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2x + 3y = 7

(a - b) x + (a + b) y = 3a + b – 2

(ii) For which value of k will the following pair of linear equations have no solution?

3x + y = 1

(2k - 1) x + (k - 1) y = 2k + 1

Answer :

(i) \(2x + 3y = 7\)

\((a - b) x + (a + b) y = 3a + b – 2\)

On comparing these equations with the general form : \(ax^2 + bx + c\)

\(a_1 = 2, b_1 = 3, c_1 = -7,a_2 = (a-b), b_2 = (a + b), c_2 = 2 - b -3a \)

Here \({{a_1}\over {a_2}} = {{b_1}\over {b_2}} = {{c_1}\over {c_2}}\) since it has infinite solutions

=>\({{2}\over {a-b}} = {{3}\over {a+b}} = {{-7}\over {2 - b -3a}}\)

=>\({{2}\over {a-b}} = {{3}\over {a+b}}\) and \({{3}\over {a+b}} = {{-7}\over {2 - b -3a}}\)

=>\(2a + 2b = 3a - 3b\) and \( 6 - 3b - 9a = -7a - 7b\)

=>\(a = 5b\).....(i) and \(-2a = -4b – 6\)......(ii)

Putting (i) in (ii), we get

\(-2 (5b) = -4b – 6\)

=>\( -10b + 4b = -6\)

=> \( -6b = –6 - b = 1\)

Putting value of b in (i), we get

\(a = 5b = 5 (1) = 5\)

Therefore, a = 5 and b = 1

Hence, for the values a = 5 and b = 1, the given pair of equations have infinite solutions.

(ii) \(3x + y = 1\)

\((2k - 1) x + (k - 1) y = 2k + 1\)

On comparing these equations with the general form : \(ax^2 + bx + c\)

\(a_1 = 3, b_1 = 1, c_1 = -1,a_2 = (2k - 1), b_2 = (k - 1), c_2 = -(2k + 1) \)

Here \({{a_1}\over {a_2}} = {{b_1}\over {b_2}} \ne {{c_1}\over {c_2}}\) since it has no solutions

=>\({{3}\over {2k - 1}} = {{1}\over {k - 1}} \ne {{-1}\over {-2k-1}}\)

=>\({{3}\over {2k - 1}} = {{1}\over {k - 1}} \)

=>\(3(k - 1) = 2k - 1 \)

=>\(3k - 3 = 2k - 1\)

=>\(k = 2\)

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