Question

is \\(\frac{2}{3}(-12b - 6 + 9b - 18)\\) equivalent to \\(2(b + 8)\\)? show your work.

Explanation:

Step1: Simplify the first expression

First, distribute \(\frac{2}{3}\) in \(\frac{2}{3}(-12b - 6 + 9b - 18)\).
Combine like terms inside the parentheses: \(-12b+9b=-3b\) and \(-6 - 18=-24\). So the expression becomes \(\frac{2}{3}(-3b - 24)\).
Now distribute \(\frac{2}{3}\): \(\frac{2}{3}\times(-3b)+\frac{2}{3}\times(-24)=-2b-16\).

Step2: Simplify the second expression

Simplify \(2(b + 8)\) by distributing 2: \(2\times b+2\times8 = 2b + 16\).

Step3: Compare the two results

The first expression simplifies to \(-2b - 16\) and the second to \(2b + 16\). Since \(-2b - 16
eq2b + 16\) (for example, if \(b = 0\), the first is \(-16\) and the second is \(16\)), they are not equivalent.

Answer:

\(\frac{2}{3}(-12b - 6 + 9b - 18)\) simplifies to \(-2b - 16\) and \(2(b + 8)\) simplifies to \(2b + 16\). These two expressions are not equivalent.