| y | 3 | |||
x= | + | |||
| 2 | 2 |
| y | 3 | |||
| | + | |=1−y | ||
| 2 | 2 |
| y | 3 | y | 3 | y | 3 | |||||||
1. | + | ≥0 ⇒ y≥−3 ⇔ | | + | |= | + | |||||||
| 2 | 2 | 2 | 2 | 2 | 2 |
| y | 3 | ||
+ | =1−y /*2 | ||
| 2 | 2 |
| y | 3 | y | 3 | y | 3 | |||||||
2. | + | <0 ⇒ y<−3 ⇔ | | + | |=− | − | |||||||
| 2 | 2 | 2 | 2 | 2 | 2 |
| y | 3 | |||
− | − | =1−y /*2 | ||
| 2 | 2 |
| ⎧ | |x|=1−y | |
| ⎩ | 2x−y=3 |
| ⎧ | x=1−y dla x≥0 | |
| ⎩ | 2(1−y)−y=3 |
| ⎧ | x=y−1 dla x<0 | |
| ⎩ | 2(y−1)−y=3 |
podaję rozwiązanie:
1/ dla x≥0 2/ x <0
x= 1−y −x= 1−y
i 2x −y= 3 i 2x −y=3
x+y=1 −x+y=1
2x −y=3 2x −y= 3
−−−−−−−−− −−−−−−−−−
3x = 4 x= 4 −−− odpada bo nie jest < 0
brak rozwiązań
x= 113 −− jest >0
to: y= 1 −113 = −13
odp: x= 113 , y= −13
| 1 | ||
2 − 2y − y = 3 => −3y = 1 => y = − | ||
| 3 |
| 1 | 1 | |||
x = 1 + | = 1 | |||
| 3 | 3 |
| 1 | 1 | |||
Odp: x = 1 | , y = − | |||
| 3 | 3 |