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Ex 3.2, 1 (iv) - Chapter 3 Class 10 Pair of Linear Equations in Two Variables
Last updated at August 17, 2026 by Teachoo
Chapter 3 Class 10 Pair of Linear Equations in Two Variables
Serial order wise
- Ex 3.1
-
Ex 3.2
- Ex 3.3
- Examples
- Case Based Questions (MCQ)
- MCQs from NCERT Exemplar
- Graph of pair of Linear Equations
- Cross Multiplication Method
- Equations reduced to pair of linear equations
- Important Linear Equation Questions
- Teachoo Questions - MCQs
- Teachoo Questions - Mix
- Teachoo Questions - Assertion Reasoning
- Teachoo Questions - Case Based
Transcript
Ex 3.2, 1 Solve the following pair of linear equations by the substitution method. (iv) 0.2x + 0.3y = 1.3 0.4x + 0.5y = 2.3 0.2x + 0.3y = 1.3 0.4x + 0.5y = 2.3 From (1) 0.2x + 0.3y = 1.3 Multiplying both side by 10 (0.2x + 0.3y ) ×10=1.3×10 2x + 3y = 13 2x = 13 – 3y x = (𝟏𝟑 − 𝟑𝒚)/𝟐 Putting value of x in (2) 0.4x + 0.5y = 2.3 0.4((13 − 3𝑦)/2) + 0.5y = 2.3 Multiplying both sides by 10 10 × 0.4((13 − 3𝑦)/2) + 10 × 0.5y = 10 × 2.3 4((13 − 3𝑦)/2)+5𝑦=23 2(13 – 3y) + 5y = 23 26 – 6y + 5y = 23 26 – y = 23 26 – 23 = y 3 = y y = 3 Putting y = 3 equation in (1) 0.2x + 0.3y = 1.3 0.2x + 0.3 (3) = 1.3 0.2x + 0.9 = 1.3 0.2x = 1.3 – 0.9 0.2x = 0.4 x = 0.4/0.2 x = 2 Hence x = 2 and y = 3 is the solution for the equation
