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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 6x + 3 < 10.
x > 6/7
x < 7/6
x < 6/7
x > 7/6

 Q 2: Solve 7x - 9 < 3.
x < 12/7
x > 12/7
x > 7/12
x < 7/12

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 Question 4: This question is available to subscribers only!


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