How many solutions will (6x-5y=14) and (12x-10y=31) have?
Answer and explanation
Correct answer: No solution
Multiplying the left-hand side of the first equation by 2 gives \(12x-10y\), but the right-hand side would be \(2\times14=28\), not 31. Thus, the coefficient ratios are equal: \(\frac{6}{12}=\frac{-5}{-10}=\frac{1}{2}\), whereas the ratio of the constants is \(\frac{14}{31}\), which is different. Therefore, the two lines are distinct and parallel, so the pair has no solution. Exam tip: If \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}\), the pair of linear equations has no solution.
Frequently asked questions
What is the correct answer to this question?
No solution
Why is this the correct answer?
Multiplying the left-hand side of the first equation by 2 gives \(12x-10y\), but the right-hand side would be \(2\times14=28\), not 31. Thus, the coefficient ratios are equal: \(\frac{6}{12}=\frac{-5}{-10}=\frac{1}{2}\), whereas the ratio of the constants is \(\frac{14}{31}\), which is different. Therefore, the two lines are distinct and parallel, so the pair has no solution. Exam tip: If \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}\), the pair of linear equations has no solution.
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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