Question
Answer
$$\dfrac{8k+9}{18}-\dfrac{2k+1}{18}=\dfrac{6k+8}{18}$$
Explanation
| $$ \begin{aligned}\frac{8k+9}{18}-\frac{2k+1}{18}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{6k+8}{18}\end{aligned} $$ | |
| ① | Subtract $ \dfrac{2k+1}{18} $ from $ \dfrac{8k+9}{18} $ to get $ \dfrac{6k+8}{18} $. To subtract expressions with the same denominators, we subtract the numerators and write the result over the common denominator. $$ \begin{aligned} \frac{8k+9}{18} - \frac{2k+1}{18} & = \frac{8k+9}{\color{blue}{18}} - \frac{2k+1}{\color{blue}{18}} =\frac{ 8k+9 - \left( 2k+1 \right) }{ \color{blue}{ 18 }} = \\[1ex] &= \frac{6k+8}{18} \end{aligned} $$ |
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Simplify Rational Expressions