Two Variables, One Line

The standard form is ax + by + c = 0. Insert the values: 3x + 0y − 1/5 = 0, or 3x − 1/5 = 0. Both are the same line.

Move all terms to the left-hand side to obtain ax + by + c = 0.

Note: Multiplying an equation by any nonzero number gives another correct standard form, so the coefficients need not be identical to the table above.

(i) Charlie is correct. “Notebook is twice the pen” translates to t = 2p. Meera reversed the relationship.

(ii) Yes. If x and y are the runs made by the two batsmen, their total is x + y = 176.

Substitute x = 4, y = 3: 5(4) − 6(3) = 20 − 18 = 2. Since LHS = RHS, (4, 3) is a solution.

(i) Let y = 0: 7x = 21 ⇒ x = 3, so (3, 0). Let x = 0: −3y = 21 ⇒ y = −7, so (0, −7).

(ii) Let x = 1: 2 + 3y = 5 ⇒ y = 1, so (1, 1). Let x = −2: −4 + 3y = 5 ⇒ y = 3, so (−2, 3). Other correct pairs also earn full credit.

From the first equation: 2m(2) + 3(−1) = 7 ⇒ 4m − 3 = 7 ⇒ 4m = 10 ⇒ m = 5/2.

From the second: 4(2) + n(−1) = −10 ⇒ 8 − n = −10 ⇒ n = 18.

Verification: (i) −5 + 12 = 7 and 10 − 3 = 7; (ii) 10 − 3 = 7 and −5 + 12 = 7; (iii) −5 + 12 = 7 and 10 − 3 = 7; (iv) 10 − 3 = 7 and −5 + 12 = 7. Plot each pair on its corresponding line.

Substitute C(2, 3): 3(2) − 7(3) = 6 − 21 = −15, not 21. Thus C does not lie on the line and does not satisfy the equation. No: for any line, a point satisfies its equation if and only if it is on that line.

(i) False. A single non-degenerate linear equation in two variables has infinitely many real ordered-pair solutions.

(ii) False. For example x + y = 1 does not pass through (0, 0).

(iii) False. For example (1, 1) is a rational solution of x + y = 2.

(iv) True. Write it as x + 0y − 3 = 0, the vertical line x = 3.

(v) True. For any x, y = (7 − 2x)/3 produces a solution.

(vi) False. 2(1) + 3(2) = 8, not 7.

(i) Multiplying the first equation by 2 gives the second. So both equations represent the same line, and every solution of one solves the other.

(ii) If ax + by = c, multiply both sides by nonzero k to get kax + kby = kc. Conversely, divide the second equation by k to recover the first. Hence they have exactly the same solutions.

Slope measures steepness (rise/run). A = 60/70 = 6/7 ≈ 0.857; B = 60/110 = 6/11 ≈ 0.545; C = 80/100 = 0.8. Least to greatest: B, C, A.

Slope = rise/run = 2.5/4 = 0.625 = 5/8. Slope = 5/8 (or 0.625).

The lines are drawn on grid paper, so we can count squares. Take one large grid square as 1 unit; any other choice gives the same ratios.

(i) The line is horizontal: rise 0 for any run, so slope = 0.

(ii) The marked points are (−2, 2) and (2, −2): slope = (−2 − 2)/(2 − (−2)) = −4/4 = −1.

(iii) The marked points are (−2, −1) and (1, 1): slope = (1 − (−1))/(1 − (−2)) = 2/3.

(iv) The line is vertical: run = 0, so the slope is undefined.

For each 1 cm rise, the run is 12 cm. For 18 cm rise: 18 × 12 = 216 cm. Horizontal length = 216 cm = 2.16 m.

Day 1: 2x + 3y = 850. Day 2: 4x + y = 1100. These are the required equations. (Extra check, not asked: solving gives x = ₹245 and y = ₹120.)

Each fare = fixed charge + per-kilometre charge × distance. Thus x + 6y = 122 and x + 8y = 160. (Extra check: subtracting gives y = ₹19/km and x = ₹8.)

Number of tickets: x + y = 200. Revenue: 150x + 100y = 25,000 (or 3x + 2y = 500). (Extra check: there were 100 adults and 100 children.)

Let x,y be the two integers. x + y = 5, x − y = −21. Add: 2x = −16 ⇒ x = −8. Then y = 5 − (−8) = 13. Integers: −8 and 13.

Call the larger number x and the smaller one y. Then x − y = 26. If x = 3y, we get 2y = 26, so the positive pair is 39 and 13. The question also allows y = 3x. This gives −2x = 26, so −13 and −39 also work: their difference is 26 and −39 is three times −13.

Let b = cost of a bat, c = cost of a ball. 7b + 6c = 8880 and 3b + 5c = 4000. Multiply the first by 5: 35b + 30c = 44400; second by 6: 18b + 30c = 24000. Subtract: 17b = 20400 ⇒ b = 1200. Then 3(1200) + 5c = 4000 ⇒ c = 80. Bat ₹1200; ball ₹80.

Let f = fixed charge, r = fee per km. f + 10r = 155; f + 15r = 220. Subtract: 5r = 65 ⇒ r = ₹13/km. Then f = 155 − 130 = ₹25. For 25 km: 25 + 25(13) = ₹350. Fixed ₹25, variable ₹13/km, 25 km fare ₹350.

Let the fraction be x/y. (x + 2)/(y + 2) = 9/11 ⇒ 11x − 9y = −4. Similarly (x + 3)/(y + 3) = 5/6 ⇒ 6x − 5y = −3. Multiply the first by 5: 55x − 45y = −20; second by 9: 54x − 45y = −27. Subtract: x = 7. Then 77 − 9y = −4 ⇒ y = 9. Fraction = 7/9. Check: 9/11 and 10/12 = 5/6.

Let fraction = x/y. (x + 1)/(y − 1) = 1 ⇒ x + 1 = y − 1 ⇒ y = x + 2. Also x/(y + 1) = 1/2 ⇒ 2x = y + 1. Substitute: 2x = x + 3 ⇒ x = 3, y = 5. Fraction = 3/5. Check: 4/4 = 1; 3/6 = 1/2.

Let present ages N and S. Five years ago: N − 5 = 3(S − 5) ⇒ N = 3S − 10. In ten years: N + 10 = 2(S + 10) ⇒ N = 2S + 10. Equate: 3S − 10 = 2S + 10 ⇒ S = 20, N = 50. Nuri 50 years; Sonu 20 years.

Let tens digit x and units digit y. x + y = 9. The original number is 10x + y and the reversed is 10y + x. Condition: 9(10x + y) = 2(10y + x) ⇒ 90x + 9y = 20y + 2x ⇒ 88x = 11y ⇒ y = 8x. Now x + 8x = 9 ⇒ x = 1 and y = 8. Number = 18. Check 9 × 18 = 162 = 2 × 81.

Let x = number of ₹50 notes, y = number of ₹100 notes. x + y = 25 and 50x + 100y = 2000, or x + 2y = 40. Subtract the first: y = 15; then x = 10. 10 notes of ₹50 and 15 notes of ₹100.

Let f = charge for first 3 days and r = additional fee per extra day. f + 4r = 27; f + 2r = 21. Subtract: 2r = 6 ⇒ r = 3. Then f = 21 − 6 = 15. Fixed first-three-day charge ₹15; each additional day ₹3.

Let g = girls and b = boys. g + b = 10 and g − b = 4. Adding gives 2g = 14 ⇒ g = 7; b = 3. The two graph lines intersect at (b, g) = (3, 7). 3 boys, 7 girls.

(i) a₁/a₂ = 5/7 while b₁/b₂ = −4/6 = −2/3. Unequal ⇒ intersect at exactly one point.

(ii) 9/18 = 3/6 = 12/24 = 1/2 ⇒ coincident, infinitely many solutions.

(iii) 6/2 = (−3)/(−1) = 3, but 10/9 ≠ 3 ⇒ parallel, no solution.

(i) Equation 2 is twice equation 1. Infinitely many solutions on x + y = 5 (for example (0,5), (5,0)). Same graphical line.

(ii) Multiplying x − y = 8 by 3 gives 3x − 3y = 24, not 16. No solution; parallel lines.

(iii) 2x + y = 6 ⇒ y = 6 − 2x. 4x − 2y = 4 ⇒ 2x − y = 2 ⇒ y = 2x − 2. Set equal: 6 − 2x = 2x − 2 ⇒ x = 2; y = 2. Unique solution (2,2).

(iv) First reduces to x − y = 1; second to x − y = 5/4. No solution; parallel lines.

If length = L, width = W, then half the perimeter is L + W, so L + W = 36. Also L − W = 4. Add: 2L = 40 ⇒ L = 20; W = 16. Length 20 m, width 16 m. Check: 2(20+16) = 72 m, whose half is 36 m.

Possible answers (many are correct):

• (i) Intersecting: x + y − 1 = 0 — coefficient ratios 2/1 and 3/1 differ.
• (ii) Parallel: 4x + 6y − 9 = 0 — same x,y coefficient ratio but different constant ratio.
• (iii) Coincident: 4x + 6y − 16 = 0 — exactly twice the original.

Let x = citron price, y = fragrant wood-apple price. 9x + 7y = 107 and 7x + 9y = 101. Add: 16x + 16y = 208 ⇒ x + y = 13. Subtract second from first: 2x − 2y = 6 ⇒ x − y = 3. Thus 2x = 16, x = 8, y = 5. Citron: 8; wood-apple: 5 (in the problem’s monetary units).

Let p = pencil cost and q = pen cost. 5p + 7q = 50; 7p + 5q = 46. Add: 12p + 12q = 96 ⇒ p + q = 8. Subtract second from first: −2p + 2q = 4 ⇒ q − p = 2. Therefore p = ₹3, q = ₹5.

Let h = height of stool, a = height of animal. First picture: h + a = 85. Second picture: h − a = 25. Add: 2h = 110 ⇒ h = 55 cm. Then a = 30 cm. Height of the stool = 55 cm. (The second measurement is a difference of heights, not the stool height.)

(i) Slope = (6 − 0)/(2 − 0) = 3.

(ii) y-intercept = 0. At x = 0 (initial time), y = 0; line passes through origin.

(iii)(a) y = 3×5 = 15 cm after 5 minutes.

(iii)(b) 21 = 3x ⇒ 7 minutes.

(iv) At x = 4, y should be 12, not 10, so (4, 10) is not on the line.

(v) If x = 3, y = 3×3 = 9. Intersection (3, 9).

(i) Plot (C, F) = (0, 32) and (10, 50), then draw a line through them; points (20,68) and (30,86) are on it too.

(ii) F = (9/5)(30) + 32 = 54 + 32 = 86°F.

(iii) 95 = (9/5)C + 32 ⇒ 63 = (9/5)C ⇒ C = 35°C.

(iv) At C = 0, F = 32°F. At F = 0, 0 = (9/5)C + 32 ⇒ C = −160/9 ≈ −17.78°C.

(v) Let the common number be t: t = (9/5)t + 32 ⇒ −(4/5)t = 32 ⇒ t = −40, so −40°C = −40°F.

Rewrite as y = 6 − 2x and y = 2x − 2. The lines intersect when 6 − 2x = 2x − 2, giving x = 2. Then y = 2. Graphical solution = (2, 2). For plotting, first line passes through (0,6) and (3,0); second through (0,−2) and (1,0).

The two light beams on the cover cross at approximately (9, 2). To read the scale, use the rising line, labelled 15x − 10y − 115 = 0. It meets the x-axis at x = 115/15 = 23/3. Comparing that distance from the origin with the position of the crossing gives x ≈ 9. Substituting gives y ≈ (15 × 9 − 115)/10 = 2. This is a graphical estimate: the second beam has no printed equation.

An x-intercept lies where y = 0. Thus 0 = mx + c, so x = −c/m, when m ≠ 0. The intercept point is (−c/m, 0). If m = 0, a horizontal line y = c has no x-intercept for c ≠ 0; y = 0 is the x-axis itself, with infinitely many points in common.

Let t = call minutes (t ≥ 0). A = 50 + 0.20t, B = 30 + 0.30t. Equality: 50 + 0.20t = 30 + 0.30t ⇒ 20 = 0.10t ⇒ t = 200 minutes, each costing ₹90.

For t > 200, Plan A is cheaper. For 0 ≤ t < 200, Plan B is cheaper. At exactly 200 they cost the same.

(i) Infinitely many: all lines y = mx + c for different c have the given finite slope m.

(ii) Exactly one: through (x₁, y₁), the line is y − y₁ = m(x − x₁). (This also holds conceptually for a specified vertical direction through a point, although its slope is undefined.)

Unique solution requires the determinant of coefficients to be nonzero: 4×2 − p×2 ≠ 0 ⇒ 8 − 2p ≠ 0 ⇒ p ≠ 4. At p = 4, the coefficient ratios match, but the constants do not, so the lines are parallel.

For infinitely many solutions, first equation must be a multiple k of the second: a + b = 3k, −2b = −k ⇒ k = 2b, and 5a + 2b + 1 = 14k = 28b. From a + b = 6b, a = 5b. Substitute into constants: 5(5b)+2b+1 = 28b ⇒ 27b+1 = 28b ⇒ b = 1; a = 5. a = 5, b = 1. Check: first becomes 6x − 2y = 28, exactly twice 3x − y = 14.

If they coincide, coefficients and constants are proportional. Since second is k − 1 times the first (from x coefficients), its y coefficient must satisfy k + 1 = 2(k − 1) ⇒ k = 3. Check constants: k + 3 = 6 and (k − 1)×3 = 2×3 = 6. k = 3.

(i) Comparing the straight lines through the completed moves: Robot 1 has rise/run = 4/3, hence 𝑦 =43⁢𝑥. Robot 2 has rise/run = 2/1, starting at (10,0), hence 𝑦 =2⁢(𝑥 −10). Equating gives (4/3)x = 2x − 20 ⇒ x = 30, y = 40. Their trend lines intersect at (30, 40).

Literal movement clarification: Each robot really traces a staircase, not a continuous diagonal. Those staircases already touch at (24, 28): Robot 1, after seven completed right-and-up cycles, goes horizontally from (21,28) to (24,28); Robot 2 after fourteen cycles reaches (24,28). They also share further points, including (30,40). The shared parts of the traced paths are the isolated points (24,28) and (33,44), and the segments from (26,32) to (27,32), (27,32) to (27,34), (28,36) to (29,36), (30,38) to (30,40), and (30,40) to (31,40). To check the range, the first path stays between y = 4x/3 − 4 and y = 4x/3. The second stays between y = 2x − 22 and y = 2x − 20. These bands can overlap only for 24 ≤ x ≤ 33. Checking the steps there gives the list above.

(ii) The straight-line trends are robot 1 y = x − 3 and robot 2 y = −(3/5)(x − 7). Those lines meet at (4.5, 1.5). That point is not on either staircase: it is left of Robot 2’s start at x = 7, and Robot 1’s first moves do not pass through it.

The traced paths do meet. Both robots start by moving right along y = 0. Robot 1 goes from (3, 0) to (8, 0), and Robot 2 goes from (7, 0) to (12, 0). Those segments overlap from (7, 0) to (8, 0), including both endpoints. After that overlap, Robot 1 moves upward and Robot 2 moves downward, so the paths do not meet again.

Since 15 minutes later crosses 3:00, b = a + 15 − 60 = a − 45. Given a = 6b, substitute: a = 6(a − 45) ⇒ a = 6a − 270 ⇒ a = 54; so b = 9. First reading 2:54, second reading 3:09. The wording “six times greater” is interpreted as a = 6b, consistent with whole minutes.

Let tens digit x, units digit y. x + y = 15. Reversing raises the number by 9: (10y + x) − (10x + y) = 9 ⇒ 9(y − x) = 9 ⇒ y − x = 1. Add both relations: 2y = 16 ⇒ y = 8; x = 7. Number = 78 (reverse 87 is 9 more).

Opposite angles of a cyclic quadrilateral total 180°. Thus A + C: (x+7)+(3y+23)=180 ⇒ x + 3y = 150. B + D: (y+8)+(4x+12)=180 ⇒ 4x + y = 160. Solve: y = 160 − 4x. Then x + 3(160 − 4x) = 150 ⇒ −11x = −330 ⇒ x = 30, y = 40. ∠A = 37°, ∠B = 48°, ∠C = 143°, ∠D = 132°. Check opposite sums: 37 + 143 = 180, 48 + 132 = 180.

Let original speed v km/h and original time t h, so distance D = vt. Faster journey: vt = (v + 6)(t − 6) ⇒ t = v + 6. Slower journey: vt = (v − 4)(t + 6) ⇒ −4t + 6v − 24 = 0 ⇒ t = (3v − 12)/2. Equate: v + 6 = (3v − 12)/2 ⇒ 2v + 12 = 3v − 12 ⇒ v = 24 km/h. Then t = 30 h and D = 24×30 = 720 km. Check faster: 720/30 = 24 h (6 less); slower: 720/20 = 36 h (6 more).

Let F be father’s present age. Present sum of four children’s ages is also F. After 20 years their combined ages rise by 4×20 = 80, to F + 80. Father becomes F + 20. Given F + 80 = 2(F + 20) ⇒ F + 80 = 2F + 40 ⇒ F = 40 years.