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Middle/High School Algebra, Geometry, and Statistics (AGS)
2.11 Solving Inequations - I

Example: Solve 3x + 2 < 10.
Solution: Given that 3x + 2 < 10
Adding -2 to both sides.
3x + 2 - 2 < 10 - 2
3x < 8
Dividing both sides by 3, we get
3x/3 < 8/3
x < 8/3
\ Any number < 8/3 is a solution.

Verification: Substituting x = 1 < 8/3 in the equation, we get
L.H.S. = 3x + 2 = 3 * 1 + 2 = 3 + 2 = 5 < 10 = R.H.S.
\ 1 < 8/3 is a root.
Similarly we can show that any number less than 8/3 is a root.

Directions: Solve the following inequality. Also write at least ten examples of your own.
Q 1: Solve 9x + 2 < 15.
x < 9/13
x > 13/9
x < 13/9
x >9/13

 Q 2: Solve 6x - 9 < 5.
x > 3/7
x > 7/3
x < 7/3
x < 3/7

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