Which condition is correct for (5x+2y=18) and (15x+ky=54) to have a unique solution?
Answer and explanation
Correct answer: \(k\ne 6\)
For two linear equations to have a unique solution, the condition is \(\frac{a_1}{a_2}\ne\frac{b_1}{b_2}\). Here, \(\frac{5}{15}\ne\frac{2}{k}\); cross-multiplication gives \(5k\ne30\), so \(k\ne6\). Therefore, option B is correct. In option A, \(k=6\) makes the corresponding coefficients proportional, so the equations do not have a unique solution. Exam tip: a unique solution requires unequal ratios of the corresponding coefficients.
Frequently asked questions
What is the correct answer to this question?
\(k\ne 6\)
Why is this the correct answer?
For two linear equations to have a unique solution, the condition is \(\frac{a_1}{a_2}\ne\frac{b_1}{b_2}\). Here, \(\frac{5}{15}\ne\frac{2}{k}\); cross-multiplication gives \(5k\ne30\), so \(k\ne6\). Therefore, option B is correct. In option A, \(k=6\) makes the corresponding coefficients proportional, so the equations do not have a unique solution. Exam tip: a unique solution requires unequal ratios of the corresponding coefficients.
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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