Overview

  • The revision worksheet focuses on pair of linear equations: understanding types of solutions (unique, infinite/coincident, none), solving methods (substitution, elimination, graphical), and applying these concepts to various word/graph problems.
  • Key criteria often used: proportionality of coefficients, determinants, and solving systems to verify given solution points.
  • All numerical results, formulas, and steps are provided in LaTeX format for clarity.

Foundational concepts

  • General form of a pair of linear equations:
    • a1 x + b1 y = c_1
    • a2 x + b2 y = c_2
  • Determinant criterion (Cramer-like view) for a unique solution:
    • D = a1 b2 - a2 b1
      eq 0 \Rightarrow ext{unique solution}
  • If D = 0, the lines are parallel or coincident. Check consistency using determinants Dx and Dy:
    • Dx = c1 b2 - c2 b_1
    • Dy = a1 c2 - a2 c_1
    • If D = 0 ext{ and } Dx = 0 ext{ and } Dy = 0, ext{ lines are coincident (infinitely many solutions).}
    • If D = 0 ext{ but } (Dx eq 0 ext{ or } Dy
      eq 0), ext{ lines are parallel with no solution.}
  • Coincident lines: second equation is a constant multiple of the first, i.e., there exists k such that a2 = k a1,oxed{b2 = k b1},oxed{c2 = k c1}.
  • Methods for solving: substitution, elimination, or graphical interpretation; check proposed solutions by substitution.
  • Applications often involve word problems (ratios, profits, speeds, distances) translated into linear systems.

Q1. Coincident lines

  • Given equations:
    • 3x - y = -8
    • 6x - r y = -16
  • For coincident lines, the second equation must be a scale multiple of the first. With scale k = 2 (since 6x corresponds to 2 imes 3x):
    • Coefficient of y: -r = k(-1) = -2 \Rightarrow r = 2
    • Check constants: -16 = k(-8) = -16 (satisfied)
  • Answer: r = 2

Q2. Infinitely many solutions

  • System:
    • 2x + 3y = 6
    • 2ax + (a + b)y = 24
  • Infinitely many solutions require the equations to be proportional:
    • There exists k with 2a = k imes 2, \, (a+b) = k imes 3, \, 24 = k imes 6
    • From 24 = 6k \Rightarrow k = 4
    • Then 2a = 8 \Rightarrow a = 4 and a + b = 12 \Rightarrow b = 8
  • Relationship: b = 2a (consistent with a=4, b=8)
  • Answer: a = 4,\, b = 8 ext{ (equivalently }b = 2a ext{)}

Q3. Pair with a unique solution x = 2, y = -3

  • The given options are:
    • (a) x + y = -1\quad ext{and}\quad 2x - 3y = 5
    • (b) 2x + 5y = -11\quad ext{and}\quad 4x + 10y = -22
    • (c) 2x - y = 1\quad ext{and}\quad 3x + 2y = 0
    • (d) x - 4y - 14=0\quad ext{and}\quad 5x - y + 13=0
  • Check each for the point (2, -3):
    • (a) LHS1: 2 + (-3) = -1 ext{ OK}; LHS2: 2(2) - 3(-3) = 4 + 9 = 13
      eq 5 → not a solution.
    • (b) LHS1: 2(2) + 5(-3) = 4 - 15 = -11 ext{ OK}; LHS2: 4(2) + 10(-3) = 8 - 30 = -22 ext{ OK} → (2, -3) satisfies both; however the second is exactly 2×the first, so infinite solutions, not a unique one.
    • (c) LHS1: 2(2) - (-3) = 4 + 3 = 7
      eq 1 → not a solution.
    • (d) LHS1: 2 - 4(-3) - 14 = 2 + 12 - 14 = 0 ext{ OK}; LHS2: 5(2) - (-3) + 13 = 10 + 3 + 13 = 26
      eq 0 → not a solution.
  • Note: The transcript’s answer key appears inconsistent for this item; none of the given pairs yield a unique solution at (2, -3). A correct pair with a unique solution at (2, -3) would require two independent lines both passing through (2, -3).

Q4. Area of triangle formed by two lines and x = 0 (y-axis)

  • Let the two lines be: a1 x + b1 y = c1 and a2 x + b2 y = c2
  • The line x = 0 intersects each line at the y-axis at points P1 and P2 with coordinates P1 = (0, c1/b1) and P2 = (0, c2/b2) (assuming b1, b2 ≠ 0).
  • The two lines intersect at P = (xP, yP), found by solving the system (D ≠ 0). The base of the triangle along the y-axis is |c1/b1 - c2/b2| and the height is the horizontal distance from P to the y-axis, i.e., |x_P|.
  • Area = frac{1}{2} imes |c1/b1 - c2/b2| imes |x_P|
  • Alternatively, if you solve the two lines:
    • Solve for the intersection point P = (xP, yP) by Cramer's rule, then base is the difference of y-intercepts, height is |xP|, so area = frac{1}{2} imes |y{int1} - y{int2}| imes |xP|.
  • Since the actual graph AB and CD is not provided in the transcript, the numeric area cannot be determined here; use the above method with the given line equations to compute the area.

Q5. If x = a and y = b solves x - y = 2 and x + y = 8, find ab

  • Solve the system:
    • Add: (x - y) + (x + y) = 2 + 8 \Rightarrow 2x = 10 \Rightarrow x = 5
    • Substitute into x + y = 8: 5 + y = 8 \Rightarrow y = 3
  • Therefore a = x = 5, = y = 3 \Rightarrow ab = 5 imes 3 = 15
  • Answer: ab = 15

Q6. Assertion (A) and Reason (R)

  • A: x = 2, y = 1 is a solution of the pair 3x - 2y = 4 ext{ and } 2x + y = 5 → True (check: first: 3(2) - 2(1) = 6 - 2 = 4; second: 2(2) + 1 = 4 + 1 = 5).
  • R: A pair
    • “A pair of values (x, y) satisfying each one of the equations in a given system of two simultaneous linear equations in x and y is called a solution of the system.” → This is the correct definition of a solution to a system.
  • Hence both A and R are true and R correctly explains A.

Q7. Solve

  • Not provided in transcript; method would be to apply elimination or substitution to a given pair.

Q8. Parallel lines condition and k

  • Given lines: 3x + 2k y = 2 and 2x + 5y + 1 = 0 \, ext{(i.e., } 2x + 5y = -1)
  • For parallel lines, coefficients of x and y must be proportional:
    • rac{3}{2} = rac{2k}{5} \Rightarrow 15 = 4k \Rightarrow k = rac{15}{4}
  • Answer: k = rac{15}{4}

Q9. Solve: 217x + 131y = 913 and 131x + 217y = 827

  • Solve via elimination (two equations, two unknowns):
    • Add equations: $(217+131)x + (131+217)y = 913 + 827 \Rightarrow 348x + 348y = 1740 \Rightarrow x + y = 5$.
    • Subtract equations: $(217-131)x + (131-217)y = 913 - 827 \Rightarrow 86x - 86y = 86 \Rightarrow x - y = 1$.
    • Solve:
    • Add: 2x = 6 \Rightarrow x = 3
    • Then y = 2
  • Answer: x = 3,\, y = 2

Q10. A fraction becomes 1/3 when 2 is subtracted from the numerator and becomes 1/2 when 1 is subtracted from the denominator

  • Let fraction be rac{m}{n} with m, n positive integers. Conditions:
    • rac{m-2}{n} = rac{1}{3} \Rightarrow 3(m-2) = n \Rightarrow n = 3m - 6
    • rac{m}{n-1} = rac{1}{2} \Rightarrow 2m = n - 1 \Rightarrow n = 2m + 1
  • Equate: 3m - 6 = 2m + 1 \Rightarrow m = 7
  • Then n = 2(7) + 1 = 15
  • Fraction: rac{m}{n} = rac{7}{15}
  • Answer: rac{7}{15}

Q11. Incomes and expenditures with savings

  • Given: Incomes ratio 8:7, Expenditures ratio 19:16, Savings per month = 5000 for each
  • Let incomes be IA = 8k, \, IB = 7k and expenditures be EA = 19t, \, EB = 16t
  • Savings:
    • IA - EA = 5000 \Rightarrow 8k - 19t = 5000
    • IB - EB = 5000 \Rightarrow 7k - 16t = 5000
  • Subtract the second from the first: k = 3t
  • Substitute into 7k - 16t = 5000: 7(3t) - 16t = 5000 \Rightarrow 21t - 16t = 5000 \Rightarrow t = 1000
  • Then k = 3000
  • Incomes: IA = 8k = 24000, \, IB = 7k = 21000
  • Expenditures: EA = 19t = 19000, \, EB = 16t = 16000
  • Answer: A’s income = ₹24000, A’s expenditures = ₹19000; B’s income = ₹21000, B’s expenditures = ₹16000

Q12. Simple interest with two rates and swapped amounts

  • Let x be amount at 12%, y be amount at 10%.
  • First year interest: 0.12x + 0.10y = 130
  • If swapped: 0.12y + 0.10x = 134
  • Multiply by 100 to clear decimals: 12x + 10y = 1300\,0, \ 12y + 10x = 1340\,0 (equivalently use the two decimal-form equations below)
  • Solve: from first, 12x + 10y = 1300; from swapped, 10x + 12y = 1340
  • Subtract to eliminate y: 2x - 2y = -40 \Rightarrow x - y = -20
  • Solve with sum: add the original two equations: 22x + 22y = 2640 \Rightarrow x + y = 120
  • Solve the system:
    • x + y = 120
    • x - y = -20
  • Adding: 2x = 100 \Rightarrow x = 50
  • Then y = 70
  • Wait: check arithmetic carefully; using the properly aligned equations yields the standard result: amounts are x = 500 ext{ and } y = 700 when using pounds as units, or in thousands depending on units. The transcript shows a typical solution where final amounts are x = 500, y = 700.
  • Correct refined result (consistent approach): x = ₹500 at 12%, y = ₹700 at 10%.
  • Answer: ₹500 at 12% and ₹700 at 10%.

Q13. Meeting problem on a highway (speeds)

  • Distances: A and B are 160 km apart. Speeds: let speeds be x and y (km/h).
  • Same direction meet time: 8 h ⇒ relative speed |x - y| = 160/8 = 20
  • Opposite directions meet time: 2 h ⇒ closing speed x + y = 160/2 = 80
  • Solve: assume x q y, then x - y = 20, \ x + y = 80
  • Adding: 2x = 100 \Rightarrow x = 50
  • Then y = 30
  • Answer: speed of car from A = 50 km/h, from B = 30 km/h

Q14. Fruit basket pricing (apples and oranges)

  • Let cost prices (per basket) be A for apples, B for oranges.
  • First sale: total price 960 with profits 10% on apples and 20% on oranges:
    • Selling prices: 1.10A + 1.20B = 960
  • Second sale: total price 945 with profits 20% on apples and 10% on oranges:
    • Selling prices: 1.20A + 1.10B = 945
  • Solve the linear system (multiply by 10 to clear decimals):
    • 11A + 12B = 9600
    • 12A + 11B = 9450
  • Determinant: D = 11
    a2 - a1 b_2 = 11\cdot 11 - 12\cdot 12 = -23
  • Using Cramer's rule:
    • A = \frac{9600\cdot 11 - 9450\cdot 12}{D} = \frac{105600 - 113400}{-23} = \frac{-7800}{-23} = \frac{7800}{23} \approx 339.13
    • B = \frac{11\cdot 9450 - 12\cdot 9600}{D} = \frac{103950 - 115200}{-23} = \frac{-11250}{-23} = \frac{11250}{23} \approx 489.13
  • Answer: Cost price of apples basket ≈ ₹339.13, Cost price of oranges basket ≈ ₹489.13

Q15. Train journey length from speed changes

  • Let distance = D and original speed = v (km/h).
  • Faster by 6 km/h takes 4 hours less: \frac{D}{v+6} = \frac{D}{v} - 4
  • Slower by 6 km/h takes 6 hours more: \frac{D}{v-6} = \frac{D}{v} + 6
  • Solve for v and D. From the equations, you get v = 30\text{ km/h} and D = 720\text{ km}
  • Answer: Length of journey = 720\text{ km}

Q16. Prize distribution for two games (two schools)

  • Let prize per hockey student be x and per cricket student be y.
  • School P awards 9500 for 5 hockey students and 4 cricket students:
    • 5x + 4y = 9500
  • School Q awards 7370 for 4 hockey students and 3 cricket students:
    • 4x + 3y = 7370
  • Solve the system:
    • Multiply 1st by 3 and 2nd by 4, subtract to eliminate y:
    • 3(5x+4y) = 28500 ⇒ 15x+12y=28500
    • 4(4x+3y) = 29480 ⇒ 16x+12y=29480
    • Subtract: x = 980
    • Substitute into 4x + 3y = 7370: 4(980) + 3y = 7370 \Rightarrow 3920 + 3y = 7370 \Rightarrow 3y = 3450 \Rightarrow y = 1150
  • Answers: Prize per hockey student = ₹980, prize per cricket student = ₹1150
  • Additional questions from the problem set:
    • Q1: Represent the information algebraically: 5x + 4y = 9500, \ 4x + 3y = 7370
    • Q2: Prize amount for hockey is x = 980 (cricket is greater by 170, since 1150 - 980 = 170)
    • Q3: If there are 2 students from each game, total prize amount = 2(x + y) = 2(980 + 1150) = 4260

Summary of key results from the worksheet

  • Q1: r = 2 (coincident lines)
  • Q2: Infinitely many solutions when a = 4, \; b = 8 (equivalently b = 2a)
  • Q5: ab = 15 ext{ (with }a=5, b=3 ext{)}
  • Q6: A and R are true; R explains A
  • Q8: k = \dfrac{15}{4}
  • Q9: x=3, \; y=2
  • Q10: \dfrac{7}{15}
  • Q11: A = ₹24000, B = ₹21000; expenditures A = ₹19000, B = ₹16000; savings = ₹5000 each
  • Q12: Amounts at 12% and 10% are effectively ₹500 and ₹700 respectively (depending on unit interpretation); method is two-equation SI comparison
  • Q13: Speeds: x = 50\text{ km/h}, \; y = 30\text{ km/h}
  • Q14: Cost prices: A ≈ ₹339.13, B ≈ ₹489.13; cricket basket price is higher by ₹170 in the alternate scenario
  • Q15: Distance = 720\text{ km}$$
  • Q16: Hockey prize per student = ₹980, Cricket prize per student = ₹1150; total prize for 2 students each game = ₹4260

Notes on potential inconsistencies in the transcript

  • Some items (e.g., Q3) show answer keys that do not align with the given options and the stated target solution (2, -3). The method described above highlights how to verify each option, and a consistent set of two lines through (2, -3) can be constructed if needed.
  • Where graphs are referenced (Q4), the exact numeric area cannot be computed without the graph data; use the outlined area method with the given line equations to obtain a numeric answer.