CBSE 10th Grade Mathematics > Pair of Linear Equations in Two Variables
Pair of Linear Equations in Two Variables
Question 1 of 151 / 15
Method 1: Elimination (Eliminating y) Multiply second equation by 3: 12x - 3y = 15 ... (3) Add first equation to (3): (2x + 3y) + (12x - 3y) = 11 + 15. 14x = 26 → x = 13/7. Substitute x = 13/7 in first equation: 2(13/7) + 3y = 11. 26/7 + 3y = 11 → 3y = 77/7 - 26/7 = 51/7 → y = 17/7. Solution: x = 13/7, y = 17/7. Method 2: Elimination (Eliminating x) Multiply first equation by 2: 4x + 6y = 22 ... (3) Subtract second equation from (3): (4x + 6y) - (4x - y) = 22 - 5. 7y = 17 → y = 17/7. Substitute y = 17/7 in second equation: 4x - 17/7 = 5. 4x = 5 + 17/7 = 35/7 + 17/7 = 52/7 → x = 13/7. Solution: x = 13/7, y = 17/7. Method 3: Substitution From 4x - y = 5, we get y = 4x - 5. Substitute y in 2x + 3y = 11: 2x + 3(4x - 5) = 11. 2x + 12x - 15 = 11 → 14x = 26 → x = 13/7. Substitute x in y = 4x - 5: y = 4(13/7) - 5 = 52/7 - 35/7 = 17/7. Solution: x = 13/7, y = 17/7. User can do any one of the methods to solve the equation.In substitution method, user can do any one of the equations to substitute in the other equation.**Method 1: Substitution** From second equation: x = 9 - 5y. Substitute in first: 3(9 - 5y) - 2y = 7. 27 - 15y - 2y = 7 → -17y = -20 → y = 20/17. Substitute back: x = 9 - 5(20/17) = 9 - 100/17 = (153 - 100)/17 = 53/17. Solution: x = 53/17, y = 20/17. **Method 2: Elimination** Multiply second equation by 3: 3x + 15y = 27 ... (3) Subtract first equation (3x - 2y = 7) from (3): (3x + 15y) - (3x - 2y) = 27 - 7. 17y = 20 → y = 20/17. Substitute y = 20/17 in x + 5y = 9: x + 5(20/17) = 9 → x = 9 - 100/17 = (153 - 100)/17 = 53/17. Solution: x = 53/17, y = 20/17.we can eliminate x by multiplying the first equation by 5 and the second equation by 2.User can do any one of the methods to solve the equation.In substitution method, user can do any one of the equations to substitute in the other equation.For x + y = 4: When x = 0, y = 4; When x = 4, y = 0. Here user can take any values for x or y and get y or x respectively to plot the graph. For x - y = 2: When x = 2, y = 0; When x = 0, y = -2. Here user can take any values for x or y and get y or x respectively to plot the graph. Plot these points and draw the lines. Lines intersect at (3, 1). Graph Description: The line x + y = 4 passes through (0, 4) and (4, 0). The line x - y = 2 passes through (2, 0) and (0, -2). The two lines intersect at the point (3, 1). Solution: x = 3, y = 1. Nature: Unique solution (intersecting lines).Compare ratios: a₁/a₂ = 3/6 = 1/2, b₁/b₂ = 2/4 = 1/2, c₁/c₂ = 6/12 = 1/2. Since a₁/a₂ = b₁/b₂ = c₁/c₂, the lines are coincident. Nature: Infinitely many solutions (Dependent/Consistent system).Compare ratios: a₁/a₂ = 2/4 = 1/2, b₁/b₂ = -3/-6 = 1/2, c₁/c₂ = 4/9. Since a₁/a₂ = b₁/b₂ ≠ c₁/c₂, the lines are parallel. The system is Inconsistent (no solution).Let father's present age = x years and son's present age = y years. Five years ago: x - 5 = 4(y - 5) → x - 5 = 4y - 20 → x - 4y = -15 ... (1) Five years hence: x + 5 = 2(y + 5) → x + 5 = 2y + 10 → x - 2y = 5 ... (2) Subtract (1) from (2): (x - 2y) - (x - 4y) = 5 - (-15). 2y = 20 → y = 10. Substitute in (2): x - 2(10) = 5 → x = 25. Father's present age = 25 years, Son's present age = 10 years.Let cost of one pen = ₹x and one notebook = ₹y. Equations: 4x + 2y = 38 ... (1) and 3x + 5y = 67 ... (2) From (1): 2x + y = 19 → y = 19 - 2x ... (3) Substitute in (2): 3x + 5(19 - 2x) = 67. 3x + 95 - 10x = 67 → -7x = -28 → x = 4. From (3): y = 19 - 2(4) = 11. Cost of one pen = ₹4, Cost of one notebook = ₹11.Let speed of swimmer = x m/min and speed of current = y m/min. Upstream speed = (x - y) m/min, Downstream speed = (x + y) m/min. Equations: 900/(x-y) + 1050/(x+y) = 35 ... (1) and 600/(x-y) + 900/(x+y) = 25 ... (2) Let u = 1/(x-y) and v = 1/(x+y). 900u + 1050v = 35 and 600u + 900v = 25. Multiply (2) by 1.5: 900u + 1350v = 37.5 ... (3) Subtract (1) from (3): 350v = 2.5 → v = 2.5/350 = 1/140. Substitute v = 1/140 in (2): 600u + 900(1/140) = 25 → 600u + 9 = 25. 600u = 16 → u = 16/600 = 2/75. So x + y = 140 and x - y = 75/2 = 37.5. Adding both: 2x = 177.5 → x = 88.75 m/min. Subtracting: 2y = 102.5 → y = 51.25 m/min. Speed of swimmer = 88.75 m/min = 5.325 km/h. Speed of current = 51.25 m/min = 3.075 km/h. while equations can be solved by elimination method. Also user can represent unkowns with any alphabet.Let u = 1/x and v = 1/y. Equations become: 2u + 3v = 9 ... (1) and 4u + 9v = 21 ... (2) Multiply (1) by 2: 4u + 6v = 18 ... (3) Subtract (3) from (2): (4u + 9v) - (4u + 6v) = 21 - 18. 3v = 3 → v = 1. Substitute in (1): 2u + 3(1) = 9 → 2u = 6 → u = 3. Since u = 1/x = 3, so x = 1/3. Since v = 1/y = 1, so y = 1. Solution: x = 1/3, y = 1.Let 20% solution = x liters and 60% solution = y liters. Equations: x + y = 12 ... (1) Acid amount: 0.20x + 0.60y = 0.40(12) = 4.8 Multiply by 10: 2x + 6y = 48 → x + 3y = 24 ... (2) Subtract (1) from (2): (x + 3y) - (x + y) = 24 - 12. 2y = 12 → y = 6. Substitute in (1): x + 6 = 12 → x = 6. 20% solution = 6 liters, 60% solution = 6 liters.For no solution: a₁/a₂ = b₁/b₂ ≠ c₁/c₂. Here: k/16 = 4/k and (k-4)/k ≠ 4/k. From k/16 = 4/k: k² = 64 → k = ±8. Check k = 8: 8/16 = 1/2, 4/8 = 1/2, (8-4)/8 = 4/8 = 1/2. Since all ratios equal, k = 8 gives infinitely many solutions, not no solution. Check k = -8: -8/16 = -1/2, 4/-8 = -1/2, (-8-4)/-8 = -12/-8 = 3/2. Since -1/2 = -1/2 ≠ 3/2, k = -8 gives no solution. ✓ Therefore k = -8.Let correct answers = x and wrong answers = y. Equations: x + y = 25 and 4x - y = 60. Add both equations: 5x = 85 → x = 17. Then y = 8. Correct answers = 17, Wrong answers = 8.Let Miguel's present age = x years and grandfather's present age = y years. In 3 years, grandfather's age = y + 3. Last year, Miguel's age = x - 1. According to condition: y + 3 = 6(x - 1) → y + 3 = 6x - 6 → 6x - y = 9 ... (1) Sum of ages: x + y = 68 ... (2) Add (1) and (2): (6x - y) + (x + y) = 9 + 68. 7x = 77 → x = 11. Substitute x = 11 in (2): 11 + y = 68 → y = 57. Miguel is 11 years old, Grandfather is 57 years old.Rewrite: 2ax - 2by + a + 4b = 0 ... (1) and 2bx + 2ay + b - 4a = 0 ... (2) Multiply (1) by a: 2a²x - 2aby + a² + 4ab = 0 ... (3) Multiply (2) by b: 2b²x + 2aby + b² - 4ab = 0 ... (4) Add (3) and (4): 2a²x + 2b²x + a² + b² = 0. 2x(a² + b²) + (a² + b²) = 0 → (a² + b²)(2x + 1) = 0. Since a² + b² ≠ 0 (unless both a,b = 0), we have 2x + 1 = 0 → x = -1/2. Multiply (1) by b: 2abx - 2b²y + ab + 4b² = 0 ... (5) Multiply (2) by a: 2abx + 2a²y + ab - 4a² = 0 ... (6) Subtract (5) from (6): 2a²y + 2b²y - 4a² - 4b² = 0. 2y(a² + b²) = 4(a² + b²) → y = 2. Solution: x = -1/2, y = 2.No, a pair of linear equations cannot have exactly two solutions. Graphical reasoning: Two straight lines in a plane can either: (i) Intersect at exactly ONE point → Unique solution (ii) Be parallel (never meet) → No solution (iii) Be coincident (same line) → Infinitely many solutions Since two straight lines cannot intersect at more than one point (by definition of a line), it is impossible to have exactly two solutions. Therefore, the answer is NO - exactly two solutions is not possible.
