Question : On comparing the ratios \(a_1\over a_2\), \(b_1\over b_2\) and \(c_1\over c_2\), find out whether the following pair of linear equations are consistent, or inconsistent.
(i) 3x + 2y = 5; 2x – 3y = 7
(ii) 2x – 3y = 8; 4x – 6y = 9
(iii) \(3\over 2\)x + \(5\over 3\)y = 7; 9x – 10y = 14
(iv) 5x – 3y = 11; -10x + 6y = -22
(v) \(4\over 3\)x + 2y = 8; 2x + 3y = 12
Solution :
(i) Rewrite the given equations as:
3x + 2y – 5 = 0; 2x – 3y – 7 = 0
\(a_1\) = 3, \(b_1\) = 2, \(c_1\) = 5
\(a_2\) = 2, \(b_2\) = -3, \(c_2\) = -7
\(a_1\over a_2\) = \(3\over 2\), \(b_1\over b_2\) = \(2\over -3\)
Thus, \(3\over 2\) \(\ne\) \(2\over -3\), i.e. \(a_1\over a_2\) \(\ne\) \(b_1\over b_2\)
Hence, the pair of linear equations is consistent.
(ii) Rewrite the given equations as:
2x – 3y – 8 = 0; 4x – 6y – 9 = 0
\(a_1\) = 2, \(b_1\) = -3, \(c_1\) = -8
\(a_2\) = 4, \(b_2\) = -6, \(c_2\) = -9
\(a_1\over a_2\) = \(2\over 4\) = \(1\over 2\), \(b_1\over b_2\) = \(-3\over -6\) = \(1\over 2\), \(c_1\over c_2\) = \(-8\over -9\) = \(8\over 9\)
Thus, \(1\over 2\) = \(1\over 2\) \(\ne\) \(8\over 9\), i.e. \(a_1\over a_2\) = \(b_1\over b_2\) \(\ne\) \(c_1\over c_2\)
Hence, the pair of linear equations is inconsistent.
(iii) Rewrite the given equations as:
\(3\over 2\)x + \(5\over 3\)y – 7 = 0; 9x – 10y – 14 = 0
\(a_1\) = \(3\over 2\), \(b_1\) = \(5\over 3\), \(c_1\) = -7
\(a_2\) = 9, \(b_2\) = -10, \(c_2\) = -14
\(a_1\over a_2\) = \(1\over 6\), \(b_1\over b_2\) = \(-1\over 6\)
Thus, \(1\over 6\) \(\ne\) \(-1\over 6\), i.e. \(a_1\over a_2\) \(\ne\) \(b_1\over b_2\)
Hence, the pair of linear equations is consistent.
(iv) Rewrite the given equations as:
5x – 3y – 11 = 0; -10x + 6y + 22 = 0
\(a_1\) = 5, \(b_1\) = -3, \(c_1\) = -11
\(a_2\) = -10, \(b_2\) = 6, \(c_2\) = 22
\(a_1\over a_2\) = \(-1\over 2\), \(b_1\over b_2\) = \(-1\over 2\), \(c_1\over c_2\) = \(-1\over 2\)
Thus, \(-1\over 2\) = \(-1\over 2\) = \(-1\over 2\), i.e. \(a_1\over a_2\) = \(b_1\over b_2\) = \(c_1\over c_2\)
Hence, the pair of linear equations is consistent (or dependent).
(v) Rewrite the given equations as:
\(4\over 3\)x + 2y – 8 = 0; 2x + 3y – 12 = 0
\(a_1\) = \(4\over 3\), \(b_1\) = 2, \(c_1\) = -8
\(a_2\) = 2, \(b_2\) = 3, \(c_2\) = -12
\(a_1\over a_2\) = \(2\over 3\), \(b_1\over b_2\) = \(2\over 3\), \(c_1\over c_2\) = \(2\over 3\)
Thus, \(2\over 3\) = \(2\over 3\) = \(2\over 3\), i.e. \(a_1\over a_2\) = \(b_1\over b_2\) = \(c_1\over c_2\)
Hence, the pair of linear equations is consistent (or dependent).
