– 2x < 4 or x > – 2
i.e., all the real numbers which are greater than –2, are the solutions of the given inequality. Hence, the solution set is (–2, ∞).
The marks obtained by a student of Class XI in first and second terminal examination are 62 and 48, respectively. Find the number of minimum marks he should get in the annual examination to have an average of at least 60 marks.
Let x be the marks obtained by student in the annual examination.
(62+48+x)/3 ≥ 60
The student must obtain a minimum of 70 marks to get an average of at least 60 marks.
3x + 2y > 6 graphically.
This line divides the xy-plane in two half planes I and II. We select a point (not on the line), say (0, 0), which lies in one of the half planes (Fig 6.8) and determine if this point satisfies the given inequality, we note that
3 (0) + 2 (0) > 6
0 > 6, which is false.
Hence, half plane I is not the solution region of the given inequality. Clearly, any point on the line does not satisfy the given strict inequality. In other words, the shaded half plane II excluding the points on the line is the solution region of the inequality.