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What will (q) be for (7x+qy=29) and (14x+10y=58) to have infinitely many solutions?

Answer and explanation

Correct answer: \(5\)

For infinitely many solutions, the two linear equations must represent the same line, so one equation must be a multiple of the other. Here, \(14x+10y=58\) must be twice the first equation. Thus, \(2q=10\), giving \(q=5\). With a close option such as \(q=4\), the coefficients of \(y\) would not be in the same ratio, so infinitely many solutions would not occur. Exam tip: for infinitely many solutions, check \(a_1/a_2=b_1/b_2=c_1/c_2\).

Related tags

Pair Of Linear EquationsInfinitely Many SolutionsDependent EquationsCoefficient RatiosParameter QConditions For SolvabilityPair Of Linear Equations In Two VariablesMathematics

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