1988 United States presidential election in New York

Ronald Reagan Republican George H. W. Bush Republican Pre-consolidation: Post-consolidation: Pre-consolidation: Post-consolidation: The 1988 United States presidential election in New York took place on November 8, 1988, as part of the 1988 United States presidential election.

New York was won by Democratic Governor Michael Dukakis of Massachusetts with 51.62% of the popular vote over Republican Vice President George H. W. Bush of Texas, who took 47.52%, a victory margin of 4.10%.

Dukakis’ statewide victory is largely attributable to winning four of five boroughs of New York City overall with 66.2% of the vote.

This was the last election in which a Republican presidential nominee won heavily populated Westchester County, as well as Monroe, Onondaga, and Ulster Counties,[2] and also the last election in which New York was decided by a single-digit margin.

Beginning in 1992, the Democrats would make substantial inroads in the suburbs around New York City as well as parts of upstate, making New York a solid blue state that has gone Democratic by double-digit margins in every election since.

International policy with the buckling Soviet Union was a critical component of the political landscape in the late 1980s. Vice President, President-Elect Bush can be seen here standing with United States President Ronald Reagan and Soviet General Secretary Mikhail Gorbachev , on the New York waterfront, 1988.