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Darren and Quincy wanted to justify that the expression [tex]\frac{1}{6}(6x + 12) - \frac{1}{2}(4x + 2)[/tex] is equivalent to [tex]-x + 1[/tex]. Their justifications are shown below.

Quincy's Method: Substitute 6 into both expressions.
[tex]\[
\begin{array}{|c|}
\hline
\text{Quincy's Method: Substitute 6 into both expressions} \\
\hline
\frac{1}{6}(6 \cdot 6 + 12) - \frac{1}{2}(4 \cdot 6 + 2) \\
\hline
-x + 1 \rightarrow -6 + 1 \\
\hline
\end{array}
\][/tex]

Is Quincy's method correct? Compare the simplified values.

Sagot :

GinnyAnsweridn
To justify that the expression [tex]\(\frac{1}{6}(6x + 12) - \frac{1}{2}(4x + 2)\)[/tex] is equivalent to [tex]\(-x + 1\)[/tex], Darren and Quincy can substitute [tex]\(x = 6\)[/tex] into both expressions and evaluate them. Let's go through the detailed steps for each expression.

### Step-by-Step Solution

#### Evaluate [tex]\(\frac{1}{6}(6 x + 12) - \frac{1}{2}(4 x + 2)\)[/tex]:

1. Substitute [tex]\(x = 6\)[/tex] into the expression:
[tex]\[ \frac{1}{6}(6(6) + 12) - \frac{1}{2}(4(6) + 2) \][/tex]

2. Simplify the terms inside the parentheses:
[tex]\[ (6(6) + 12) = (36 + 12) = 48 \][/tex]
[tex]\[ (4(6) + 2) = (24 + 2) = 26 \][/tex]

3. Continue with the expression:
[tex]\[ \frac{1}{6}(48) - \frac{1}{2}(26) \][/tex]

4. Simplify each fraction:
[tex]\[ \frac{1}{6}(48) = 8 \][/tex]
[tex]\[ \frac{1}{2}(26) = 13 \][/tex]

5. Subtract the two results:
[tex]\[ 8 - 13 = -5 \][/tex]

So, the value of [tex]\(\frac{1}{6}(6x + 12) - \frac{1}{2}(4x + 2)\)[/tex] when [tex]\(x = 6\)[/tex] is [tex]\(-5\)[/tex].

#### Evaluate [tex]\(-x + 1\)[/tex]:

1. Substitute [tex]\(x = 6\)[/tex] into the expression:
[tex]\[ -6 + 1 \][/tex]

2. Simplify:
[tex]\[ -6 + 1 = -5 \][/tex]

So, the value of [tex]\(-x + 1\)[/tex] when [tex]\(x = 6\)[/tex] is also [tex]\(-5\)[/tex].

#### Conclusion:

Since both expressions result in [tex]\(-5\)[/tex] when [tex]\(x = 6\)[/tex], we can conclude that [tex]\(\frac{1}{6}(6x + 12) - \frac{1}{2}(4x + 2)\)[/tex] is equivalent to [tex]\(-x + 1\)[/tex].
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