1988 United States presidential election in Texas

1988 United States presidential election in Texas
Turnout	66.17% (of registered voters) ; 44.34% (of voting age population)
Bush 40–50% 50–60% 60–70% 70–80% 80–90%	Dukakis 40–50% 50–60% 60–70% 70–80% 80–90%
President before election; Ronald Reagan; Republican	Elected President ; George H. W. Bush; Republican

The 1988 United States presidential election in Texas took place on November 8, 1988. All fifty states and the District of Columbia, were part of the 1988 United States presidential election. Texas voters chose 29 electors to the Electoral College, which selected the president and vice president.

Incumbent Vice President George H. W. Bush won his home state against Massachusetts Governor Michael Dukakis. Bush ran with Indiana Senator Dan Quayle as Vice President, and Dukakis ran with Texas Senator Lloyd Bentsen.

Texas voted for Bush by a solid 12.6% margin. Texas had been considered a competitive, albeit Republican-leaning, state in the run-up to the election.^[2]

^ "Turnout and Voter Registration Figures (1970-current)".
^ Sentinel, Walter Manley II, Special to The. "NATION'S FATE RESTS WITH 7 STATES THE VOTE OF THE ELECTORAL COLLEGE IS THE ONLY ONE THAT COUNTS". OrlandoSentinel.com. Retrieved January 17, 2021.{{cite web}}: CS1 maint: multiple names: authors list (link)

[1] "Turnout and Voter Registration Figures (1970-current)".

[2] Sentinel, Walter Manley II, Special to The. "NATION'S FATE RESTS WITH 7 STATES THE VOTE OF THE ELECTORAL COLLEGE IS THE ONLY ONE THAT COUNTS". OrlandoSentinel.com. Retrieved January 17, 2021.{{cite web}}: CS1 maint: multiple names: authors list (link)

[1]

[2]