Which condition is correct for the equations (6x+ky=24) and (3x+5y=17) to have a unique solution?
Answer and explanation
Correct answer: \(k\ne10\)
Two linear equations have a unique solution when \(\frac{a_1}{a_2}\ne\frac{b_1}{b_2}\). Here, \(a_1=6, a_2=3, b_1=k, b_2=5\), so \(\frac{6}{3}\ne\frac{k}{5}\), or \(2\ne\frac{k}{5}\). This gives \(k\ne10\). If \(k=10\), the ratios of the coefficients of \(x\) and \(y\) would be equal, so the equations would not have a unique solution. Exam tip: For a unique solution, check that the ratios of corresponding coefficients are unequal.
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What is the correct answer to this question?
\(k\ne10\)
Why is this the correct answer?
Two linear equations have a unique solution when \(\frac{a_1}{a_2}\ne\frac{b_1}{b_2}\). Here, \(a_1=6, a_2=3, b_1=k, b_2=5\), so \(\frac{6}{3}\ne\frac{k}{5}\), or \(2\ne\frac{k}{5}\). This gives \(k\ne10\). If \(k=10\), the ratios of the coefficients of \(x\) and \(y\) would be equal, so the equations would not have a unique solution. Exam tip: For a unique solution, check that the ratios of corresponding coefficients 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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