When will (13x-ty=5) and (26x-10y=14) have a unique solution?
Answer and explanation
Correct answer: \(t\ne 5\)
A pair of linear equations has a unique solution when \(\frac{a_1}{a_2}\ne\frac{b_1}{b_2}\). Here, \(\frac{13}{26}=\frac12\) and \(\frac{-t}{-10}=\frac{t}{10}\). Therefore, a unique solution requires \(\frac12\ne\frac{t}{10}\), which gives \(t\ne5\). At \(t=5\), the coefficients on the left-hand sides are proportional, so the solution cannot be unique. Exam tip: include the negative signs while comparing the coefficients of \(y\).
Frequently asked questions
What is the correct answer to this question?
\(t\ne 5\)
Why is this the correct answer?
A pair of linear equations has a unique solution when \(\frac{a_1}{a_2}\ne\frac{b_1}{b_2}\). Here, \(\frac{13}{26}=\frac12\) and \(\frac{-t}{-10}=\frac{t}{10}\). Therefore, a unique solution requires \(\frac12\ne\frac{t}{10}\), which gives \(t\ne5\). At \(t=5\), the coefficients on the left-hand sides are proportional, so the solution cannot be unique. Exam tip: include the negative signs while comparing the coefficients of \(y\).
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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