Questions: Linear Inequalities
Linear Inequalities24 min readLesson 4 of 5
Q1. Solve 3x - 5 < 7 for real x.
Answer: Add 5 to both sides to get 3x less than 12. Divide by the positive 3 without changing the sign to get x less than 4. The solution set is the open interval from minus infinity to 4.
Hinglish: Dono taraf 5 add karo: 3x < 12. Positive 3 se divide karo, sign same rahega: x < 4. Solution set (-infinity, 4).
Q2. Solve -2x + 7 ≥ 1 for real x.
Answer: Subtract 7 from both sides to get minus 2x at least minus 6. Now divide by the negative number minus 2 and flip the sign: x at most 3. The solution set is the closed interval from minus infinity to 3.
Hinglish: 7 subtract karo: -2x >= -6. Ab -2 se divide karo aur sign palto: x <= 3. Solution set (-infinity, 3].
Q3. Solve 5x - 3 ≤ 3x + 1 for real x and show the solution on a number line.
Answer: Subtract 3x from both sides to get 2x minus 3 at most 1, then add 3 to get 2x at most 4, so x at most 2. On the number line, put a filled circle at 2 and shade to the left with an arrow.
Hinglish: 3x subtract karo: 2x - 3 <= 1, phir 2x <= 4, yaani x <= 2. Number line par 2 par filled circle aur left side shading.
Q4. Solve 4x + 3 < 6x + 7.
Answer: Bring variable terms to one side: 4x minus 6x less than 7 minus 3, so minus 2x less than 4. Dividing by minus 2 flips the sign, giving x greater than minus 2. The solution set is the open interval from minus 2 to infinity.
Hinglish: 4x - 6x < 7 - 3, yaani -2x < 4. Minus 2 se divide karte waqt sign palto: x > -2. Solution set (-2, infinity).
Q5. Solve the double inequality -3 ≤ 2x + 1 ≤ 7.
Answer: Apply the same operation to all three parts. Subtract 1 throughout to get minus 4 at most 2x at most 6. Divide by the positive 2 to get minus 2 at most x at most 3. The solution set is the closed interval from minus 2 to 3.
Hinglish: Teeno parts par same operation karo: 1 subtract karo to -4 <= 2x <= 6. 2 se divide karo to -2 <= x <= 3. Solution set [-2, 3].
Q6. Solve the system x + 3 > 2 and 2x - 1 < 9.
Answer: The first inequality gives x greater than minus 1. The second gives 2x less than 10, so x less than 5. Both must hold, so the solution is the intersection, the open interval from minus 1 to 5. Testing x equal to 0 satisfies both, and testing x equal to 6 fails the second, which confirms the overlap rather than a union.
Hinglish: Pehli se x > -1, doosri se x < 5. Dono saath chahiye, to intersection (-1, 5) hai.
Q7. Ravi scored 70 and 75 in two tests. What minimum score in the third test gives an average of at least 60?
Answer: Let the third score be x. The average condition is (70 + 75 + x)/3 at least 60. Multiplying by 3 gives 145 plus x at least 180, so x at least 35. The minimum score is 35. Since marks are counted in whole numbers, 35 is attainable and is the least value; any smaller score would pull the average below 60, which the check (70 + 75 + 34)/3 confirms.
Hinglish: Average ki condition: (70 + 75 + x)/3 >= 60. 3 se multiply karo to 145 + x >= 180, isliye x >= 35. Minimum score 35.
Q8. Two consecutive positive odd integers, both below 10, add up to more than 12. List the possible pairs.
Answer: Let the smaller odd integer be x and the next be x plus 2. Both are positive, the larger is below 10, and the sum exceeds 12. The options are (1, 3), (3, 5), (5, 7) and (7, 9); the sums are 4, 8, 12 and 16. The sum must be strictly more than 12, so the pair (5, 7) is out and only (7, 9) survives.
Hinglish: Possible pairs (1,3), (3,5), (5,7), (7,9) ke sums 4, 8, 12, 16 hain. Sum 12 se strictly zyada chahiye, isliye sirf (7, 9) bachta hai.
Q9. Solve (x - 1)/2 ≤ (x + 2)/3 for real x.
Answer: Multiply both sides by the positive 6 to clear the denominators, giving 3(x minus 1) at most 2(x plus 2). Expanding gives 3x minus 3 at most 2x plus 4, so x at most 7. The solution set is the closed interval ending at 7. A quick check with x equal to 7 satisfies the original, while x equal to 8 fails, so the boundary is correct and it is included.
Hinglish: Positive 6 se multiply karo: 3(x - 1) <= 2(x + 2). Expand karo: 3x - 3 <= 2x + 4, isliye x <= 7. Solution set (-infinity, 7].
Q10. A triangle has its longest side three times the shortest and its third side 2 cm shorter than the longest. If the perimeter is at least 61 cm, find the minimum shortest side.
Answer: Let the shortest side be x. Then the longest is 3x and the third side is 3x minus 2. The perimeter is 7x minus 2, which must be at least 61, so 7x at least 63 and x at least 9. The minimum shortest side is 9 cm. Checking the triangle inequality with x equal to 9 gives sides 9, 27 and 25, and 9 plus 25 is greater than 27, so the triangle is valid. For a smaller x such as 8 the perimeter is 54, which fails the bound, so 9 is indeed the least possible value in whole centimetres.
Hinglish: Shortest x, longest 3x, teesri side 3x - 2. Perimeter 7x - 2 >= 61 se x >= 9. Minimum shortest side 9 cm. Triangle inequality check karo: 9 + 25 > 27 sahi hai.
Common mistakes on this chapter
Flipping the sign only sometimes is the first error: the flip is needed exactly when multiplying or dividing by a negative number. Ignoring the universe word is the second. Writing an open endpoint for a slack inequality is the third. Missing the overlap in a system is the fourth. Read the final answer once with these four checks.
Quiz
Quiz below: five multiple choice questions on solving rules, solution sets, number lines and word problems. Attempt them after the notes above, then move to the previous year page.