Question

5 are \\(\frac{1}{2}x + \frac{3}{4} - \frac{5}{8}x - \frac{7}{8}\\) and \\(\frac{1}{8}(x + 1)\\) equivalent expressions? show your work.

Explanation:

Step1: Simplify the first expression

First, combine the \(x\)-terms and the constant terms in \(\frac{1}{2}x+\frac{3}{4}-\frac{5}{8}x - \frac{7}{8}\).
For the \(x\)-terms: \(\frac{1}{2}x-\frac{5}{8}x=\frac{4}{8}x-\frac{5}{8}x=-\frac{1}{8}x\)
For the constant terms: \(\frac{3}{4}-\frac{7}{8}=\frac{6}{8}-\frac{7}{8}=-\frac{1}{8}\)
So the first expression simplifies to \(-\frac{1}{8}x-\frac{1}{8}\)

Step2: Simplify the second expression

Simplify \(\frac{1}{8}(x + 1)\) using the distributive property \(a(b + c)=ab+ac\), where \(a=\frac{1}{8}\), \(b = x\), \(c = 1\).
\(\frac{1}{8}(x + 1)=\frac{1}{8}x+\frac{1}{8}\)

Step3: Compare the two simplified expressions

The first expression is \(-\frac{1}{8}x-\frac{1}{8}\) and the second is \(\frac{1}{8}x+\frac{1}{8}\). Since the coefficients of \(x\) and the constant terms are not the same (one has \(-\frac{1}{8}x\) and the other has \(\frac{1}{8}x\), one has \(-\frac{1}{8}\) and the other has \(\frac{1}{8}\)), the two expressions are not equivalent.

Answer:

The two expressions \(\frac{1}{2}x+\frac{3}{4}-\frac{5}{8}x-\frac{7}{8}\) and \(\frac{1}{8}(x + 1)\) are not equivalent.