The American easily won 6-1, 6-3
Xinyu Wang came from behind and defeated Avanesyan 1-6, 6-2 and 6-3, while Udvardy beat Niemeier 7-6 (1) and 7-6 (5)
The American tennis player Sloane Stephenssecond favourite, won this Tuesday 6-1 y 6-3 French Leolia Jeanjean, in the first round of the Mérida Open, a Mexican WTA 250 tournament that takes place on a hard court. Stephens, 41 in the world, will face the winner between the German Tatjana Maria and the Russian Varvara Gracheva in the next round. The 2017 US Open champion broke her opponent’s serve twice in the first set and saved her own twice to win it 6-1. The North American was accurate, winning 77.8% of the points on the first serve and 63.6% of the second returns.
In the second set, the Frenchwoman was more competitive, however she succumbed to the aggressive tennis of Stephens, who broke her rival’s serve two more times and hit two aces. Jeanjean committed two double faults and missed returns, before a Stephens who was supported by the Mexican fans in the main stadium of the Mérida Open. “I’m happy with the way I played, obviously the conditions favored me. I enjoy playing in places with high humidity. I love playing in Mexico, it’s always a good experience,” Stephens explained after the victory.
In another result, the Chinese Xinyu Wang came back and defeated the Russian Elina Avanesyan 1-6, 6-2 and 6-3 in three sets, to advance to the next phase in which she will face the Italian Elisabetta Cocciaretto, seventh. Classified in Merida. Similarly, the Hungarian Panna Udvardy beat the German Jule Niemeier 7-6 (1) and 7-6 (5). Udvardy will challenge the winner later in the duel between the Colombian Camila Osorio who will seek to eliminate the Polish Magda Linette, the first favorite to win the tournament in southern Mexico.
The Argentine Nadia Podoroska will also challenge the Swedish Rebecca Peterson and the Mexican Fernanda Contreras the American Alycia Parks, in the contest that distributes a bag of 259,303 dollars.