Question
Answer
$$n\dfrac{(k-2)n-(k-4)}{2}=\dfrac{kn^2-kn-2n^2+4n}{2}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}n\frac{(k-2)n-(k-4)}{2}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}n\frac{kn-2n-(k-4)}{2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}n\frac{kn-2n-k+4}{2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{kn^2-kn-2n^2+4n}{2}\end{aligned} $$ | |
| ① | $$ \left( \color{blue}{k-2}\right) \cdot n = kn-2n $$ |
| ② | Remove the parentheses by changing the sign of each term within them. $$ - \left( k-4 \right) = -k+4 $$ |
| ③ | Multiply $n$ by $ \dfrac{kn-2n-k+4}{2} $ to get $ \dfrac{kn^2-kn-2n^2+4n}{2} $. Step 1: Write $ n $ as a fraction by putting $ \color{red}{1} $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} n \cdot \frac{kn-2n-k+4}{2} & \xlongequal{\text{Step 1}} \frac{n}{\color{red}{1}} \cdot \frac{kn-2n-k+4}{2} \xlongequal{\text{Step 2}} \frac{ n \cdot \left( kn-2n-k+4 \right) }{ 1 \cdot 2 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ kn^2-2n^2-kn+4n }{ 2 } = \frac{kn^2-kn-2n^2+4n}{2} \end{aligned} $$ |
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