2018 Illinois Senate election

2018 Illinois Senate election

← 2016 November 6, 2018 (2018-11-06) 2020 →

39 of 59 seats in the Illinois Senate
30 seats needed for a majority
  Majority party Minority party
 
Leader John Cullerton Bill Brady
Party Democratic Republican
Leader's seat 6th-Chicago 44th-Bloomington
Last election 37 22
Seats won 40 19
Seat change Increase3 Decrease3
Popular vote 1,613,380 1,162,385
Percentage 57.45% 41.39%
Swing Decrease2.00% Increase0.84%

Results:
     Democratic gain
     Democratic hold      Republican hold
     No election
Vote Share:
     50–60%      60–70%      70–80%      >90%
     50–60%      60–70%      >90%

Senate President before election

John Cullerton
Democratic

Elected Senate President

John Cullerton
Democratic

The 2018 elections for the Illinois Senate took place on November 6, 2018 to elect senators from 39 of the state's 59 Senate districts to serve in the 101st General Assembly, with seats apportioned among the states based on the 2010 United States census. Under the Illinois Constitution of 1970, senators are divided into three groups, each group having a two-year term at a different part of the decade between censuses, with the rest of the decade being taken up by two four-year terms.[1] The Democratic Party has held a majority in the Senate since 2003, and gained a net of 3 seats.

The elections for Illinois's 18 congressional districts, Governor, statewide constitutional officers, and all 118 seats in the Illinois House of Representatives were also held on this date.

The Republicans needed to win eight seats in order to become the majority party. However, the Democratic Party picked up three additional seats and increased the party's supermajority to 40 seats.[2]

  1. ^ "Illinois Constitution Article IV, Section 2(a)". ilga.gov. Retrieved December 6, 2022.
  2. ^ Miller, Rich (November 20, 2018). "Sen. Connelly concedes". Capitol Fax. Retrieved November 24, 2018.

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