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 ...
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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 ...