Question
Gabriel determined that his total cost would be represented by 2.5x + 2y – 2. His sister states that the expression should be x + x + 0.5x + y + y – 2. Who is correct? Explain.
Answers
GPT-4o mini
To determine who is correct, we need to simplify both expressions and see if they are equivalent.
Gabriel's expression is:
\[ 2.5x + 2y - 2 \]
Now, let's look at his sister's expression:
\[ x + x + 0.5x + y + y - 2 \]
First, let's simplify his sister's expression step by step:
1. Combine like terms for \(x\):
- \( x + x = 2x \)
- \( 2x + 0.5x = 2.5x \)
So, the \(x\) terms combine to \(2.5x\).
2. Combine like terms for \(y\):
- \( y + y = 2y \)
So, the \(y\) terms combine to \(2y\).
3. Now we can put together the simplified parts:
\[ 2.5x + 2y - 2 \]
Now, we compare the simplified version of his sister's expression with Gabriel's expression:
- Gabriel's expression: \( 2.5x + 2y - 2 \)
- Sister's simplified expression: \( 2.5x + 2y - 2 \)
Since both expressions are identical, we can conclude that both Gabriel and his sister are correct in their representations of cost – they just arrived at it through different approaches.
Thus, **both expressions are equivalent**.
Gabriel's expression is:
\[ 2.5x + 2y - 2 \]
Now, let's look at his sister's expression:
\[ x + x + 0.5x + y + y - 2 \]
First, let's simplify his sister's expression step by step:
1. Combine like terms for \(x\):
- \( x + x = 2x \)
- \( 2x + 0.5x = 2.5x \)
So, the \(x\) terms combine to \(2.5x\).
2. Combine like terms for \(y\):
- \( y + y = 2y \)
So, the \(y\) terms combine to \(2y\).
3. Now we can put together the simplified parts:
\[ 2.5x + 2y - 2 \]
Now, we compare the simplified version of his sister's expression with Gabriel's expression:
- Gabriel's expression: \( 2.5x + 2y - 2 \)
- Sister's simplified expression: \( 2.5x + 2y - 2 \)
Since both expressions are identical, we can conclude that both Gabriel and his sister are correct in their representations of cost – they just arrived at it through different approaches.
Thus, **both expressions are equivalent**.