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.
Frequently asked questions
What is the correct answer to this question?
\(\frac{6}{18}=\frac{11}{33}=\frac{-47}{-141}\)
Why is this the correct answer?
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.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Pair of Linear Equations in Two Variables. Topic: Conditions for solvability.
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