For what values of a and b, the following system of equations have infinite number of solutions ?

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

Leave a Comment

Your email address will not be published.