+26387 Are there two ways to solve the problem "4n-7+2n=7"?
Second way to solve:
\(\small{ \begin{array}{rcl} 4n-7+2n &=& 7 \\ 4n - 7 + 2n - 7 &=& 0 \\ 2n + 2n -7+2n-7 &=& 0 \\ 2n + (2n-7)+(2n-7) &=& 0 \\ 2n + 2(2n-7) &=& 0 \\ 2(2n-7) &=& - 2n \\ 2n-7 &=& -n \\ 3n &=& 7 \\ \mathbf{n} & \mathbf{=} & \mathbf{ \dfrac{7}{3} } \end{array} } \)
+6251 Not really. However you slice it you're going to end up with
\(6n = 14\\ n= \dfrac {14}{6} = \dfrac 7 3\)
+26387 Are there two ways to solve the problem "4n-7+2n=7"?
Second way to solve:
\(\small{ \begin{array}{rcl} 4n-7+2n &=& 7 \\ 4n - 7 + 2n - 7 &=& 0 \\ 2n + 2n -7+2n-7 &=& 0 \\ 2n + (2n-7)+(2n-7) &=& 0 \\ 2n + 2(2n-7) &=& 0 \\ 2(2n-7) &=& - 2n \\ 2n-7 &=& -n \\ 3n &=& 7 \\ \mathbf{n} & \mathbf{=} & \mathbf{ \dfrac{7}{3} } \end{array} } \)
