Infinitely many solutions require (a + b)/3 = −2b/(−1) = (5a + 2b + 1)/14. Equating gives a = 5b and 5a − 26b = −1. Substituting a = 5b yields b = 1, which gives a = 5.
Class 9 Maths (Ganita Manjari part two) Chapter 13 questions answer
Class 9 Maths (Ganita Manjari part two) Two Variables, One Line Question Answer
For a system to possess infinitely many solutions, the ratio of coefficients must satisfy a₁/a₂ = b₁/b₂ = c₁/c₂. Comparing gives (a + b)/3 = 2b = (5a + 2b + 1)/14. From (a + b)/3 = 2b, we get a = 5b. From (5a + 2b + 1)/14 = 2b, we get 5a − 26b = −1. Substituting a = 5b gives 25b − 26b = −1, yielding b = 1 and a = 5.
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