Fdfdfd
1) Ecuaciones lineales con una incógnita:
a) 2x + 1 = 7
2x = 7 – 1
2x = 6
x = 3
Comprobamos: sustituimos el valor hallado de x enla ecuación original: 2(3) + 1 = 7
6 + 1 = 7
→ 7 = 7 identidad.
b) 6 ( x – ½ ) = 2x – 1
6x – 6(1/2) = 2x – 1
6x – 3 = 2x – 1
6x – 2x = -1 + 3
4x = 2
x = ½Comprobamos: sustituimos el valor hallado de x en la ecuación original: 6(1/2 – ½) = 2(1/2) – 1
(6/2 – 6/2) = 2/2 – 1
→ 0 = 0 identidad.
c) 5 ( x + 1 ) - x = 4x + 15
5 ( x + 1 ) - x = 4x + 155x + 5 – x = 4x + 15
( 5 – 1 ) x +5 = 4x + 15
4x + 5 = 4x + 15
0 x = 10
x = 10 / 0 La ecuación no tiene solución.
En las ecuaciones dadas compruebe el resultado:
|a) |[pic]|
| [pic] |= |[pic] |
|[pic] |= |[pic] |
|[pic] |= |[pic] |
|b) |[pic] |
|[pic]|= |[pic] |
|[pic] |= |[pic] |
|[pic] |= |[pic] |
|[pic] |= |[pic]|
|[pic] |= |[pic] |
|[pic] |= |[pic] |
| |[pic] |
| | |
| ||
| | |
|c) | |
| | |
| | |
|[pic] |= |[pic] ||[pic] |= |[pic] |
|[pic] |= |[pic] |
|[pic] |= |[pic] |
|[pic] |= |[pic] |
|[pic]|= |[pic] |
|d) |[pic] |
|[pic] |= |[pic] |
|[pic] |= |[pic] |
|[pic] |= |[pic]...
Regístrate para leer el documento completo.