Moldova’s best tennis player Radu Albot qualified for the second preliminary round at the Australian Open, the first Grand Slam tournament of the year. In the morning of January 9, Radu Albot beat South Africa’s Rik De Voest in three sets. He lost the first set 4-6, but then won the next sets 7-6 and 6-3, Info-Prim Neo reports. Radu Albot’s next opponent in the qualifiers will be Amir Weintraub from Israel, who defeated American Wayne Odesnik in the inaugural round. Radu Albot must go through three qualifying rounds. In the 2012 edition, the Moldovan tennis player was eliminated in the second qualifying round. Albot comes 220th in the world ATP rankings.