Mathematical and Physical Journal
for High Schools
Issued by the MATFUND Foundation
Already signed up?
New to KöMaL?
I want the old design back!!! :-)

Problem C. 1182. (October 2013)

C. 1182. Solve the following simultaneous equations:

\frac{x^2 +2xy +3y^2}{3x^2 + 2xy +y^2} + \frac{3x^2 + 2xy +y^2}{x^2 +2xy +3y^2} =2;

3x-2y=1.

(5 pont)

Deadline expired on November 11, 2013.


Sorry, the solution is available only in Hungarian. Google translation

Megoldás. Legyen \(\displaystyle a=\frac{x^2+2xy+3y^2}{3x^2+2xy+y^2}\). Az első egyenlet ekkor \(\displaystyle a+\frac 1a=2\), amiből \(\displaystyle a^2+1=2a\), vagyis \(\displaystyle a^2-2a+1=0\). Ez utóbbiból \(\displaystyle (a-1)^2=0\). Ebből pedig \(\displaystyle a=1\) következik.

Tehát \(\displaystyle x^2+2xy+3y^2=3x^2+2xy+y^2\), amiből \(\displaystyle 2y^2=2x^2\), vagyis \(\displaystyle y=\pm x\) következik. Ezt írjuk be a második egyenletbe:

I. eset: \(\displaystyle y=x\). Ekkor \(\displaystyle 3x-2x=1\), vagyis \(\displaystyle x=1\), \(\displaystyle y=1\).

II. eset: \(\displaystyle y=-x\). Ekkor \(\displaystyle 3x-2\cdot(-x)=1\), amiből \(\displaystyle x=1/5\), \(\displaystyle y=-1/5\).

Mindkét megoldás kielégíti az eredeti két egyenletet.


Statistics:

199 students sent a solution.
5 points:120 students.
4 points:43 students.
3 points:15 students.
2 points:2 students.
1 point:3 students.
0 point:3 students.
Unfair, not evaluated:13 solutions.

Problems in Mathematics of KöMaL, October 2013