Question
For given pair of equations, how many solutions are
possible? 3x + 4y = 15 and 6x + 8y = 10Solution
For a pair of linear equations a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, no solution is possible when (a1/a2) = (b1/b2) ≠(c1/c2) In the given pair of equations, a1 = 3, a2 = 6, b1 = 4, b2 = 8, c1 = −15 and c2 = −10 Since, (3/6) = (4/8) ≠(−15/−10), therefore, no solution is possible.
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