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}{ | + | ::<math>\begin{array}{rclcl} x \, + \, (x \, + \, 14) & = & 18 & & \\ |
| − | + | x \, + \, x \, + \, 14 & = & 18 & & \\ | |
| − | + | 2\,x \, + \, 14 & = & 18 & \qquad | & {\color{Red} {- \, 14}} \\ | |
| − | + | 2\,x \, + \, 14 \, {\color{Red} {- \, 14}} & = & 18 \, {\color{Red} {- \, 14}} & & \\ | |
| − | + | 2\,x \, & = & 4 & \qquad | & {\color{Red} {/ \; 2}} \\ | |
| − | \end{array}</math> | + | \displaystyle \frac{2\,x}{{\color{Red} {2}}} & = & \displaystyle \frac{4}{{\color{Red} {2}}} & & \\ |
| − | + | x \, & = & 2 & & | |
| − | + | \end{array}</math> | |
Versionen från 14 maj 2016 kl. 17.25
- \[\begin{array}{rclcl} x \, + \, (x \, + \, 14) & = & 18 & & \\ x \, + \, x \, + \, 14 & = & 18 & & \\ 2\,x \, + \, 14 & = & 18 & \qquad | & {\color{Red} {- \, 14}} \\ 2\,x \, + \, 14 \, {\color{Red} {- \, 14}} & = & 18 \, {\color{Red} {- \, 14}} & & \\ 2\,x \, & = & 4 & \qquad | & {\color{Red} {/ \; 2}} \\ \displaystyle \frac{2\,x}{{\color{Red} {2}}} & = & \displaystyle \frac{4}{{\color{Red} {2}}} & & \\ x \, & = & 2 & & \end{array}\]
