Rational & Modulus Inequalities: Theory & Concepts

Rational inequalities:
The inequalities of the form (ax+b/cx+d) < k or (ax+b/cx+d) > k are called rational inequalities where ax + b and cx + d are linear algebraic expressions. In order to solve rational inequalities the following steps should be followed.
  • Make the right hand side equal to zero by transposing the constant term to the left side.
  • Solve the left hand side so that it becomes of the form mx+1/nx+p <0 or mx+1/nx+p >0 Make sure that ‘m’ and ‘n’ which are the coefficients of ‘x’ in the numerator and denominator are positive.
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Put the numerator and denominator equal to zero to get the critical points. Put these points on the number line. The number line will be divided in three parts. The rightmost part will give solution for the positive inequalities, the middle part will give solution for the negative inequalities and the leftmost part will give solution for the positive inequalities.
Let us solve some rational inequality problems:
Illustration 1: Solve 3x+5/5x-2<0
Solution: We have 3x+5/5x-2>0
Here the right hand side is already zero. So here we do not need to do any calculations. To get the critical points, put the numerator and denominator equal to zero.
We have 3x + 5 = 0 ⇒ x=(-5/3) and 5x – 2 = 0 ⇒ x = 2/5
Plot these points on the number line.
Since the given inequality is negative, so the solution is (-5/3)< x < (2/5)
Modulus inequalities or Absolute value inequalities
Inequalities of the form |ax + b| < k or |ax + b| > k are called the modulus or absolute inequalities.
To solve these inequalities, keep the following rules in mind:
  • If |x| < a, then – a < x < a
  • If |x| > a, then either x > a or x < - a
  • If |x – l| < a, then l – a < x < l + a
  • If |x – l| > a, then either x > l + a or x < l – a.
Let us try some modulus inequality questions:
Illustration 2: Solve |x – 3| < 5.
Solution: We have |x – 3| < 5
⇒ - 5 < x – 3 < 5 ( If |x| < a, then – a < x < a)
⇒ 3 – 5 < x < 3 + 5 (Same number can be added on both sides of the inequality)
⇒ -2 < x < 8 which is the required solution.
Illustration 3: Solve |8x + 5| < 9.
Solution: We have |8x + 5| < 9.
⇒ - 9 < 8x + 5 < 9 ( If |x| < a, then – a < x < a)
⇒ - 5 – 9 < 8x < 9 – 5 (Same number can be added on both sides of the inequality)
⇒ - 14 < 8x < 4
⇒ -14/8 < x < 4/8
⇒ -7/4 < x < 1/2 which is the required solution.
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