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### Grade 7 - Mathematics6.35 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 7x - 9 < 3.x < 7/12x > 7/12x < 12/7x > 12/7 Q 2: Solve 5x - 15 < 5.x < 4None of thesex < -4x > 4 Q 3: Solve 9x + 2 < 15.x >9/13x < 9/13x < 13/9x > 13/9 Q 4: Solve 6x + 3 < 10.x > 6/7x > 7/6x < 6/7x < 7/6 Q 5: Solve 15x - 10 < 5.x > 1x < -1None of thesex < 1 Q 6: Solve 4x - 8 < 3.x > 11/4x < 4/11x > 4/11x < 11/4 Question 7: This question is available to subscribers only! Question 8: This question is available to subscribers only!