Women's doubles
2019 Launceston Tennis International
Champions	Chang Kai-chen Hsu Ching-wen
Runners-up	Alexandra Bozovic Isabelle Wallace
Score	6–2, 6–4

Jessica Moore and Ellen Perez were the defending champions, but Moore chose to participate at the 2019 St. Petersburg Ladies' Trophy instead. Perez partnered alongside compatriot Arina Rodionova, but lost in the semifinals to Chang Kai-chen and Hsu Ching-wen.^[1]

Chang and Hsu won the title, defeating Alexandra Bozovic and Isabelle Wallace in the final, 6–2, 6–4.

Seeds

Draw

Draw

	First round							Quarterfinals							Semifinals							Final
	1	E Perez Ar Rodionova	6	5	[11]
		M Inglis K McPhee	1	7	[9]			1	E Perez Ar Rodionova	3	6	[12]
	WC	A Berezov I Bozicevic	2	0					N Podoroska Xiy Wang	6	4	[10]
		N Podoroska Xiy Wang	6	6											1	E Perez Ar Rodionova	3	2
	3	L Cabrera M Zanevska	5	6	[7]											K-c Chang C-w Hsu	6	6
		K-c Chang C-w Hsu	7	1	[10]				K-c Chang C-w Hsu	77	6
		G García Pérez E Rybakina	6	6					G García Pérez E Rybakina	65	3
	WC	M Jovanovic A Parnaby	2	2																			K-c Chang C-w Hsu	6	6
		S Akita M Kobori	4	6	[11]																		A Bozovic I Wallace	2	4
		M Minokoshi E Sema	6	4	[13]				M Minokoshi E Sema	3	5
		A Bozovic I Wallace	7	0	[10]				A Bozovic I Wallace	6	7
	4	N Hibino K Okamura	5	6	[7]											A Bozovic I Wallace	77	7
		D Aiava N Bains	6	6												D Aiava N Bains	65	5
	WC	G Da Silva-Fick T Patterson	3	3					D Aiava N Bains	w/o
		A Bai G Taylor	2	2				2	I Bara E Gorgodze
	2	I Bara E Gorgodze	6	6

References

• Main Draw