Question

factoring expressions continued
11 determine which of the following expressions are equivalent. show your work.

• $\frac{1}{6}(x - 3)$
• $\frac{1}{4}x - \frac{3}{5} - \frac{1}{12}x + \frac{1}{10}$
• $\frac{1}{18}x + \frac{1}{9}x - \frac{1}{2}$

12 explain a different method you could use to solve problem 11.

Explanation:

Step1: Simplify the first expression

Expand \(\frac{1}{6}(x - 3)\) using the distributive property \(a(b - c)=ab - ac\).
\(\frac{1}{6}(x - 3)=\frac{1}{6}x-\frac{1}{6}\times3=\frac{1}{6}x-\frac{1}{2}\)

Step2: Simplify the second expression

Combine like terms for \(\frac{1}{4}x-\frac{3}{5}-\frac{1}{12}x+\frac{1}{10}\).
First, combine the \(x\)-terms: \(\frac{1}{4}x-\frac{1}{12}x=\frac{3}{12}x-\frac{1}{12}x=\frac{2}{12}x=\frac{1}{6}x\).
Then, combine the constant terms: \(-\frac{3}{5}+\frac{1}{10}=-\frac{6}{10}+\frac{1}{10}=-\frac{5}{10}=-\frac{1}{2}\).
So the second expression simplifies to \(\frac{1}{6}x-\frac{1}{2}\).

Step3: Simplify the third expression

Combine like terms for \(\frac{1}{18}x+\frac{1}{9}x-\frac{1}{2}\).
Convert \(\frac{1}{9}x\) to eighteenths: \(\frac{1}{9}x=\frac{2}{18}x\).
Then, \(\frac{1}{18}x+\frac{2}{18}x=\frac{3}{18}x=\frac{1}{6}x\).
So the third expression is \(\frac{1}{6}x-\frac{1}{2}\).

One different method is to substitute a specific value for \(x\) (e.g., \(x = 0\), \(x = 6\)) into each expression and check if the results are equal. If for a chosen \(x\), all three expressions give the same output, it suggests they are equivalent (this is based on the principle that equivalent expressions have the same value for all \(x\), so testing specific values can be a quick check).

Answer:

All three expressions \(\boldsymbol{\frac{1}{6}(x - 3)}\), \(\boldsymbol{\frac{1}{4}x-\frac{3}{5}-\frac{1}{12}x+\frac{1}{10}}\), and \(\boldsymbol{\frac{1}{18}x+\frac{1}{9}x-\frac{1}{2}}\) are equivalent as they all simplify to \(\frac{1}{6}x-\frac{1}{2}\).

For problem 12: