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NCERT Exemplar · Q16

Q.If −3x+17<−13-3x + 17 < -13, then
(A) x∈(10,∞)x \in (10, \infty)
(B) x∈[10,∞)x \in [10, \infty)
(C) x∈(−∞,10]x \in (-\infty, 10]
(D) x∈[−10,10)x \in [-10, 10)

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Isolate xx by reversing operations in order, remembering that dividing by a negative flips the inequality sign. The solution is x∈(10,∞)x \in (10, \infty).

Understanding Linear Inequalities

When we solve a linear inequality, we're finding all values of xx that make the statement true. The process mirrors equation-solving—undo operations to isolate the variable—with one critical difference: multiplying or dividing both sides by a negative number reverses the inequality direction.

The inequality −3x+17<−13-3x + 17 < -13 asks: for which xx does this relationship hold? We'll manipulate it step by step, maintaining the truth of the statement.

Solution

1. Isolate the term containing xx

Subtract 1717 from both sides:

−3x+17−17<−13−17-3x + 17 - 17 < -13 - 17

−3x<−30-3x < -30

2. Solve for xx by dividing by the coefficient

The coefficient of xx is −3-3. When we divide both sides by −3-3, we must reverse the inequality sign:

−3x−3>−30−3\frac{-3x}{-3} > \frac{-30}{-3}

x>10x > 10

Watch out

The most common error here is forgetting to flip the inequality when dividing by −3-3. If you wrote x<10x < 10, you'd have the opposite of the correct answer. …

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