Question :
(i) For what values of a and b, the following system of equations have infinite number of solutions ?
2x + 3y = 7
(a – b)x + (a + b)y = 3a + b – 2
(ii) For what value of k will the following pair of linear equations have no solution ?
3x + y = 1
(2k – 1)x + (k – 1)y = 2k + 1
Solution :
(i) The given linear equations can be written as
2x + 3y – 7 = 0
(a – b)x + (a + b)y – (3a + b – 2) = 0
Above equations is of the form
\(a_1x + b_1y + c_1\) = 0
\(a_2x + b_2y + c_2\) = 0
where,
\(a_1\) = 2, \(b_1\) = 3, \(c_1\) = -7
\(a_2\) = (a – b), \(b_2\) = (a + b), \(c_2\) = -(3a + b – 2)
For the linear equations to have infinitely many solutions,
\(a_1\over a_2\) = \(b_1\over b_2\) = \(c_1\over c_2\)
Here, \(a_1\over a_2\) = \(2\over a – b\), \(b_1\over b_2\) = \(3\over a + b\), \(c_1\over c_2\) = \(-7\over -(3a + b – 2)\) = \(7\over 3a + b – 2\)
\(\implies\) \(2\over a – b\) = \(3\over a + b\) = \(7\over 3a + b – 2\)
\(\implies\) \(2\over a – b\) = \(3\over a + b\) and \(3\over a + b\) = \(7\over 3a + b – 2\)
\(\implies\) 2a + 2b = 3a – 3b and 9a + 3b – 6 = 7a + 7b
\(\implies\) 2a – 3a = -3b – 2b and 9a – 7a = 7b – 3b + 6
\(\implies\) a = 5b ……(1) and a = 2b + 3 ….(2)
Solving equation (1) and (2), we get
5b = 2b + 3 \(\implies\) b = 1
Substituting the value of b in equation (1), we get
a = 5
Thus, a = 5 and b = 1.
(ii) The given linear equations can be written as
3x + y – 1 = 0
(2k – 1)x + (k – 1)y – (2k + 1) = 0
Above equations is of the form
\(a_1x + b_1y + c_1\) = 0
\(a_2x + b_2y + c_2\) = 0
where,
\(a_1\) = 3, \(b_1\) = 1, \(c_1\) = -1
\(a_2\) = (2k – 1), \(b_2\) = (k – 1), \(c_2\) = -(2k + 1)
For the linear equations to have no solutions,
\(a_1\over a_2\) = \(b_1\over b_2\) \(\ne\) \(c_1\over c_2\)
\(\implies\) \(3\over 2k -1\) = \(1\over k – 1\) and \(1\over k – 1\) \(\ne\) \(-1\over -(2k + 1\) and \(3\over 2k -1\) \(\ne\) \(1\over (2k + 1\)
\(\implies\) 3k – 3 = 2k – 1 and 2k + 1 \(\ne\) k – 1 and 6k + 3 \(\ne\) 2k – 1
\(\implies\) k = 2 and k\(\ne\) -2 and k \(\ne\) -1
Hence, the given linear equations has no solutions for k = 2 and k\(\ne\) -2 and k \(\ne\) -1
