When will (hx+12y=6) and (10x+15y=20) have a unique solution?
Answer and explanation
Correct answer: \(h\ne 8\)
For two linear equations \(a_1x+b_1y=c_1\) and \(a_2x+b_2y=c_2\), a unique solution exists when \(\frac{a_1}{a_2}\ne\frac{b_1}{b_2}\). Here, \(\frac{h}{10}\ne\frac{12}{15}=\frac45\), which gives \(h\ne 8\). Thus, option B is correct. If \(h=8\), the coefficients on the left-hand sides become proportional, so the pair cannot have a unique solution. Exam tip: for a unique solution, compare only \(a_1/a_2\) and \(b_1/b_2\).
Frequently asked questions
What is the correct answer to this question?
\(h\ne 8\)
Why is this the correct answer?
For two linear equations \(a_1x+b_1y=c_1\) and \(a_2x+b_2y=c_2\), a unique solution exists when \(\frac{a_1}{a_2}\ne\frac{b_1}{b_2}\). Here, \(\frac{h}{10}\ne\frac{12}{15}=\frac45\), which gives \(h\ne 8\). Thus, option B is correct. If \(h=8\), the coefficients on the left-hand sides become proportional, so the pair cannot have a unique solution. Exam tip: for a unique solution, compare only \(a_1/a_2\) and \(b_1/b_2\).
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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