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Which ratio relation is correct for the equations (6x+11y-47=0) and (18x+33y-141=0)?

Answer and explanation

Correct answer: \(\frac{6}{18}=\frac{11}{33}=\frac{-47}{-141}\)

Here, \(\frac{6}{18}=\frac{1}{3}\), \(\frac{11}{33}=\frac{1}{3}\), and \(\frac{-47}{-141}=\frac{1}{3}\). Thus, \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\), so the two lines are coincident and the pair has infinitely many solutions. In an exam, retain the signs of the constant terms while forming the ratios.

Related tags

Linear EquationsSolvability ConditionsCoincident LinesInfinite SolutionsRatio ComparisonConditions For SolvabilityPair Of Linear Equations In Two VariablesMathematics

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