Linear Inequalities — Practice

Question 1

Solve: −3x + 9 > 0 and represent the solution on a number line.

Accepted Answer

−3x > −9. Divide by −3 (flip sign): x < 3. Solution: x < 3, i.e., (−∞, 3). Number line: open circle at 3, arrow pointing left.

Question 2

Solve: −5 ≤ 3x − 2 < 10 and write the solution in interval notation.

Accepted Answer

Split into two: (i) −5 ≤ 3x−2 → −3 ≤ 3x → −1 ≤ x. (ii) 3x−2 < 10 → 3x < 12 → x < 4. Intersection: −1 ≤ x < 4. Interval notation: [−1, 4).

Question 3

Solve graphically: 2x + 3y ≤ 12, x ≥ 0, y ≥ 0.

Accepted Answer

Step 1: Draw the line 2x+3y=12. x-intercept: (6,0); y-intercept: (0,4). Step 2: Line is solid (≤ includes equality). Step 3: Test (0,0): 2(0)+3(0)=0 ≤ 12 ✓, so shade the side containing the origin. Step 4: Also x≥0 (right of y-axis) and y≥0 (above x-axis). Step 5: The feasible region is the triangle with vertices (0,0), (6,0), and (0,4) — shaded.

Practice — Linear Inequalities

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