< 2 Subtract 2 from both sides to get 0 on the right: - 2 < 0 Get a common denominator of 2x. Multiply the 2 by - 2· < 0 - < 0 < 0 < 0 < 0 Factor the numerator < 0 Divide both sides by 3 < 0 Multiply both sides by 2 < 0 Find all critical values by setting numerator and denominator = 0 1-x=0 x=0 -x=-1 x=1 So the critical values are 0 and 1 Put these on a number line: ----------o--o--------- -3 -2 -1 0 1 2 3 4 test a value to the left of 0, say x=-1 Substitute it in < 0 < 0 < 0 < 0 -2 < 0 This is true so we shade the portion of the number line left of 0 < ==========o--o--------- -3 -2 -1 0 1 2 3 4 test a value between 0 and 1 say x=0.5 Substitute it in < 0 < 0 1 < 0 That is false so we do not shade the portion of the number line between 0 and 1. So we still have < ==========o--o--------- -3 -2 -1 0 1 2 3 4 We test a value to the right of 1, say x=2 Substitute it in < 0 < 0 < 0 < 0 This is true so we shade the portion of the number line right of 1 < =========o--o======== > -3 -2 -1 0 1 2 3 4 That's the graph of the solution set, in interval notation that is Edwin