$$r-\dfrac{6}{2}(r+1)+2\dfrac{r}{2}=-r-3$$
Tap the blue circles to see an explanation.
| $$ \begin{aligned}r-\frac{6}{2}(r+1)+2\frac{r}{2}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}r - \frac{ 6 : \color{orangered}{ 2 } }{ 2 : \color{orangered}{ 2 }} \cdot \left(r+1\right) + r \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}r-\frac{3}{1}(r+1)+r \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}r-3(r+1)+r \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} \htmlClass{explanationCircle explanationCircle7}{\textcircled {7}} } }}}r-(3r+3)+r \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle8}{\textcircled {8}} } }}}r-3r-3+r \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle9}{\textcircled {9}} } }}}-r-3\end{aligned} $$ | |
| ① | Divide both the top and bottom numbers by $ \color{orangered}{ 2 } $. |
| ② | Multiply $2$ by $ \dfrac{r}{2} $ to get $ r$. Step 1: Write $ 2 $ as a fraction by putting $ \color{red}{1} $ in the denominator. Step 2: Cancel $ \color{blue}{ 2 } $ in first and second fraction. Step 3: Multiply numerators and denominators. Step 4: Simplify numerator and denominator. $$ \begin{aligned} 2 \cdot \frac{r}{2} & \xlongequal{\text{Step 1}} \frac{2}{\color{red}{1}} \cdot \frac{r}{2} \xlongequal{\text{Step 2}} \frac{\color{blue}{1}}{1} \cdot \frac{r}{\color{blue}{1}} = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 1 \cdot r }{ 1 \cdot 1 } \xlongequal{\text{Step 4}} \frac{ r }{ 1 } =r \end{aligned} $$ |
| ③ | Multiply $2$ by $ \dfrac{r}{2} $ to get $ r$. Step 1: Write $ 2 $ as a fraction by putting $ \color{red}{1} $ in the denominator. Step 2: Cancel $ \color{blue}{ 2 } $ in first and second fraction. Step 3: Multiply numerators and denominators. Step 4: Simplify numerator and denominator. $$ \begin{aligned} 2 \cdot \frac{r}{2} & \xlongequal{\text{Step 1}} \frac{2}{\color{red}{1}} \cdot \frac{r}{2} \xlongequal{\text{Step 2}} \frac{\color{blue}{1}}{1} \cdot \frac{r}{\color{blue}{1}} = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 1 \cdot r }{ 1 \cdot 1 } \xlongequal{\text{Step 4}} \frac{ r }{ 1 } =r \end{aligned} $$ |
| ④ | Remove 1 from denominator. |
| ⑤ | Multiply $2$ by $ \dfrac{r}{2} $ to get $ r$. Step 1: Write $ 2 $ as a fraction by putting $ \color{red}{1} $ in the denominator. Step 2: Cancel $ \color{blue}{ 2 } $ in first and second fraction. Step 3: Multiply numerators and denominators. Step 4: Simplify numerator and denominator. $$ \begin{aligned} 2 \cdot \frac{r}{2} & \xlongequal{\text{Step 1}} \frac{2}{\color{red}{1}} \cdot \frac{r}{2} \xlongequal{\text{Step 2}} \frac{\color{blue}{1}}{1} \cdot \frac{r}{\color{blue}{1}} = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 1 \cdot r }{ 1 \cdot 1 } \xlongequal{\text{Step 4}} \frac{ r }{ 1 } =r \end{aligned} $$ |
| ⑥ | Multiply $ \color{blue}{3} $ by $ \left( r+1\right) $ $$ \color{blue}{3} \cdot \left( r+1\right) = 3r+3 $$ |
| ⑦ | Multiply $2$ by $ \dfrac{r}{2} $ to get $ r$. Step 1: Write $ 2 $ as a fraction by putting $ \color{red}{1} $ in the denominator. Step 2: Cancel $ \color{blue}{ 2 } $ in first and second fraction. Step 3: Multiply numerators and denominators. Step 4: Simplify numerator and denominator. $$ \begin{aligned} 2 \cdot \frac{r}{2} & \xlongequal{\text{Step 1}} \frac{2}{\color{red}{1}} \cdot \frac{r}{2} \xlongequal{\text{Step 2}} \frac{\color{blue}{1}}{1} \cdot \frac{r}{\color{blue}{1}} = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 1 \cdot r }{ 1 \cdot 1 } \xlongequal{\text{Step 4}} \frac{ r }{ 1 } =r \end{aligned} $$ |
| ⑧ | Remove the parentheses by changing the sign of each term within them. $$ - \left( 3r+3 \right) = -3r-3 $$ |
| ⑨ | Combine like terms: $$ \color{blue}{r} \color{red}{-3r} -3+ \color{red}{r} = \color{red}{-r} -3 $$ |