For ((z+2)x+3y=5) and (6x+(z-1)y=7) to be non-unique, what are the possible values of (z)?
Answer and explanation
Correct answer: \(z=4,-5\)
A pair of linear equations does not have a unique solution when the determinant of the coefficient matrix is zero: \((z+2)(z-1)-3\times6=0\). Thus, \(z^2+z-20=0\,\), or \((z+5)(z-4)=0\). Therefore, \(z=4\) or \(z=-5\), so option C is correct. Exam tip: to test uniqueness, first set \(a_1b_2-a_2b_1=0\).
Frequently asked questions
What is the correct answer to this question?
\(z=4,-5\)
Why is this the correct answer?
A pair of linear equations does not have a unique solution when the determinant of the coefficient matrix is zero: \((z+2)(z-1)-3\times6=0\). Thus, \(z^2+z-20=0\,\), or \((z+5)(z-4)=0\). Therefore, \(z=4\) or \(z=-5\), so option C is correct. Exam tip: to test uniqueness, first set \(a_1b_2-a_2b_1=0\).
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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