Which ratio relation is correct for (4x+9y-31=0) and (12x+27y-93=0)?
Answer and explanation
Correct answer: \(\frac{4}{12}=\frac{9}{27}=\frac{-31}{-93}\)
For the first equation, a₁=4, b₁=9, c₁=-31, and for the second equation, a₂=12, b₂=27, c₂=-93. Therefore, \(\frac{a_1}{a_2}=\frac{4}{12}=\frac{1}{3}\), \(\frac{b_1}{b_2}=\frac{9}{27}=\frac{1}{3}\), and \(\frac{c_1}{c_2}=\frac{-31}{-93}=\frac{1}{3}\). Since all three ratios are equal, the two lines are coincident and the pair has infinitely many solutions. Hence, option C is correct; options A and D incorrectly make one ratio unequal, while B incorrectly states that the first two ratios are unequal. Exam tip: When comparing the three ratios, check \(c_1/c_2\) with its signs included.
Frequently asked questions
What is the correct answer to this question?
\(\frac{4}{12}=\frac{9}{27}=\frac{-31}{-93}\)
Why is this the correct answer?
For the first equation, a₁=4, b₁=9, c₁=-31, and for the second equation, a₂=12, b₂=27, c₂=-93. Therefore, \(\frac{a_1}{a_2}=\frac{4}{12}=\frac{1}{3}\), \(\frac{b_1}{b_2}=\frac{9}{27}=\frac{1}{3}\), and \(\frac{c_1}{c_2}=\frac{-31}{-93}=\frac{1}{3}\). Since all three ratios are equal, the two lines are coincident and the pair has infinitely many solutions. Hence, option C is correct; options A and D incorrectly make one ratio unequal, while B incorrectly states that the first two ratios are unequal. Exam tip: When comparing the three ratios, check \(c_1/c_2\) with its signs included.
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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