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Write 2y − 3 = x/2 in standard form and give a, b, c.
Multiply by 2: 4y − 6 = x, so x − 4y + 6 = 0.
a = 1, b = −4, c = 6.
Answer: x − 4y + 6 = 0
Is (3, −2) a solution of 4x + 5y = 2?
4(3) + 5(−2) = 12 − 10 = 2. Yes.
Answer: yes
(1, 2) is a solution of 3x + ky = 11. Find k.
3 + 2k = 11, so k = 4.
Answer: k = 4
Find the intercepts of 3x + 4y = 12.
x = 0: y = 3. y = 0: x = 4.
y-intercept 3, x-intercept 4.
Answer: 3 and 4
Find the slope and y-intercept of 2x + 3y − 6 = 0.
y = −(2/3)x + 2: slope −2/3, intercept 2.
Answer: −2/3, 2
Find the slope of the line through (−2, 3) and (4, −9).
(−9 − 3) ÷ (4 + 2) = −12/6 = −2.
Answer: −2
Solve x + y = 7 and 2x − y = 2.
y = 7 − x: 2x − 7 + x = 2, 3x = 9, x = 3, y = 4.
Answer: (3, 4)
Solve 3x + 2y = 12 and 5x − 2y = 4.
Add: 8x = 16, x = 2; 6 + 2y = 12, y = 3.
Answer: (2, 3)
Solve 3x + 2y = 18 and 2x + 3y = 21.
Multiply: 9x + 6y = 54 and 4x + 6y = 42. Subtract: 5x = 12, x = 12/5.
Then 2y = 18 − 36/5 = 54/5, so y = 27/5.
Answer: (12/5, 27/5)
Pen costs ₹x and notebook ₹y. 2 pens + 3 notebooks = ₹60; 3 pens + 2 notebooks = ₹65. Find x and y.
2x + 3y = 60, 3x + 2y = 65. Add: x + y = 25; subtract: x − y = 5. So x = 15, y = 10.
Answer: ₹15, ₹10
The sum of the digits of a two-digit number is 11 and reversing the digits increases it by 27. Find the number.
x + y = 11; 9(y − x) = 27 gives y − x = 3. y = 7, x = 4. Number 47.
Answer: 47
Ravi is 4 times as old as Tara now. In 6 years he will be 2.5 times as old as she will be. Find their ages.
Let Tara be t, Ravi 4t: 4t + 6 = 2.5(t + 6) gives 1.5t = 9, t = 6 and Ravi is 24.
Answer: 6 and 24
Classify x + 2y = 4 and 2x + 4y = 8.
1/2 = 2/4 = 4/8: coincident lines, infinitely many solutions.
Answer: infinite
Classify 3x − y = 5 and 6x − 2y = 7.
3/6 = −1/−2 = 1/2 but 5/7 ≠ 1/2: parallel, no solution.
Answer: none
For what p does 2x + py = 4, x + 3y = 5 have a unique solution?
2/1 ≠ p/3, so p ≠ 6.
Answer: p ≠ 6
Find k for kx + 2y = 6 and 3x + 6y = 18 to be coincident.
k/3 = 2/6 = 6/18 = 1/3, so k = 1.
Answer: k = 1
Solve x + y = 5, x − y = 1 graphically.
Lines through (0, 5), (5, 0) and (0, −1), (1, 0) meet at (3, 2).
Answer: (3, 2)
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(i) 5x + 2y − 2.7 = 0: a = 5, b = 2, c = −2.7. (ii) x/2 − 3y/2 − 4 = 0: a = 1/2, b = −3/2, c = −4 (the text writes x/2 − 4 = 3y/2). (iii) 1.5x − 1.3y − 4 = 0: a = 1.5, b = −1.3, c = −4. (iv) 0x + 3y − 5 = 0: a = 0, b = 3, c = −5.
(1, 9/2): 3 + 9 = 12 yes. (0, 6): 12 yes. (4, 1): 14 no. (1/2, 21/4): 3/2 + 21/2 = 12 yes. There are infinitely many (choose x, solve for y).
x = 0 gives (0, 7); y = 0 gives (7/2, 0). From the line: (1, 5), (2, 3), (3, 1) and (3/2, 4).
(i) (0, 4) and (3, 0). (ii) (0, 0), (1, −2/5) and (5, −2): the line passes through the origin.
−9a + 16 = −2 ⇒ a = 2. −6 + 4b = 14 ⇒ b = 5.
Yes. Substituting x = 0, y = 0 in ax + by = 0 gives 0 = 0 for every a and b (including the cases b = 0, the y-axis, and a = 0, the x-axis).
Slope 3/2: y = 1.5(x − 2). x = 8 gives y = 9: (8, 9). y = 6 gives x = 6: (6, 6).
Example 5: (5 − 2) ÷ (4 − 1) = 1. Example 6: (2 − 5) ÷ (−1 + 4) = −1 and the same either way.
(−2 − 3) ÷ (1 − 0) = −5 and (8 − 3) ÷ (−1 − 0) = −5.
From the first row T + R = 14. The last column gives 2C + T + R = 22 ⇒ 2C + 14 = 22 ⇒ C = 4. The other shapes follow from the remaining rows.
x = 3 − 2y; 7(3 − 2y) − 15y = 2 ⇒ −29y = −19 ⇒ y = 19/29, x = 3 − 38/29 = 49/29. Writing y = (3 − x)/2 gives the same solution.
x + 4x + 50 = 180 ⇒ x = 26°, y = 104°.
9x − 4y = 2000, 7x − 3y = 2000 ⇒ x = 2000, y = 4000. Incomes ₹18 000 and ₹14 000. Eliminating x instead: multiply by 7 and 9 and subtract gives y = 4000 as well.
11(x + y) = 66 ⇒ x + y = 6 and x − y = 2 ⇒ x = 4, y = 2: the number is 42 (with the larger digit in the tens place; 24 comes with the other assumption).
Both reduce to 2x − y = 4: infinitely many solutions. A rule: a₁/a₂ = b₁/b₂ = c₁/c₂ gives infinitely many; a₁/a₂ = b₁/b₂ ≠ c₁/c₂ none; a₁/a₂ ≠ b₁/b₂ exactly one. No: a pair need not have exactly one or infinitely many — it may have none.
11: x + 3y = 6 and 2x − 3y = 12 meet at (6, 0): unique solution (a₁/a₂ ≠ b₁/b₂). 12: x + 2y − 4 = 0 and 2x + 4y − 12 = 0: 1/2 = 2/4 ≠ −4/−12 so parallel, no solution. 13: 2x + 3y = 9 and 4x + 6y = 18 coincide: infinitely many solutions.
y = mx + d₁ and y = mx + d₂ with d₁ ≠ d₂: mx + d₁ = mx + d₂ would give d₁ = d₂, impossible, so the lines have no common point.
3x + 0·y − 1/5 = 0, i.e. 3x = 1/5.
y − 15 = 2x: −2x + y − 15 = 0 (a = −2, b = 1, c = −15). 3y − 2x = 0: −2x + 3y + 0 = 0 (a = −2, b = 3, c = 0). 5x = 3y: 5x − 3y + 0 = 0. x = 8: x + 0y − 8 = 0 (a = 1, b = 0, c = −8). 3y = 1: 0x + 3y − 1 = 0 (a = 0, b = 3, c = −1).
(i) t = 2p (Charlie) is correct: the notebook costs twice the pen. (ii) Yes, x + y = 176 is correct.
5(4) − 6(3) = 20 − 18 = 2. Yes.
(i) 7x − 3y = 21: (0, −7), (3, 0). (ii) 2x + 3y = 5: (1, 1), (0, 5/3).
4m − 3 = 7 ⇒ m = 5/2. 8 − n = −10 ⇒ n = 18.
(i) 5x + 3y = 7: (2, −1) in IV, (−1, 4) in II. (ii) 5x − 3y = 7: (2, 1) in I, (−1, −4) in III. (iii) −5x + 3y = 7: (1, 4) in I, (−2, −1) in III. (iv) −5x − 3y = 7: (−2, 1) in II, (1, −4) in IV.
3(2) − 7(3) = −15 ≠ 21: C is not on the line and does not satisfy the equation. A point off the line cannot satisfy the equation, since every solution is a point on the line.
(i) False (infinitely many). (ii) False (only if c = 0). (iii) False (e.g. (1/2, 2) for 2x + 3y = 7). (iv) True: x + 0y − 3 = 0. (v) True. (vi) False: 2 + 6 = 8 ≠ 7.
(i) The second is twice the first, so any (x, y) satisfying one satisfies the other. (ii) Multiply by k ≠ 0 or divide by k.
Slopes: A = 60/70 ≈ 0.857, B = 60/110 ≈ 0.545, C = 80/100 = 0.8. Least to greatest: B, C, A.
2.5 ÷ 4 = 0.625 (5/8).
Read from the grids (approximate): (i) horizontal line: slope 0. (ii) falling line: slope −1. (iii) rising line: slope about 1/2. (iv) vertical line: slope undefined.
Horizontal length = 18 × 12 = 216 cm.
2x + 3y = 850 and 4x + y = 1100.
x + 6y = 122 and x + 8y = 160 (so y = 19, x = 8).
x + y = 200 and 150x + 100y = 25 000 (x = 100, y = 100).
p + 2d = 90, 2p + 7d = 260. Subtract 2 × first: 3d = 80, d = 80/3. That gives p = 90 − 160/3 = 110/3. (The numbers are not whole.)
(i) x + y = 5, x − y = −21: −8 and 13. (ii) x − y = 26, x = 3y: 39 and 13. (iii) 7b + 6c = 8880, 3b + 5c = 4000: bat ₹1200, ball ₹80. (iv) f + 10k = 155, f + 15k = 220: fixed ₹25, ₹13/km; 25 km costs ₹350. (v) 11n − 9d = −4 and 6n − 5d = −3: n = 7, d = 9: fraction 7/9.
(vi) n + 1 = d − 1 and 2n = d + 1: n = 3, d = 5: 3/5. (vii) N − 5 = 3(S − 5), N + 10 = 2(S + 10): Sonu 20, Nuri 50. (viii) x + y = 9 and 9(10x + y) = 2(10y + x): number 18. (ix) a + b = 25, 50a + 100b = 2000: 10 notes of ₹50 and 15 of ₹100. (x) f + 4e = 27, f + 2e = 21: fixed ₹15, ₹3 per extra day.
b + g = 10, g = b + 4. The lines meet at (3, 7): 3 boys and 7 girls.
(i) 5/7 ≠ −4/6: intersect. (ii) 9/18 = 3/6 = 12/24: coincident. (iii) 6/2 = −3/−1 = 3 ≠ 10/9: parallel.
(i) coincident: infinitely many, the line x + y = 5. (ii) 1/3 = 1/3 ≠ 8/16: parallel, none. (iii) 2/4 ≠ 1/−2: unique, (2, 2). (iv) 2/4 = 1/2 ≠ 2/5: parallel, none.
l + w = 36, l = w + 4: 20 m by 16 m.
(i) intersecting: x − y − 1 = 0 (2/1 ≠ 3/−1). (ii) parallel: 4x + 6y − 9 = 0. (iii) coincident: 4x + 6y − 16 = 0.
9x + 7y = 107, 7x + 9y = 101. Adding: 16(x + y) = 208 ⇒ x + y = 13. Subtracting: 2(x − y) = 6 ⇒ x − y = 3. So citron ₹8, wood-apple ₹5.
Add: 12(p + n) = 96 ⇒ p + n = 8; subtract: 2(p − n) = −4 ⇒ p − n = −2. Pencil ₹3, pen ₹5.
Let the stool be h cm and the cat c cm. Cat on the stool: h + c = 85. Cat under the stool: h − c = 25. So h = 55 cm (and c = 30 cm).
(i) Slope = 6 ÷ 2 = 3. (ii) The y-intercept is 0: the tank starts empty. (iii) (a) 15 cm after 5 min. (b) 21 ÷ 3 = 7 min. (iv) 3 × 4 = 12 ≠ 10: (4, 10) is not on the line. (v) x = 3 gives y = 9: the point (3, 9).
(i) Plot (0, 32), (100, 212), (−40, −40). (ii) 30°C = 86°F. (iii) 95°F ⇒ C = (95 − 32) × 5/9 = 35°C. (iv) 0°C = 32°F; 0°F = −160/9 ≈ −17.8°C. (v) Yes: C = F = −40.
The lines meet at (2, 2).
The cover figure is not reproduced in the text of the PDF; to find an intersection point, write the two line equations from two points on each and solve the pair.
Put y = 0: x = −c/m (for m ≠ 0).
50 + 0.2t = 30 + 0.3t ⇒ t = 200. Plan B is cheaper for fewer than 200 minutes, Plan A for more than 200 minutes, equal at 200 minutes.
(i) Infinitely many (all parallel). (ii) Exactly one.
4/2 ≠ p/2 ⇒ p ≠ 4.
(a + b)/3 = 2b ⇒ a = 5b. (a + b)/3 = (5a + 2b + 1)/14 ⇒ 8b = a + 3 ⇒ 3b = 3: b = 1, a = 5.
1/(k − 1) = 2/(k + 1) ⇒ k = 3; check 1/2 = 2/4 = 3/6 ✓. k = 3.
(i) Robot 1 follows y = 4x/3; Robot 2 follows y = 2(x − 10). They meet where 4x/3 = 2x − 20, x = 30, y = 40: both are at a corner there, so the paths meet at (30, 40). (ii) The slopes are 1 and −3/5 with Robot 2 going down from y = 0: Robot 1 stays at y ≥ 0 and runs along y = 0 from x = 3 to 8, while Robot 2 runs along y = 0 from x = 7 to 12: the paths overlap on the segment from (7, 0) to (8, 0). (The straight lines y = x − 3 and y = −0.6(x − 7) would meet at (4.5, 1.5), but that point is not on Robot 2’s path.)
a + 15 = 60 + b and a = 6b ⇒ b = 9, a = 54. The second time is 3:09.
x + y = 15 and y − x = 1 ⇒ y = 8, x = 7. The number is 78.
Opposite angles add to 180°: x + 7 + 3y + 23 = 180 ⇒ x + 3y = 150; y + 8 + 4x + 12 = 180 ⇒ 4x + y = 160. So x = 30, y = 40. ∠A = 37°, ∠B = 48°, ∠C = 143°, ∠D = 132°.
With speed v and time t (distance vt): (v + 6)(t − 6) = vt ⇒ t − v = 6; (v − 4)(t + 6) = vt ⇒ 3v − 2t = 12. So v = 24 km/h, t = 30 h, distance 720 km.
F = C and C + 4 × 20 = 2(F + 20) ⇒ F + 80 = 2F + 40 ⇒ F = 40 years.