can see that, for this choice of values, the expressions give different results. Note that we can choose infinitely many pairs of values for x and y. Here we are only showing one possibility.
d
We are given two algebraic expressions which are claimed to be equivalent.
(x+y)^2 and x^2+y^2
Let's see whether this is true by recalling the obtained results.
x
y
(x+y)^2
x^2+y^2
Match
1
0
1
1
✓
1
2
9
5
*
3
2
25
13
*
Note that the expressions have different results for some values of x and y. Since the expressions do not produce the same result for all values of x and y, then they are not equivalent.
Alternative Solution
Expand (x+y)^2
We can also determine if both expressions are equivalent by expanding (x+y)^2. Recall that, in a power, the exponent tells us how many times to multiply the base. Then, since the exponent is 2, the base (x+y) appears twice in the multiplication. (x+y)^2 = (x+y)(x+y)
We can calculate this using the FOIL Method.
Let's simplify the right-hand side of the above formula.
This means that (x+y)^2 is equal to x^2+y^2 plus the term 2xy. ( x+y )^2_(First Expression) = x^2 + y^2_(Second Expression) + 2xy
That is why we obtained different results for x^2+y^2 for some values of x and y. Furthermore, because 2xy equals 0 when either x or y is equal to 0, the expressions given produce same results when x or y are equal to 0.
