Exercise 5.1 answers: Linear Inequalities

Open NCERT p. 95Matches NCERT’s answer

(i) x is a natural number.

1. 24x < 100. Divide both sides by 24 — 24 is positive, so the inequality sign does not change.
2. x < 100/24, which simplifies to x < 25/6 (that is 4 and 1/6).
3. Natural numbers are 1, 2, 3, 4, ... — keep only the ones below 4 1/6.

Answerx = 1, 2, 3 or 4.

(ii) x is an integer.

1. The same working gives x < 25/6, i.e. x < 4 1/6.
2. Integers include negative numbers, zero and positive numbers, with no smallest one.
3. Every integer up to 4 satisfies x < 4 1/6.

Answerx can be any integer up to 4: ..., 1, 2, 3, 4 (no smallest value).

Watch this explained “Reading the answer off”, 12:05 into Why testing values one by one is not a method, and what replaces it

Open NCERT p. 95Matches NCERT’s answer

(i) x is a natural number.

1. −12x > 30. Divide both sides by −12.
2. Dividing (or multiplying) an inequality by a negative number flips the sign, so > becomes <.
3. x < 30/(−12) = −5/2, i.e. x < −2 1/2.
4. Natural numbers start at 1 and only go up, so none of them can be less than −2 1/2.

AnswerNo natural number works — the solution set is empty.

(ii) x is an integer.

1. The same working gives x < −2 1/2.
2. Integers less than −2 1/2 are ..., −5, −4, −3.
3. There is no smallest one; the list continues forever downward.

Answerx can be any integer less than or equal to −3: ..., −5, −4, −3.

Watch this explained “The answer that is empty”, 13:13 into Why testing values one by one is not a method, and what replaces it

Open NCERT p. 95Matches NCERT’s answer

(i) x is an integer.

1. 5x − 3 < 7. Add 3 to both sides: 5x < 10.
2. Divide both sides by 5 — 5 is positive, so the sign stays the same: x < 2.
3. Integers less than 2 are ..., −1, 0, 1.

Answerx can be any integer less than 2: ..., −1, 0, 1 (no smallest value).

(ii) x is a real number.

1. The same working gives x < 2.
2. Every real number smaller than 2 satisfies this, and there is no smallest one.

Answerx can be any real number less than 2, written as the interval (−∞, 2).

Watch this explained “Reading the answer off”, 12:05 into Why testing values one by one is not a method, and what replaces it

Open NCERT p. 95Matches NCERT’s answer

(i) x is an integer.

1. 3x + 8 > 2. Subtract 8 from both sides: 3x > −6.
2. Divide both sides by 3 — 3 is positive, so the sign stays the same: x > −2.
3. Integers greater than −2 are −1, 0, 1, 2, ...

Answerx can be any integer greater than −2: −1, 0, 1, 2, ... (no largest value).

(ii) x is a real number.

1. The same working gives x > −2.
2. Every real number greater than −2 satisfies this, with no largest one.

Answerx can be any real number greater than −2, written as the interval (−2, ∞).

Watch this explained “Reading the answer off”, 12:05 into Why testing values one by one is not a method, and what replaces it

Open NCERT p. 95Matches NCERT’s answer

1. Subtract 4x from both sides (subtracting never flips an inequality): 3 < x + 7.
2. Subtract 7 from both sides: −4 < x, i.e. x > −4.

Answerx > −4, i.e. the interval (−4, ∞).

Watch this explained “The same answer twice”, 9:15 into The one rule that differs from equation-solving, and the reason it differs

Open NCERT p. 95Matches NCERT’s answer

1. Subtract 3x from both sides: −7 > 2x − 1.
2. Add 1 to both sides: −6 > 2x.
3. Divide both sides by 2 — 2 is positive, so the sign stays the same: −3 > x, i.e. x < −3.

Answerx < −3, i.e. the interval (−∞, −3).

Watch this explained “The same answer twice”, 9:15 into The one rule that differs from equation-solving, and the reason it differs

Open NCERT p. 95Matches NCERT’s answer

1. Multiply out both brackets: 3x − 3 ≤ 2x − 6.
2. Subtract 2x from both sides: x − 3 ≤ −6.
3. Add 3 to both sides: x ≤ −3.

Answerx ≤ −3, i.e. the interval (−∞, −3].

Watch this explained “Reading the answer off”, 12:05 into Why testing values one by one is not a method, and what replaces it

Open NCERT p. 95Matches NCERT’s answer

1. Multiply out both brackets: 6 − 3x ≥ 2 − 2x.
2. Add 3x to both sides (this keeps a positive x term): 6 ≥ 2 + x.
3. Subtract 2 from both sides: 4 ≥ x, i.e. x ≤ 4.

Answerx ≤ 4, i.e. the interval (−∞, 4].

Watch this explained “The same answer twice”, 9:15 into The one rule that differs from equation-solving, and the reason it differs

Open NCERT p. 95Matches NCERT’s answer

1. The denominators are 2 and 3, so multiply every term by 6, their LCM — 6 is positive, so the sign stays the same: 6x + 3x + 2x < 66.
2. Combine like terms: 11x < 66.
3. Divide both sides by 11: x < 6.

Answerx < 6, i.e. the interval (−∞, 6).

Watch this explained “Clearing first, then flipping”, 10:23 into The one rule that differs from equation-solving, and the reason it differs

Open NCERT p. 95Matches NCERT’s answer

1. Subtract x/3 from both sides: 0 > x/2 − x/3 + 1.
2. x/2 − x/3 = x/6 (common denominator 6), so 0 > x/6 + 1.
3. Subtract 1 from both sides: −1 > x/6.
4. Multiply both sides by 6 — 6 is positive, so the sign stays the same: −6 > x, i.e. x < −6.

Answerx < −6, i.e. the interval (−∞, −6).

Watch this explained “Clearing first, then flipping”, 10:23 into The one rule that differs from equation-solving, and the reason it differs

Open NCERT p. 95Matches NCERT’s answer

1. Multiply both sides by 15, the LCM of 5 and 3 — 15 is positive, so the sign stays the same: 9(x − 2) ≤ 25(2 − x).
2. Multiply out both sides: 9x − 18 ≤ 50 − 25x.
3. Add 25x to both sides: 34x − 18 ≤ 50.
4. Add 18 to both sides: 34x ≤ 68.
5. Divide both sides by 34: x ≤ 2.

Answerx ≤ 2, i.e. the interval (−∞, 2].

Watch this explained “Clearing first, then flipping”, 10:23 into The one rule that differs from equation-solving, and the reason it differs

Open NCERT p. 95Matches NCERT’s answer

1. Multiply both sides by 30, the LCM of all the denominators — 30 is positive, so the sign stays the same: 9x + 60 ≥ 10x − 60.
2. Subtract 9x from both sides: 60 ≥ x − 60.
3. Add 60 to both sides: 120 ≥ x, i.e. x ≤ 120.

Answerx ≤ 120, i.e. the interval (−∞, 120].

Watch this explained “Clearing first, then flipping”, 10:23 into The one rule that differs from equation-solving, and the reason it differs

Open NCERT p. 95Matches NCERT’s answer

1. Multiply out: 4x + 6 − 10 < 6x − 12.
2. Simplify the left side: 4x − 4 < 6x − 12.
3. Subtract 4x from both sides: −4 < 2x − 12.
4. Add 12 to both sides: 8 < 2x.
5. Divide both sides by 2: 4 < x, i.e. x > 4.

Answerx > 4, i.e. the interval (4, ∞).

Watch this explained “The same answer twice”, 9:15 into The one rule that differs from equation-solving, and the reason it differs

Open NCERT p. 95Matches NCERT’s answer

1. Remove the brackets: 37 − 3x − 5 ≥ 9x − 8x + 24.
2. Simplify each side: 32 − 3x ≥ x + 24.
3. Add 3x to both sides: 32 ≥ 4x + 24.
4. Subtract 24 from both sides: 8 ≥ 4x.
5. Divide both sides by 4: 2 ≥ x, i.e. x ≤ 2.

Answerx ≤ 2, i.e. the interval (−∞, 2].

Watch this explained “Reading the answer off”, 12:05 into Why testing values one by one is not a method, and what replaces it

Open NCERT p. 95Matches NCERT’s answer

1. Multiply every term by 60, the LCM of 4, 3 and 5 — 60 is positive, so the sign stays the same: 15x < 20(5x − 2) − 12(7x − 3).
2. Multiply out the right side: 15x < 100x − 40 − 84x + 36.
3. Combine like terms on the right: 15x < 16x − 4.
4. Subtract 15x from both sides: 0 < x − 4.
5. Add 4 to both sides: 4 < x, i.e. x > 4.

Answerx > 4, i.e. the interval (4, ∞).

Watch this explained “Clearing first, then flipping”, 10:23 into The one rule that differs from equation-solving, and the reason it differs

Open NCERT p. 95Matches NCERT’s answer

1. Multiply every term by 60, the LCM of 3, 4 and 5 — 60 is positive, so the sign stays the same: 20(2x − 1) ≥ 15(3x − 2) − 12(2 − x).
2. Multiply out both sides: 40x − 20 ≥ 45x − 30 − 24 + 12x.
3. Combine like terms on the right: 40x − 20 ≥ 57x − 54.
4. Subtract 40x from both sides: −20 ≥ 17x − 54.
5. Add 54 to both sides: 34 ≥ 17x.
6. Divide both sides by 17: 2 ≥ x, i.e. x ≤ 2.

Answerx ≤ 2, i.e. the interval (−∞, 2].

Watch this explained “Clearing first, then flipping”, 10:23 into The one rule that differs from equation-solving, and the reason it differs

Open NCERT p. 95Matches NCERT’s answer

1. Subtract 2x from both sides: x − 2 < 1.
2. Add 2 to both sides: x < 3.

3. On the number line	What to draw
   at x = 3	an open (hollow) circle — 3 itself is not included
   to the left of 3	shade the line and put an arrow, since it goes on forever

Answerx < 3, i.e. the interval (−∞, 3) — a hollow circle at 3 with the line shaded to the left.

Watch this explained “Three questions, three answers”, 0:57 into Drawing the answer as a piece of the number line, hollow circle or solid

Open NCERT p. 95Matches NCERT’s answer

1. Subtract 3x from both sides: 2x − 3 ≥ −5.
2. Add 3 to both sides: 2x ≥ −2.
3. Divide both sides by 2: x ≥ −1.

4. On the number line	What to draw
   at x = −1	a filled (solid) circle — −1 is included
   to the right of −1	shade the line and put an arrow, since it goes on forever

Answerx ≥ −1, i.e. the interval [−1, ∞) — a solid circle at −1 with the line shaded to the right.

Watch this explained “The pair that isolates the mark”, 6:17 into Drawing the answer as a piece of the number line, hollow circle or solid

Open NCERT p. 95Matches NCERT’s answer

1. Multiply out both brackets: 3 − 3x < 2x + 8.
2. Add 3x to both sides (this keeps a positive x term): 3 < 5x + 8.
3. Subtract 8 from both sides: −5 < 5x.
4. Divide both sides by 5: −1 < x, i.e. x > −1.

5. On the number line	What to draw
   at x = −1	an open (hollow) circle — −1 itself is not included
   to the right of −1	shade the line and put an arrow, since it goes on forever

Answerx > −1, i.e. the interval (−1, ∞) — a hollow circle at −1 with the line shaded to the right.

Watch this explained “The pair that isolates the mark”, 6:17 into Drawing the answer as a piece of the number line, hollow circle or solid

Open NCERT p. 95Matches NCERT’s answer

1. Multiply every term by 30, the LCM of 2, 3 and 5 — 30 is positive, so the sign stays the same: 15x ≥ 10(5x − 2) − 6(7x − 3).
2. Multiply out the right side: 15x ≥ 50x − 20 − 42x + 18.
3. Combine like terms on the right: 15x ≥ 8x − 2.
4. Subtract 8x from both sides: 7x ≥ −2.
5. Divide both sides by 7: x ≥ −2/7.

6. On the number line	What to draw
   at x = −2/7	a filled (solid) circle — −2/7 is included
   to the right of −2/7	shade the line and put an arrow, since it goes on forever

Answerx ≥ −2/7, i.e. the interval [−2/7, ∞) — a solid circle at −2/7 with the line shaded to the right.

Watch this explained “An endpoint off the ticks”, 8:27 into Drawing the answer as a piece of the number line, hollow circle or solid

Open NCERT p. 95Matches NCERT’s answer

1. Let the marks Ravi gets in the third test be x.
2. The average of the three tests is (70 + 75 + x) ÷ 3.
3. "At least 60" means this average must be ≥ 60.
4. So (70 + 75 + x) ÷ 3 ≥ 60, which gives 145 + x ≥ 180.
5. Subtracting 145 from both sides, x ≥ 35.

AnswerRavi must score at least 35 marks in the third test.

Watch this explained “Sam at the shop”, 0:53 into Everyday constraints that fix a range rather than a value

Open NCERT p. 95Matches NCERT’s answer

1. Let the marks Sunita gets in the fifth exam be x.
2. The average of the five exams is (87 + 92 + 94 + 95 + x) ÷ 5.
3. "90 or more" means this average must be ≥ 90.
4. So (87 + 92 + 94 + 95 + x) ÷ 5 ≥ 90, which gives 368 + x ≥ 450.
5. Subtracting 368 from both sides, x ≥ 82.

AnswerSunita must score at least 82 marks in the fifth examination.

Watch this explained “Sam at the shop”, 0:53 into Everyday constraints that fix a range rather than a value

Open NCERT p. 95Matches NCERT’s answer

1. Let x be the smaller odd positive integer, so the next odd integer is x + 2.
2. Both integers are smaller than 10, so x + 2 < 10, which gives x < 8.
3. Their sum is more than 11, so x + (x + 2) > 11, which gives 2x > 9, so x > 4.5.
4. So x is an odd positive integer with 4.5 < x < 8.
5. The odd integers in this range are 5 and 7.
6. For x = 5, the pair is (5, 7). For x = 7, the pair is (7, 9).

AnswerThe pairs are (5, 7) and (7, 9).

Watch this explained “Two constraints at once”, 9:10 into Drawing the answer as a piece of the number line, hollow circle or solid

Open NCERT p. 95Matches NCERT’s answer

1. Let x be the smaller even positive integer, so the next even integer is x + 2.
2. Both integers are larger than 5, so x > 5.
3. Their sum is less than 23, so x + (x + 2) < 23, which gives 2x < 21, so x < 10.5.
4. So x is an even positive integer with 5 < x < 10.5.
5. The even integers in this range are 6, 8 and 10.
6. This gives the pairs (6, 8), (8, 10) and (10, 12).

AnswerThe pairs are (6, 8), (8, 10) and (10, 12).

Watch this explained “Two constraints at once”, 9:10 into Drawing the answer as a piece of the number line, hollow circle or solid

Open NCERT p. 96Matches NCERT’s answer

1. Let the shortest side be x cm.
2. The longest side is 3x, and the third side is 3x − 2.
3. The perimeter is x + 3x + (3x − 2) = 7x − 2.
4. "At least 61" means 7x − 2 ≥ 61.
5. Adding 2 to both sides, 7x ≥ 63.
6. Dividing both sides by 7, x ≥ 9.

AnswerThe shortest side must be at least 9 cm.

Watch this explained “Sam at the shop”, 0:53 into Everyday constraints that fix a range rather than a value

Open NCERT p. 96Matches NCERT’s answer

1. Let x be the length of the shortest board. The pieces are x, (x + 3) and 2x, as the hint says.
2. All three pieces come from a 91 cm board, so x + (x + 3) + 2x ≤ 91, which gives 4x + 3 ≤ 91, so x ≤ 22.
3. The third piece is at least 5 cm longer than the second, so 2x ≥ (x + 3) + 5, which gives x ≥ 8.
4. Putting both together, x lies between 8 and 22, both included.

AnswerThe shortest board can be any length from 8 cm to 22 cm.

Watch this explained “Two constraints at once”, 9:10 into Drawing the answer as a piece of the number line, hollow circle or solid