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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 \(\displaystyle x+y\le 0\), 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