What is the correct ratio relation for the equations (8x-3y+22=0) and (16x-6y+47=0)?
Answer and explanation
Correct answer: \(\frac{8}{16}=\frac{-3}{-6}\ne\frac{22}{47}\)
For the first equation, \(a_1=8, b_1=-3, c_1=22\), and for the second equation, \(a_2=16, b_2=-6, c_2=47\). Thus, \(\frac{a_1}{a_2}=\frac{8}{16}=\frac{1}{2}\) and \(\frac{b_1}{b_2}=\frac{-3}{-6}=\frac{1}{2}\), whereas \(\frac{c_1}{c_2}=\frac{22}{47}\), which is not equal to \(\frac{1}{2}\). Therefore, \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}\), so the 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?
\(\frac{8}{16}=\frac{-3}{-6}\ne\frac{22}{47}\)
Why is this the correct answer?
For the first equation, \(a_1=8, b_1=-3, c_1=22\), and for the second equation, \(a_2=16, b_2=-6, c_2=47\). Thus, \(\frac{a_1}{a_2}=\frac{8}{16}=\frac{1}{2}\) and \(\frac{b_1}{b_2}=\frac{-3}{-6}=\frac{1}{2}\), whereas \(\frac{c_1}{c_2}=\frac{22}{47}\), which is not equal to \(\frac{1}{2}\). Therefore, \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}\), so the 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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