United States presidential election in Texas, 1988

United States presidential election in Texas, 1988

← 1984 November 8, 1988 1992 →
Nominee George H. W. Bush Michael Dukakis Party Republican Democratic Home state Texas Massachusetts Running mate Dan Quayle Lloyd Bentsen Electoral vote 29 0 Popular vote 3,036,829 2,352,748 Percentage 55.95% 43.35%
County Results   Dukakis—80-90%   Dukakis—70-80%   Dukakis—60-70%   Dukakis—50-60%   Dukakis—<50%   Bush—<50%   Bush—50-60%   Bush—60-70%   Bush—70-80%   Bush—80-90%
President before election Ronald Reagan Republican Elected President George H. W. Bush Republican

Ronald Reagan
Republican

George H. W. Bush
Republican

The 1988 United States presidential election in Texas took place on November 8, 1988. All 50 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.

Texas was won by incumbent United States Vice President George H. W. Bush of Texas, who was running against Massachusetts Governor Michael Dukakis. Bush ran with Indiana Senator Dan Quayle as Vice President, and Dukakis ran with Texas Senator Lloyd Bentsen.

Texas weighed in for this election as 2% more Republican than the national average.

The presidential election of 1988 was a very partisan election for Texas, with more than 99% of the electorate voting for either the Democratic or Republican parties. This is one of the last elections in Texas where you see many rural counties (including a large Democratic stronghold along the border with Louisiana) voting for the Democratic Party, and the populated counties, such as Houston's Harris County, voting mainly for the Republican candidate. Since the Presidential election of 1996, this demographic trend in Texas has reversed to its current form of 'urban Democrats' and 'rural Republicans' in the state of Texas, and elsewhere in the United States.