Let's analyze each step in the given proof to understand the justification for Step 5.
### Step-by-Step Analysis
Step 1:
[tex]\[
(x+y)^2 - (x-y)^2 = (x+y)(x+y) - (x-y)(x-y)
\][/tex]
This step rewrites the squares as products of binomials.
Step 2:
[tex]\[
(x+y)(x+y) - (x-y)(x-y) = \left(x^2 + xy + xy + y^2\right) - \left(x^2 - xy - xy + y^2\right)
\][/tex]
This step expands the binomials using the distributive property (or FOIL method).
Step 3:
[tex]\[
\left(x^2 + xy + xy + y^2\right) - \left(x^2 - xy - xy + y^2\right) = \left(x^2 + 2xy + y^2\right) - \left(x^2 - 2xy + y^2\right)
\][/tex]
This step simplifies the expressions inside the parentheses by combining like terms.
Step 4:
[tex]\[
\left(x^2 + 2xy + y^2\right) - \left(x^2 - 2xy + y^2\right) = x^2 + 2xy + y^2 - x^2 + 2xy - y^2
\][/tex]
This step distributes the subtraction over each term inside the parentheses.
Step 5:
[tex]\[
x^2 + 2xy + y^2 - x^2 + 2xy - y^2 = 4xy
\][/tex]
In this step, we observe that [tex]\(x^2\)[/tex] cancels out with [tex]\(-x^2\)[/tex] and [tex]\(y^2\)[/tex] cancels out with [tex]\(-y^2\)[/tex]. What remains is [tex]\(2xy + 2xy = 4xy\)[/tex].
The justification for this step is combining like terms, as we combined [tex]\(2xy\)[/tex] and [tex]\(2xy\)[/tex] to get [tex]\(4xy\)[/tex].
### Conclusion
The appropriate justification for Step 5 of Darlene's proof is:
B. Combining like terms
