Important Questions Class 10 Maths Chapter 3: Pair Of Linear Equations In Two Variables With Solutions

A pair of linear equations in two variables contains two equations written using the same two unknowns. Its solution is the ordered pair that satisfies both equations at the same time.

When two lines meet at a point, that point solves both equations simultaneously. This is the core logic of Chapter 3: two equations, two unknowns, one ordered-pair answer. CBSE 2026 tests Important Questions Class 10 Maths Chapter 3 through the graphical method, substitution method, and elimination method. The cross-multiplication method does not form part of the current CBSE 2026 syllabus. Chapter 3 usually appears through one word problem and one algebraic question in board-style papers, based on the NCERT Reprint 2026-27.

Key Takeaways

• Pair Of Linear Equations: Chapter 3 focuses on graphical method, substitution method, elimination method, and word problems.
• Main Rule: a1/a2 ≠ b1/b2 means the pair has one unique solution.
• Exam Trap: Compare a1/a2, b1/b2, and c1/c2 in the same order.
• CBSE 2026 Update: Cross-Multiplication Method is not included in the current syllabus.

Important Questions Class 10 Maths Chapter 3 Structure 2026

Important Questions Class 10 Maths Chapter 3: Consistency Conditions

Consistency conditions help identify the number of solutions before solving the equations. These conditions decide whether two lines intersect, run parallel, or coincide.

Q1. What Is The Condition For A Unique Solution In Pair Of Linear Equations?

A pair has a unique solution when a1/a2 ≠ b1/b2. The two lines intersect at exactly one point.

1. Given Data:
   Equations:
   a1x + b1y + c1 = 0
   a2x + b2y + c2 = 0
2. Formula Used:
   a1/a2 ≠ b1/b2
3. Graph Meaning:
   The lines intersect at one point.
4. Final Result:
   The pair has one solution

Q2. What Is The Condition For No Solution In Linear Equations In Two Variables Class 10?

A pair has no solution when a1/a2 = b1/b2 ≠ c1/c2. The two lines are parallel.

1. Given Data:
   Equations:
   a1x + b1y + c1 = 0
   a2x + b2y + c2 = 0
2. Formula Used:
   a1/a2 = b1/b2 ≠ c1/c2
3. Graph Meaning:
   The lines are parallel.
4. Final Result:
   The pair is inconsistent

Q3. What Is The Condition For Infinitely Many Solutions In Pair Of Linear Equations?

A pair has infinitely many solutions when a1/a2 = b1/b2 = c1/c2. The two equations represent the same line.

1. Given Data:
   Equations:
   a1x + b1y + c1 = 0
   a2x + b2y + c2 = 0
2. Formula Used:
   a1/a2 = b1/b2 = c1/c2
3. Graph Meaning:
   The lines coincide.
4. Final Result:
   The pair has infinitely many solutions

Linear Equations Class 10 MCQ With Answers

These linear equations class 10 MCQ questions test consistency conditions, solution types, and coefficient-based reasoning. Each question follows the CBSE 2026 board pattern.

Q1. When Is A System Of Two Linear Equations Inconsistent?

A system is inconsistent when the lines are parallel. Parallel lines have no common point.

1. Given Data:
   Two straight lines
2. Rule Used:
   Parallel lines never meet.
3. Final Result:
   Answer: (B) Parallel

Q2. What Type Of Solution Does 2x + 3y = 5 And 4x + 6y = 10 Have?

The pair has infinitely many solutions. Both equations represent the same line.

1. Given Data:
   2x + 3y = 5
   4x + 6y = 10
2. Formula Used:
   a1/a2 = b1/b2 = c1/c2
3. Calculation:
   2/4 = 3/6 = 5/10
   1/2 = 1/2 = 1/2
4. Final Result:
   Answer: (C) Infinitely many solutions

Q3. Find k If 3x + ky = 5 And 6x + 10y = 9 Have No Solution.

The value of k is 5. For no solution, coefficient ratios of x and y must match.

1. Given Data:
   3x + ky = 5
   6x + 10y = 9
2. Formula Used:
   a1/a2 = b1/b2 ≠ c1/c2
3. Calculation:
   3/6 = k/10
   1/2 = k/10
   k = 5
4. Final Result:
   Answer: (A) 5

Q4. If a1/a2 ≠ b1/b2, What Type Of Solution Does The Pair Have?

The pair has a unique solution. Unequal ratios mean the lines intersect at one point.

1. Given Data:
   a1/a2 ≠ b1/b2
2. Rule Used:
   Intersecting lines have one solution.
3. Final Result:
   Answer: (B) Unique solution

Q5. What Type Of Pair Is x - 2y = 0 And 3x + 4y = 20?

The pair has a unique solution. The coefficient ratios are unequal.

1. Given Data:
   x - 2y = 0
   3x + 4y = 20
2. Formula Used:
   Compare a1/a2 and b1/b2
3. Calculation:
   a1/a2 = 1/3
   b1/b2 = -2/4 = -1/2
   1/3 ≠ -1/2
4. Final Result:
   Answer: (B) Consistent with unique solution

Q6. How Many Solutions Can A Pair Of Linear Equations Have?

A pair can have no solution, one solution, or infinitely many solutions. Two lines can be parallel, intersecting, or coincident.

1. Given Data:
   Two linear equations in two variables
2. Rule Used:
   Straight lines have three possible positions.
3. Final Result:
   Answer: (C) No solution, one solution, or infinitely many

Q7. What Type Of Lines Are 5x - 4y + 8 = 0 And 7x + 6y - 9 = 0?

The lines are intersecting. The ratios a1/a2 and b1/b2 are unequal.

1. Given Data:
   5x - 4y + 8 = 0
   7x + 6y - 9 = 0
2. Formula Used:
   a1/a2 ≠ b1/b2
3. Calculation:
   a1/a2 = 5/7
   b1/b2 = -4/6 = -2/3
   5/7 ≠ -2/3
4. Final Result:
   Answer: (C) Intersecting

Q8. Find k If 2x + 3y = 7 And 4x + ky = 14 Have A Unique Solution.

The condition is k ≠ 6. A unique solution needs a1/a2 ≠ b1/b2.

1. Given Data:
   2x + 3y = 7
   4x + ky = 14
2. Formula Used:
   a1/a2 ≠ b1/b2
3. Calculation:
   2/4 ≠ 3/k
   1/2 ≠ 3/k
   k ≠ 6
4. Final Result:
   Answer: (B) k ≠ 6

Q9. What Does 9x + 3y + 12 = 0 And 18x + 6y + 24 = 0 Represent?

The pair represents coincident lines. All three coefficient ratios are equal.

1. Given Data:
   9x + 3y + 12 = 0
   18x + 6y + 24 = 0
2. Formula Used:
   a1/a2 = b1/b2 = c1/c2
3. Calculation:
   9/18 = 3/6 = 12/24
   1/2 = 1/2 = 1/2
4. Final Result:
   Answer: (C) Coincident lines

Q10. A Two-Digit Number And Its Reversed Number Have Sum 66. If Digits Are x And y, Find The Equation.

The equation is x + y = 6. Write both numbers using place value.

1. Given Data:
   Ten’s digit = x
   Unit’s digit = y
2. Formula Used:
   Number = 10x + y
   Reversed number = 10y + x
3. Calculation:
   (10x + y) + (10y + x) = 66
   11x + 11y = 66
   x + y = 6
4. Final Result:
   Answer: (B) x + y = 6

Consistent And Inconsistent Equations Class 10 Questions

These consistent and inconsistent equations class 10 questions ask students to compare coefficient ratios. The answer must state the condition and solution type.

Q1. Check Whether 3x + 2y = 5 And 2x - 3y = 7 Are Consistent.

The pair is consistent with a unique solution. The ratios a1/a2 and b1/b2 are unequal.

1. Given Data:
   3x + 2y = 5
   2x - 3y = 7
2. Formula Used:
   Compare a1/a2 and b1/b2.
3. Calculation:
   a1/a2 = 3/2
   b1/b2 = 2/(-3) = -2/3
   3/2 ≠ -2/3
4. Final Result:
   Consistent with unique solution

Q2. Check Whether 2x - 3y = 8 And 4x - 6y = 9 Are Consistent.

The pair is inconsistent. The lines are parallel because a1/a2 = b1/b2 ≠ c1/c2.

1. Given Data:
   2x - 3y = 8
   4x - 6y = 9
2. Formula Used:
   a1/a2 = b1/b2 ≠ c1/c2
3. Calculation:
   a1/a2 = 2/4 = 1/2
   b1/b2 = -3/(-6) = 1/2
   c1/c2 = 8/9
   1/2 = 1/2 ≠ 8/9
4. Final Result:
   Inconsistent

Q3. Check Whether 3x/2 + 5y/3 = 7 And 9x - 10y = 14 Are Consistent.

The pair is consistent with a unique solution. Convert the fractional equation first.

1. Given Data:
   3x/2 + 5y/3 = 7
   9x - 10y = 14
2. Formula Used:
   Compare a1/a2 and b1/b2.
3. Calculation:
   3x/2 + 5y/3 = 7
   Multiply by 6.
   9x + 10y = 42
   Compare with 9x - 10y = 14.
   a1/a2 = 9/9 = 1
   b1/b2 = 10/(-10) = -1
4. Final Result:
   Consistent with unique solution

Q4. Check Whether 5x - 3y = 11 And -10x + 6y = -22 Are Consistent.

The pair is consistent with infinitely many solutions. All three ratios are equal.

1. Given Data:
   5x - 3y = 11
   -10x + 6y = -22
2. Formula Used:
   a1/a2 = b1/b2 = c1/c2
3. Calculation:
   a1/a2 = 5/(-10) = -1/2
   b1/b2 = -3/6 = -1/2
   c1/c2 = 11/(-22) = -1/2
4. Final Result:
   Consistent with infinitely many solutions

Q5. Find k If 3x - y = 7 And 6x - 2y = k Have Infinitely Many Solutions.

The value of k is 14. Infinitely many solutions require all three ratios to be equal.

1. Given Data:
   3x - y = 7
   6x - 2y = k
2. Formula Used:
   a1/a2 = b1/b2 = c1/c2
3. Calculation:
   3/6 = -1/(-2) = 7/k
   1/2 = 7/k
   k = 14
4. Final Result:
   k = 14

Q6. Find k If 3x - 4y = 7 And kx + 3y - 5 = 0 Have No Solution.

The value of k is -9/4. No solution requires equal variable-coefficient ratios.

1. Given Data:
   3x - 4y = 7
   kx + 3y - 5 = 0
2. Formula Used:
   a1/a2 = b1/b2
3. Calculation:
   3/k = -4/3
   9 = -4k
   k = -9/4
4. Final Result:
   k = -9/4

Graphical Method Class 10 Linear Equations Questions

These graphical method class 10 linear equations questions show solutions visually. Intersecting lines give one solution, parallel lines give no solution, and coincident lines give infinitely many solutions.

Q1. Check Graphically Whether x + 3y = 6 And 2x - 3y = 12 Are Consistent.

The pair is consistent, and the solution is x = 6, y = 0. Both lines intersect at (6, 0).

1. Given Data:
   x + 3y = 6
   2x - 3y = 12
2. Points For First Line:
   When x = 0, y = 2
   When y = 0, x = 6
   Points: (0, 2), (6, 0)
3. Points For Second Line:
   When x = 0, y = -4
   When y = 0, x = 6
   Points: (0, -4), (6, 0)
4. Final Result:
   x = 6, y = 0

Q2. Show Graphically That 2x + 3y = 8 And 4x + 6y = 16 Have Infinitely Many Solutions.

The pair has infinitely many solutions. Both equations represent the same line.

1. Given Data:
   2x + 3y = 8
   4x + 6y = 16
2. Formula Used:
   a1/a2 = b1/b2 = c1/c2
3. Calculation:
   2/4 = 3/6 = 8/16
   1/2 = 1/2 = 1/2
4. Example Points:
   (1, 2), (4, 0), and (-2, 4) satisfy the same line.
5. Final Result:
   Infinitely many solutions

Q3. Ten Students Took Part In A Quiz. Girls Are 4 More Than Boys. Find Boys And Girls Graphically.

The answer is 3 boys and 7 girls. The two graph lines intersect at (3, 7).

1. Given Data:
   Total students = 10
   Girls are 4 more than boys.
2. Equation Formation:
   Let boys = x and girls = y.
   x + y = 10
   y - x = 4
3. Graph Points:
   For x + y = 10: (0, 10), (5, 5)
   For y - x = 4: (0, 4), (3, 7)
4. Final Result:
   Boys = 3, Girls = 7

Q4. Draw x - y + 1 = 0 And 3x + 2y - 12 = 0. Find The Triangle With The X-Axis.

The triangle vertices are (-1, 0), (4, 0), and (2, 3). These points come from x-axis intercepts and intersection.

1. Given Data:
   x - y + 1 = 0
   3x + 2y - 12 = 0
2. X-Axis Intercepts:
   First line cuts x-axis at (-1, 0).
   Second line cuts x-axis at (4, 0).
3. Intersection:
   From x - y + 1 = 0, y = x + 1.
   Substitute in 3x + 2y = 12.
   3x + 2(x + 1) = 12
   5x + 2 = 12
   x = 2, y = 3
4. Final Result:
   Triangle vertices are (-1, 0), (4, 0), and (2, 3)

Substitution Method Class 10 Important Questions With Solutions

These substitution method class 10 important questions work best when one equation allows easy isolation of a variable. Replace that variable in the other equation.

Q1. Solve 7x - 15y = 2 And x + 2y = 3 By Substitution.

The solution is x = 49/29, y = 19/29. Substitute x from the second equation.

1. Given Data:
   7x - 15y = 2
   x + 2y = 3
2. Formula Used:
   x = 3 - 2y
3. Calculation:
   7(3 - 2y) - 15y = 2
   21 - 14y - 15y = 2
   -29y = -19
   y = 19/29
4. Find x:
   x = 3 - 2(19/29)
   x = 49/29
5. Final Result:
   x = 49/29, y = 19/29

Q2. Solve x + y = 14 And x - y = 4.

The solution is x = 9, y = 5. Substitute x = 14 - y in the second equation.

1. Given Data:
   x + y = 14
   x - y = 4
2. Formula Used:
   x = 14 - y
3. Calculation:
   (14 - y) - y = 4
   14 - 2y = 4
   2y = 10
   y = 5
4. Find x:
   x + 5 = 14
   x = 9
5. Final Result:
   x = 9, y = 5

Q3. Solve 2x + 3y = 11 And 2x - 4y = -24. Find m If y = mx + 3.

The solution is x = -2, y = 5, m = -1. Eliminate x first, then substitute in y = mx + 3.

1. Given Data:
   2x + 3y = 11
   2x - 4y = -24
   y = mx + 3
2. Calculation:
   (2x + 3y) - (2x - 4y) = 11 - (-24)
   7y = 35
   y = 5
3. Find x:
   2x + 3(5) = 11
   2x + 15 = 11
   x = -2
4. Find m:
   5 = m(-2) + 3
   2 = -2m
   m = -1
5. Final Result:
   x = -2, y = 5, m = -1

Q4. Aftab Was Seven Times His Daughter’s Age Seven Years Ago. After Three Years, He Will Be Three Times Her Age. Find Ages.

Aftab is 42 years old, and his daughter is 12 years old. Form two equations from age conditions.

1. Given Data:
   Let Aftab’s age = s
   Let daughter’s age = t
2. Equation Formation:
   Seven years ago:
   s - 7 = 7(t - 7)
   s - 7t + 42 = 0
3. Second Condition:
   Three years from now:
   s + 3 = 3(t + 3)
   s - 3t = 6
4. Calculation:
   From s - 3t = 6, s = 3t + 6.
   Substitute in s - 7t + 42 = 0.
   3t + 6 - 7t + 42 = 0
   -4t = -48
   t = 12
   s = 42
5. Final Result:
   Aftab = 42 years, Daughter = 12 years

Q5. A Fraction Becomes 9/11 If 2 Is Added To Numerator And Denominator. If 3 Is Added, It Becomes 5/6. Find The Fraction.

The fraction is 7/9. Form equations from both changed fractions.

1. Given Data:
   Fraction = x/y
2. Equation Formation:
   (x + 2)/(y + 2) = 9/11
   11x + 22 = 9y + 18
   11x - 9y = -4
3. Second Equation:
   (x + 3)/(y + 3) = 5/6
   6x + 18 = 5y + 15
   6x - 5y = -3
4. Calculation:
   Multiply 11x - 9y = -4 by 5.
   55x - 45y = -20
   Multiply 6x - 5y = -3 by 9.
   54x - 45y = -27
   Subtract: x = 7
5. Find y:
   6(7) - 5y = -3
   42 - 5y = -3
   y = 9
6. Final Result:
   Fraction = 7/9

Elimination Method Class 10 Important Questions With Solutions

These elimination method class 10 important questions require equal coefficients for one variable. Add or subtract equations to remove that variable.

Q1. Incomes Are In Ratio 9 : 7 And Expenditures Are In Ratio 4 : 3. Each Saves ₹2000. Find Monthly Incomes.

The monthly incomes are ₹18,000 and ₹14,000. Use income minus expenditure equals saving.

1. Given Data:
   Incomes = 9x and 7x
   Expenditures = 4y and 3y
   Saving = ₹2000 each
2. Equation Formation:
   9x - 4y = 2000
   7x - 3y = 2000
3. Calculation:
   Multiply first equation by 3.
   27x - 12y = 6000
   Multiply second equation by 4.
   28x - 12y = 8000
   Subtract: x = 2000
4. Final Result:
   Monthly incomes are ₹18,000 and ₹14,000

Q2. A Two-Digit Number And Its Reversed Number Have Sum 66. Digits Differ By 2. Find The Numbers.

The numbers are 42 and 24. Both digit orders satisfy the given conditions.

1. Given Data:
   Ten’s digit = x
   Unit’s digit = y
2. Equation Formation:
   Number = 10x + y
   Reversed number = 10y + x
   x + y = 6
3. Difference Condition:
   x - y = 2 or y - x = 2
4. Calculation:
   If x - y = 2, then x = 4 and y = 2.
   Number = 42
   If y - x = 2, then y = 4 and x = 2.
   Number = 24
5. Final Result:
   The numbers are 42 and 24

Q3. Solve 3x + 4y = 10 And 2x - 2y = 2 By Elimination.

The solution is x = 2, y = 1. Make y-coefficients equal and add equations.

1. Given Data:
   3x + 4y = 10
   2x - 2y = 2
2. Formula Used:
   Eliminate one variable.
3. Calculation:
   Multiply second equation by 2.
   4x - 4y = 4
   Add with first equation.
   7x = 14
   x = 2
4. Find y:
   2(2) - 2y = 2
   4 - 2y = 2
   y = 1
5. Final Result:
   x = 2, y = 1

Q4. Solve 3x - 5y - 4 = 0 And 9x = 2y + 7 By Elimination.

The solution is x = 9/13, y = -5/13. Rewrite both equations in standard form.

1. Given Data:
   3x - 5y - 4 = 0
   9x = 2y + 7
2. Standard Form:
   3x - 5y = 4
   9x - 2y = 7
3. Calculation:
   Multiply first equation by 3.
   9x - 15y = 12
   Subtract this from 9x - 2y = 7.
   13y = -5
   y = -5/13
4. Find x:
   3x - 5(-5/13) = 4
   3x + 25/13 = 4
   3x = 27/13
   x = 9/13
5. Final Result:
   x = 9/13, y = -5/13

Q5. Five Years Ago, Nuri Was Thrice As Old As Sonu. Ten Years Later, Nuri Will Be Twice Sonu’s Age. Find Ages.

Nuri is 50 years old, and Sonu is 20 years old. Form two equations from age conditions.

1. Given Data:
   Nuri’s age = x
   Sonu’s age = y
2. Equation Formation:
   Five years ago:
   x - 5 = 3(y - 5)
   x - 3y = -10
3. Second Condition:
   Ten years later:
   x + 10 = 2(y + 10)
   x - 2y = 10
4. Calculation:
   Subtract x - 3y = -10 from x - 2y = 10.
   y = 20
   x - 2(20) = 10
   x = 50
5. Final Result:
   Nuri = 50 years, Sonu = 20 years

Word Problems On Linear Equations Class 10 With Solutions

These word problems on linear equations class 10 questions test equation formation and solving. They cover age, digits, fractions, geometry, money, and real-life conditions.

Q1. Half The Perimeter Of A Rectangular Garden Is 36 m. Length Is 4 m More Than Width. Find Dimensions.

The dimensions are 20 m and 16 m. Half the perimeter equals length plus width.

1. Given Data:
   Half perimeter = 36 m
   Length = width + 4
2. Equation Formation:
   Let length = l and width = w.
   l + w = 36
   l = w + 4
3. Calculation:
   w + 4 + w = 36
   2w = 32
   w = 16
   l = 20
4. Final Result:
   Length = 20 m, Width = 16 m

Q2. The Difference Between Two Numbers Is 26, And One Number Is Three Times The Other. Find The Numbers.

The numbers are 39 and 13. Let the larger number be three times the smaller number.

1. Given Data:
   Difference = 26
   One number = 3 × other number
2. Equation Formation:
   Let numbers be x and y.
   x = 3y
   x - y = 26
3. Calculation:
   3y - y = 26
   2y = 26
   y = 13
   x = 39
4. Final Result:
   The numbers are 39 and 13

Q3. A Taxi Charges ₹105 For 10 km And ₹155 For 15 km. Find Fixed Charge And Charge Per km.

The fixed charge is ₹5, and the charge per km is ₹10. The fare for 25 km is ₹255.

1. Given Data:
   Fare for 10 km = ₹105
   Fare for 15 km = ₹155
2. Equation Formation:
   Fixed charge = ₹f
   Per km charge = ₹k
   f + 10k = 105
   f + 15k = 155
3. Calculation:
   Subtract the first equation from the second.
   5k = 50
   k = 10
   f + 100 = 105
   f = 5
4. Fare For 25 km:
   Fare = 5 + 25(10)
   Fare = 255
5. Final Result:
   Fixed charge = ₹5, Per km charge = ₹10, Fare = ₹255

Q4. A Library Charges A Fixed Amount For 3 Days And Extra Charge Per Day. Find Both Charges.

The fixed charge is ₹15, and the extra charge is ₹3 per day. Use extra days beyond the first 3 days.

1. Given Data:
   Saritha paid ₹27 for 7 days.
   Susy paid ₹21 for 5 days.
2. Equation Formation:
   Fixed charge = ₹f
   Extra daily charge = ₹d
   For 7 days, extra days = 4.
   f + 4d = 27
   For 5 days, extra days = 2.
   f + 2d = 21
3. Calculation:
   Subtract equations.
   2d = 6
   d = 3
   f + 2(3) = 21
   f = 15
4. Final Result:
   Fixed charge = ₹15, Extra charge = ₹3 per day

Q5. Meena Withdrew ₹2000 Using ₹50 And ₹100 Notes. She Got 25 Notes. Find Each Type.

Meena received 10 notes of ₹50 and 15 notes of ₹100. Use count and value equations.

1. Given Data:
   Total amount = ₹2000
   Total notes = 25
2. Equation Formation:
   ₹50 notes = x
   ₹100 notes = y
   x + y = 25
   50x + 100y = 2000
3. Calculation:
   Divide value equation by 50.
   x + 2y = 40
   Subtract x + y = 25.
   y = 15
   x = 10
4. Final Result:
   ₹50 notes = 10, ₹100 notes = 15

Pair Of Linear Equations Class 10 Extra Questions With Solutions

These pair of linear equations class 10 extra questions cover HOTS and application patterns. They stay within graphical method, substitution method, and elimination method.

Q1. Solve 101x + 102y = 304 And 102x + 101y = 305.

The solution is x = 2, y = 1. Add and subtract the equations for faster solving.

1. Given Data:
   101x + 102y = 304
   102x + 101y = 305
2. Calculation:
   Add both equations.
   203x + 203y = 609
   x + y = 3
3. Subtract:
   (102x + 101y) - (101x + 102y) = 305 - 304
   x - y = 1
4. Final Calculation:
   Add x + y = 3 and x - y = 1.
   2x = 4
   x = 2
   y = 1
5. Final Result:
   x = 2, y = 1

Q2. Two Rails Are Represented By x + 2y - 4 = 0 And 2x + 4y - 12 = 0. Will They Cross?

The rails will never cross. The equations represent parallel lines.

1. Given Data:
   x + 2y - 4 = 0
   2x + 4y - 12 = 0
2. Formula Used:
   a1/a2 = b1/b2 ≠ c1/c2
3. Calculation:
   a1/a2 = 1/2
   b1/b2 = 2/4 = 1/2
   c1/c2 = -4/(-12) = 1/3
   1/2 = 1/2 ≠ 1/3
4. Final Result:
   The rails will never cross

Q3. The Cost Of 2 Pencils And 3 Erasers Is ₹9. The Cost Of 4 Pencils And 6 Erasers Is ₹18. Find Each Cost.

A unique cost cannot be determined. Both equations represent the same line.

1. Given Data:
   2 pencils + 3 erasers = ₹9
   4 pencils + 6 erasers = ₹18
2. Equation Formation:
   Let pencil cost = ₹x
   Let eraser cost = ₹y
   2x + 3y = 9
   4x + 6y = 18
3. Calculation:
   The second equation is twice the first equation.
   Both equations represent the same line.
4. Final Result:
   A unique cost cannot be determined

Q4. Solve 2/x + 3/y = 4 And 4/x + 3/y = 7 By Reducing To Linear Form.

The solution is x = 2/3, y = 3. Substitute 1/x = p and 1/y = q.

1. Given Data:
   2/x + 3/y = 4
   4/x + 3/y = 7
2. Substitution:
   Let 1/x = p
   Let 1/y = q
3. Linear Equations:
   2p + 3q = 4
   4p + 3q = 7
4. Calculation:
   Subtract first equation from second.
   2p = 3
   p = 3/2
   x = 2/3
5. Find y:
   2(3/2) + 3q = 4
   3 + 3q = 4
   q = 1/3
   y = 3
6. Final Result:
   x = 2/3, y = 3

Q5. Five Years Hence, Jacob’s Age Will Be Three Times His Son’s. Five Years Ago, It Was Seven Times. Find Ages.

Jacob is 40 years old, and his son is 10 years old. Form equations from future and past conditions.

1. Given Data:
   Jacob’s age = x
   Son’s age = y
2. Equation Formation:
   Five years hence:
   x + 5 = 3(y + 5)
   x - 3y = 10
3. Second Condition:
   Five years ago:
   x - 5 = 7(y - 5)
   x - 7y = -30
4. Calculation:
   Subtract the second equation from the first.
   4y = 40
   y = 10
   x - 3(10) = 10
   x = 40
5. Final Result:
   Jacob = 40 years, Son = 10 years

Class 10 Linear Equations Previous Year Questions With Solutions

These class 10 linear equations previous year questions match recurring CBSE-style patterns. They test consistency, algebraic methods, and equation formation.

Q1. Find Whether 3x + y = 7 And 6x + 2y = 8 Represent Intersecting, Parallel, Or Coincident Lines.

The lines are parallel, so the pair has no solution. The first two ratios match, but the constant ratio differs.

1. Given Data:
   3x + y = 7
   6x + 2y = 8
2. Formula Used:
   a1/a2 = b1/b2 ≠ c1/c2
3. Calculation:
   a1/a2 = 3/6 = 1/2
   b1/b2 = 1/2
   c1/c2 = 7/8
   1/2 = 1/2 ≠ 7/8
4. Final Result:
   Parallel lines, no solution

Q2. Find k For Which 3x - 4y = 7 And kx + 3y - 5 = 0 Have No Solution.

The value of k is -9/4. Use the no-solution coefficient condition.

1. Given Data:
   3x - 4y = 7
   kx + 3y - 5 = 0
2. Formula Used:
   a1/a2 = b1/b2
3. Calculation:
   3/k = -4/3
   9 = -4k
4. Final Result:
   k = -9/4

Q3. A Father Is Three Times As Old As His Son. In 12 Years, He Will Be Twice As Old. Find Ages.

The father is 36 years old, and the son is 12 years old. Use present and future age conditions.

1. Given Data:
   Son’s present age = x
   Father’s present age = 3x
2. Equation Formation:
   After 12 years:
   3x + 12 = 2(x + 12)
3. Calculation:
   3x + 12 = 2x + 24
   x = 12
   Father’s age = 3x = 36
4. Final Result:
   Father = 36 years, Son = 12 years

Q4. Sum Of Digits Of A Two-Digit Number Is 9. Nine Times The Number Is Twice The Reversed Number. Find The Number.

The number is 18. Use place value to form the equations.

1. Given Data:
   Ten’s digit = x
   Unit’s digit = y
   x + y = 9
2. Equation Formation:
   Number = 10x + y
   Reversed number = 10y + x
   9(10x + y) = 2(10y + x)
3. Calculation:
   90x + 9y = 20y + 2x
   88x = 11y
   8x = y
4. Substitute:
   x + 8x = 9
   x = 1
   y = 8
5. Final Result:
   Number = 18

Pair Of Linear Equations Class 10 Questions And Answers: Most Repeated Variations

These pair of linear equations class 10 questions and answers cover recurring CBSE 2026 patterns. Students should practise consistency, graph reading, algebraic solving, and word problems.

Q1. What Are The Most Repeated Variations In Pair Of Linear Equations Class 10 Important Questions?

The most repeated variations are consistency check, graphical solution, substitution, elimination, digit problems, and age problems.

1. Consistency Pattern:
   Compare coefficient ratios.
2. Graph Pattern:
   Find intersection point or line type.
3. Word Problem Pattern:
   Convert conditions into two equations.
4. Final Result:
   These patterns cover most Chapter 3 board-style questions

Q2. How Should Students Solve Important Questions Class 10 Maths Chapter 3 With Solutions?

Students should identify the method before solving. The condition, coefficients, and wording decide the best method.

1. Check The Question Type:
   Consistency, graph, substitution, elimination, or word problem.
2. Choose The Method:
   Use graphical method, substitution method, or elimination method.
3. Avoid Deleted Content:
   Do not use cross-multiplication for CBSE 2026.
4. Final Result:
   Method selection decides the solution path

Useful Important Questions Class 10 Maths Links