Which condition is correct for the equations (13x+py=52) and (6x+5y=24) to have a unique solution?
Answer and explanation
Correct answer: \(p \ne \frac{65}{6}\)
Two linear equations \(a_1x+b_1y+c_1=0\) and \(a_2x+b_2y+c_2=0\) have a unique solution when \(\frac{a_1}{a_2} \ne \frac{b_1}{b_2}\). Here, we require \(\frac{13}{6} \ne \frac{p}{5}\). Thus, \(65 \ne 6p\), or \(p \ne \frac{65}{6}\). Therefore, option A is correct. In option B, the two ratios become equal, so it does not give a unique solution. Exam tip: For a unique solution, compare the ratios of the coefficients of the two variables and ensure they are unequal.
Frequently asked questions
What is the correct answer to this question?
\(p \ne \frac{65}{6}\)
Why is this the correct answer?
Two linear equations \(a_1x+b_1y+c_1=0\) and \(a_2x+b_2y+c_2=0\) have a unique solution when \(\frac{a_1}{a_2} \ne \frac{b_1}{b_2}\). Here, we require \(\frac{13}{6} \ne \frac{p}{5}\). Thus, \(65 \ne 6p\), or \(p \ne \frac{65}{6}\). Therefore, option A is correct. In option B, the two ratios become equal, so it does not give a unique solution. Exam tip: For a unique solution, compare the ratios of the coefficients of the two variables and ensure they are unequal.
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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