United States presidential election in New Mexico, 1988
|
||||||||||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
|
||||||||||||||||||||||||||
|
||||||||||||||||||||||||||
.jpg)
County Results
Dukakis—70-80%
Dukakis—60-70%
Dukakis—50-60%
Bush—<50%
Bush—50-60%
Bush—60-70%
|
||||||||||||||||||||||||||
|
||||||||||||||||||||||||||
The 1988 United States presidential election in New Mexico took place on November 8, 1988. All 50 states and the District of Columbia, were part of the 1988 United States presidential election. New Mexico voters chose 5 electors to the Electoral College, which selected the President and Vice President.
New Mexico 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.
New Mexico weighed in for this election as 2% more Democratic than the national average.
The presidential election of 1988 was a very partisan election for New Mexico, with 98% of the electorate voting for either the Democratic or Republican parties. This is one of the last Presidential elections in New Mexico in which you see the highly populated counties of Bernalillo and Los Alamos turning out primarily Republican. The Democratic stronghold in the northern part of the state is evident during this election, and remains largely intact from this point onward to current political times.
...
Wikipedia