Question
For given pair of equations, how many solutions are
possible? 4x + 6y = 16 and 8x + 12y = 32Solution
For a pair of linear equations a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, infinite solutions are possible when (a1/a2) = (b1/b2) = (c1/c2) In the given pair of equations, a1 = 4, a2 = 8, b1 = 6, b2 = 12, c1 = −16 and c2 = −32 Since, (4/8) = (6/12) = (−16/−32), therefore, infinite solutions are possible.
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