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.

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).

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