Problem C. 1067. (February 2011)
C. 1067. Solve the following simultaneous equations on the set of real numbers: |x-1|+|x+y|=6, |y-1|+|x+y+1|=4.
(5 pont)
Deadline expired on March 10, 2011.
Sorry, the solution is available only in Hungarian. Google translation
Megoldás. Bontsuk fel az abszolút értékeket a változók lehetséges értékei mellett. Megjegyezzük, hogy ha \(\displaystyle x+y\ge 0\), akkor \(\displaystyle x+y+1>0\), továbbá, ha \(\displaystyle x, \ y\ge 1\), akkor \(\displaystyle x+y>0\). Ezért a különböző felbontások szerint 10 esetünk lesz. A táblázatunkban ++, -+, – jelölje az előjeleit az \(\displaystyle x+y\), \(\displaystyle x+y+1\) kifejezéseknek (ebben a sorrendben). Az egyenletrendszer megoldása után ellenőrizzük \(\displaystyle x+y\) és \(\displaystyle x+y+1\) előjelét.
\(\displaystyle x\ge 1\) | \(\displaystyle x<1\) | ||
++ | \(\displaystyle 2x+y=7\), \(\displaystyle x+2y=4\): \(\displaystyle x=10/3\), \(\displaystyle y=1/3\)(!) | \(\displaystyle 1-x+x+y=5\): \(\displaystyle y=5\), \(\displaystyle x+2y=4\): \(\displaystyle x=-6\)(!) | |
\(\displaystyle y\ge 1\) | -+ | \(\displaystyle 1-x-x-y=6\), \(\displaystyle x+2y=4\):\(\displaystyle x=-14/3\), \(\displaystyle y=13/3\) | |
– | \(\displaystyle 1-x-x-y=6\), \(\displaystyle y-1-x-y-1=4\): \(\displaystyle x=-6\), \(\displaystyle y=7\)(!) | ||
++ | \(\displaystyle 2x+y=7\), \(\displaystyle 1-y+x+y+1=4\): \(\displaystyle x=2\), \(\displaystyle y=3\)(!) | \(\displaystyle 1-x+x+y=6\), \(\displaystyle 1-y+x+y+1=4\): \(\displaystyle x=2\)(!), \(\displaystyle y=5\)(!) | |
\(\displaystyle y< 1\) | -+ | \(\displaystyle x-1-x-y=6\), \(\displaystyle 1-y+x+y+1=4\): \(\displaystyle x=2\), \(\displaystyle y=-7\)(!) | \(\displaystyle 1-x-x-y=6\), \(\displaystyle 1-y+x+y+1=4\): \(\displaystyle x=2\)(!) |
– | \(\displaystyle x-1-x-y=6\), \(\displaystyle 1-y-x-y-1=4\): \(\displaystyle x=10\), \(\displaystyle y=-7\)(!) | \(\displaystyle 1-x-x-y=6\), \(\displaystyle 1-y-x-y-1=4\): \(\displaystyle x=-2\), \(\displaystyle y=-1\) |
Két megoldása van az egyenletrendszernek: \(\displaystyle x_1=-\frac{14}3,\ y_1=\frac{13}3\) és \(\displaystyle x_2=-2,\ y_2=-1\).
Statistics:
104 students sent a solution. 5 points: 53 students. 4 points: 3 students. 3 points: 17 students. 2 points: 12 students. 1 point: 11 students. 0 point: 7 students. Unfair, not evaluated: 1 solutions.
Problems in Mathematics of KöMaL, February 2011