What is the correct ratio relation for the equations (16x-9y+55=0) and (32x-18y+113=0)?
Answer and explanation
Correct answer: \(\frac{16}{32}=\frac{-9}{-18}\ne\frac{55}{113}\)
Here, \(a_1=16, b_1=-9, c_1=55\) and \(a_2=32, b_2=-18, c_2=113\). We get \(\frac{a_1}{a_2}=\frac{16}{32}=\frac{1}{2}\) and \(\frac{b_1}{b_2}=\frac{-9}{-18}=\frac{1}{2}\), whereas \(\frac{c_1}{c_2}=\frac{55}{113}\ne\frac{1}{2}\). Thus, \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}\), so the two lines are parallel and the pair has no solution. Exam tip: when the first two coefficient ratios are equal but the constant-term ratio is different, the pair is inconsistent.
Frequently asked questions
What is the correct answer to this question?
\(\frac{16}{32}=\frac{-9}{-18}\ne\frac{55}{113}\)
Why is this the correct answer?
Here, \(a_1=16, b_1=-9, c_1=55\) and \(a_2=32, b_2=-18, c_2=113\). We get \(\frac{a_1}{a_2}=\frac{16}{32}=\frac{1}{2}\) and \(\frac{b_1}{b_2}=\frac{-9}{-18}=\frac{1}{2}\), whereas \(\frac{c_1}{c_2}=\frac{55}{113}\ne\frac{1}{2}\). Thus, \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}\), so the two lines are parallel and the pair has no solution. Exam tip: when the first two coefficient ratios are equal but the constant-term ratio is different, the pair is inconsistent.
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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