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

Question.

(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 = 3 a + 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.

Solution:

(i) 2x + 3y – 7 = 0 ...(i)

(a – b) x + (a + b) y – (3 a + b – 2) = 0 ...(ii)

For infinite number of solutions, we have

$\frac{\mathbf{a}-\mathbf{b}}{\mathbf{2}}=\frac{\mathbf{a}+\mathbf{b}}{\mathbf{3}}=\frac{\mathbf{3 a}+\mathbf{b}-\mathbf{2}}{\mathbf{7}}$

For first and second, we have

$\frac{\mathbf{a}-\mathbf{b}}{\mathbf{2}}=\frac{\mathbf{a}+\mathbf{b}}{\mathbf{3}}$ or $3 \mathbf{a}-3 \mathbf{b}=2 \mathbf{a}+2 \mathrm{~b}$

or a = 5b ...(i)

From second and third, we have

$\frac{\mathbf{a}+\mathbf{b}}{\mathbf{3}}=\frac{\mathbf{3 a}+\mathbf{b}-\mathbf{2}}{\mathbf{7}}$

or 7a + 7b = 9a + 3b – 6 or 4b = 2a – 6

or 2b = a – 3 ...(ii)

From (i) and (ii), eliminating a,

2b = 5b – 3

b = 1

Substituting b = 1 in (i), we get a = 5

(ii) 3x + y = 1

(2k – 1)x + (k – 1)y = 2k + 1

For no, solution, $\frac{\mathbf{a}_{\mathbf{1}}}{\mathbf{a}_{\mathbf{2}}}=\frac{\mathbf{b}_{\mathbf{1}}}{\mathbf{b}_{\mathbf{2}}} \neq \frac{\mathbf{c}_{\mathbf{1}}}{\mathbf{c}_{\mathbf{2}}}$

$\frac{\mathbf{3}}{2 \mathbf{k}-\mathbf{1}}=\frac{1}{\mathbf{k}-\mathbf{1}} \neq \frac{1}{2 \mathbf{k}+1}$

So, $\frac{\mathbf{3}}{\mathbf{2 k - 1}}=\frac{\mathbf{1}}{\mathbf{k}-\mathbf{1}} \quad \& \frac{\mathbf{1}}{\mathbf{k}-\mathbf{1}} \neq \frac{\mathbf{1}}{\mathbf{2 k}+\mathbf{1}}$

3(k – 1) = 2k – 1 2k + 1 = k – 1

k = 2 k=–2