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# (i) For which values of a and b does the following pair of linear equations have an infinite number of solutions? 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

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