Wiki-Quellcode von Lösung Lösen von linearen Gleichungen
Zuletzt geändert von Stephanie Wietzorek am 2025/11/17 10:44
Verstecke letzte Bearbeiter
| author | version | line-number | content |
|---|---|---|---|
 |
1.1 | 1 | (% style="width: 100%; white-space: nowrap" class="border" %) |
| 2 | |= Gleichung |= Lösungsmenge | ||
| 3 | | 1) {{formula}}2x - 13 + 6x = 5x + 8{{/formula}} | L ={7} | ||
| 4 | | 2) {{formula}}7,3y + 5 - 2,5y - 2,8 = 6,5y - 3,2 - 1,7y + 5,4{{/formula}} | L = {{formula}}\mathbb{R}{{/formula}} | ||
| |
1.2 | 5 | | 3) {{formula}}-0,5 (3(a+2) - 5(a-2)) = a - 4{{/formula}} | L ={ } |
| 6 | | 4) {{formula}}-(-4x) + 16x = -5x + 5{{/formula}} | L ={0,2} | ||
| 7 | | 5) {{formula}}-3a + 1,25 = -1 - a{{/formula}} | L ={1,125} | ||
| 8 | | 6) {{formula}}2(0,5x + 1,5) + 0,5x = 10,5{{/formula}} | L ={5} | ||
| 9 | | 7) {{formula}}0,2 (y-2) - 3 = -1,5y{{/formula}} | L = {2} | ||
| |
2.1 | 10 | | 8) {{formula}}\frac{1}{3}(x - 2) = \frac{1}{2}x{{/formula}} | L = {-4} |
| 11 | | 9) {{formula}}3 + \frac{1}{2}b + \frac{1}{3}b - 2b = 4 + \frac{1}{6}b{{/formula}} | L = { {{formula}}-\frac{3}{4}{{/formula}} } |