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Pair of Linear Equations in Two Variables Solved Examples (10 Mathematics)

Pair of linear equations can be solved using substitution, elimination, or cross-multiplication methods. These examples demonstrate all three approaches an

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TL;DR: Pair of linear equations can be solved using substitution, elimination, or cross-multiplication methods. These examples demonstrate all three approach…

Written & reviewed by the Syllab.in Academic Team (CBSE/NCERT subject experts) · Updated Jul 23, 2026

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Pair of linear equations can be solved using substitution, elimination, or cross-multiplication methods. These examples demonstrate all three approaches an

Pair of Linear Equations in Two Variables — Solved Numerical Examples (Step by Step)

Example 1: Solve: 3x + 2y = 11 and 2x - y = 3 using elimination method.

Solution: Equation 1: 3x + 2y = 11
Equation 2: 2x - y = 3

Multiply Equation 2 by 2: 4x - 2y = 6

Add to Equation 1:
3x + 2y = 11
4x - 2y = 6
-----------
7x = 17
x = 17/7

Substitute in Equation 2:
2(17/7) - y = 3
34/7 - y = 3
y = 34/7 - 21/7 = 13/7

Example 2: Solve: x + y = 5 and 2x - 3y = 4 using substitution method.

Solution: From first equation: x = 5 - y

Substitute into second equation:
2(5 - y) - 3y = 4
10 - 2y - 3y = 4
10 - 5y = 4
-5y = -6
y = 6/5

Substitute back:
x = 5 - 6/5 = 25/5 - 6/5 = 19/5

Example 3: Solve: 4x + 3y = 18 and 2x - y = 2 using cross-multiplication.

Solution: Write as: 4x + 3y - 18 = 0 and 2x - y - 2 = 0

Using cross-multiplication formula:
x / (3 x (-2) - (-1) x (-18)) = y / ((-18) x 2 - 4 x (-2)) = 1 / (4 x (-1) - 3 x 2)

x / (-6 - 18) = y / (-36 + 8) = 1 / (-4 - 6)
x / (-24) = y / (-28) = 1 / (-10)

x = -24 / (-10) = 12/5
y = -28 / (-10) = 14/5

Example 4: The sum of two numbers is 25 and their difference is 7. Find the numbers.

Solution: Let the two numbers be x and y.

Equation 1: x + y = 25
Equation 2: x - y = 7

Add both equations:
2x = 32
x = 16

Substitute in Equation 1:
16 + y = 25
y = 9

Verification: 16 + 9 = 25 ✓ and 16 - 9 = 7 ✓

Example 5: A fraction becomes 4/5 when 1 is added to both numerator and denominator. It becomes 1/2 when 1 is subtracted from both. Find the fraction.

Solution: Let the fraction be x/y.

Condition 1: (x + 1) / (y + 1) = 4/5
5(x + 1) = 4(y + 1)
5x + 5 = 4y + 4
5x - 4y = -1 ... Equation 1

Condition 2: (x - 1) / (y - 1) = 1/2
2(x - 1) = y - 1
2x - 2 = y - 1
2x - y = 1 ... Equation 2

Multiply Equation 2 by 4: 8x - 4y = 4
Subtract Equation 1:
8x - 4y = 4
5x - 4y = -1
-----------
3x = 5
x = 5/3

From Equation 2: 2(5/3) - y = 1
10/3 - y = 1
y = 10/3 - 3/3 = 7/3

Example 6: For what values of a and b do 2x + 3y = 7 and 4x + 6y = 14 represent the same line?

Solution: For two equations a1 x + b1 y + c1 = 0 and a2 x + b2 y + c2 = 0 to represent the same line:
a1/a2 = b1/b2 = c1/c2

Rewrite equations in standard form:
2x + 3y - 7 = 0
4x + 6y - 14 = 0

Check ratios: 2/4 = 3/6 = 7/14
1/2 = 1/2 = 1/2 ✓

Yes, they represent the same line (infinitely many solutions).

Example 7: The cost of 5 pens and 3 notebooks is Rs 80. The cost of 3 pens and 5 notebooks is Rs 96. Find the cost of 1 pen and 1 notebook.

Solution: Let cost of 1 pen = x and cost of 1 notebook = y

Equation 1: 5x + 3y = 80
Equation 2: 3x + 5y = 96

Add both equations:
5x + 3y + 3x + 5y = 80 + 96
8x + 8y = 176
8(x + y) = 176
x + y = 22

To find individual values, subtract Equation 2 from Equation 1:
(5x + 3y) - (3x + 5y) = 80 - 96
2x - 2y = -16
x - y = -8

Solving x + y = 22 and x - y = -8:
2x = 14, so x = 7
y = 15

Tips

  • Elimination method works best when coefficients allow easy multiplication to make one variable cancel.
  • Substitution method is easiest when one equation already has a variable with coefficient 1.
  • Always verify your solution by substituting back into BOTH original equations.
  • If a1/a2 = b1/b2 ≠ c1/c2, the equations have no solution (parallel lines).

Frequently Asked Questions

When do a pair of equations have no solution?

A pair of linear equations has no solution when they represent parallel lines, which happens when a1/a2 = b1/b2 but c1/c2 is different.

Can a pair of linear equations have exactly 2 solutions?

No. A pair of linear equations can have either exactly one solution (intersecting lines), no solution (parallel lines), or infinitely many solutions (same line). It cannot have exactly 2 solutions.

More Mathematics Solved Examples

  • Real Numbers (HCF LCM Euclid)
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  • Linear Equations in Two Variables
  • Heron's Formula
  • Integrals (Indefinite and Definite)

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