Skillnad mellan versioner av "3.3 Lösning 1a"
Från Mathonline
Taifun (Diskussion | bidrag) m |
Taifun (Diskussion | bidrag) m |
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| Rad 1: | Rad 1: | ||
| − | ::<math>\begin{array}{rclcl} x \, + \, (x \, + \, 6) & = & 12 | + | ::<math>\begin{array}{rclcl} x \, + \, (x \, + \, 6) & = & 12 & & \\ |
| − | x \, + \, x \, + \, 6 & = & 12 | + | x \, + \, x \, + \, 6 & = & 12 & & \\ |
| − | 2\,x \, + \, 6 & = & 12 | + | 2\,x \, + \, 6 & = & 12 & \qquad | & - \, 6 \\ |
| − | + | 2\,x \, + \, 6 \, - \, 6 & = & 12 \, - \, 6 & & \\ | |
| − | 2\,x \, & = & 6 | + | 2\,x \, & = & 6 & \qquad | & / \; 2 \\ |
| − | + | \displaystyle \frac{2\,x}{2} & = & \displaystyle \frac{6}{2} & & \\ | |
| − | x \, & = & 3 | + | x \, & = & 3 & & |
\end{array}</math> | \end{array}</math> | ||
Versionen från 14 maj 2016 kl. 17.44
- \[\begin{array}{rclcl} x \, + \, (x \, + \, 6) & = & 12 & & \\ x \, + \, x \, + \, 6 & = & 12 & & \\ 2\,x \, + \, 6 & = & 12 & \qquad | & - \, 6 \\ 2\,x \, + \, 6 \, - \, 6 & = & 12 \, - \, 6 & & \\ 2\,x \, & = & 6 & \qquad | & / \; 2 \\ \displaystyle \frac{2\,x}{2} & = & \displaystyle \frac{6}{2} & & \\ x \, & = & 3 & & \end{array}\]
