Which condition is correct for (qx+6y=18) and (5x+3y=12) to have a unique solution?
Answer and explanation
Correct answer: \(\frac{q}{5}\ne\frac{6}{3}\)
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, \(a_1=q, b_1=6, a_2=5, b_2=3\), so the required condition is \(\frac{q}{5}\ne\frac{6}{3}\), or \(q\ne10\). Option A gives equal ratios, so it does not represent a unique solution. Exam tip: for a unique solution, the ratios of the coefficients of the two variables must be unequal.
Frequently asked questions
What is the correct answer to this question?
\(\frac{q}{5}\ne\frac{6}{3}\)
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, \(a_1=q, b_1=6, a_2=5, b_2=3\), so the required condition is \(\frac{q}{5}\ne\frac{6}{3}\), or \(q\ne10\). Option A gives equal ratios, so it does not represent a unique solution. Exam tip: for a unique solution, the ratios of the coefficients of the two variables must be 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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