Question

2 essay 5 points is \\(\frac{d + c}{cd} = \frac{1}{c} + \frac{1}{d}\\)? explain. (2 points) correct answer(s) (3 points) correct work shown

Explanation:

Step1: Simplify the right - hand side

To add the fractions \(\frac{1}{c}\) and \(\frac{1}{d}\), we need a common denominator. The common denominator of \(c\) and \(d\) is \(cd\).
Using the rule for adding fractions \(\frac{a}{b}+\frac{e}{f}=\frac{af + be}{bf}\), for \(\frac{1}{c}+\frac{1}{d}\), we have \(\frac{1\times d+1\times c}{c\times d}=\frac{d + c}{cd}\) (which is the same as \(\frac{c + d}{cd}\)).

Step2: Compare with the left - hand side

The left - hand side of the equation is \(\frac{d + c}{cd}\). Since the right - hand side simplifies to \(\frac{d + c}{cd}\), which is equal to the left - hand side.

Answer:

Yes, \(\frac{d + c}{cd}=\frac{1}{c}+\frac{1}{d}\) because when we simplify the right - hand side \(\frac{1}{c}+\frac{1}{d}\) using the common denominator \(cd\), we get \(\frac{d + c}{cd}\), which is equal to the left - hand side.