Which of the following pairs of linear equations has unique solution, no solution, or infi nitely many solutions.

Solution:

(i) x – 3y – 3 = 0, 3x – 9y – 2 = 0

$\frac{a_{1}}{a_{2}}=\frac{1}{3}, \frac{h_{1}}{b_{2}}=\frac{-3}{-9}=\frac{1}{3}, \frac{c_{1}}{c_{2}}=\frac{3}{2}$

$\Rightarrow \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$

Hence, no solution.

(ii) 2x + y = 5 ...(i) and 3x + 2y = 8 ...(ii)

$\frac{a_{1}}{a_{2}} \neq \frac{h_{1}}{h_{2}}\left(\frac{a_{1}}{a_{2}}=\frac{2}{3}, \frac{h_{1}}{b_{2}}=\frac{1}{2}\right)$

Here, we have a unique solution. By cross multiplication, we have



$\Rightarrow \frac{x}{\{(1)(-8)-(2)(-5)\}}=\frac{y}{\{(-5)(3)-(-8)(2)\}}$

$=\frac{1}{\{(Q)(2)-(3)(1)\}}$

$\Rightarrow \frac{x}{(-8+10)}=\frac{y}{(-15+16)}=\frac{1}{(4-3)}$

$\Rightarrow \frac{x}{2}=\frac{y}{1}=\frac{1}{1} \Rightarrow \frac{x}{2}=\frac{1}{1}$ and $\frac{y}{1}=\frac{1}{1}$

$\Rightarrow x=2$ and $y=1$

(iii) 3x – 5y = 20 ........(i)

6x – 10y = 40 .......(ii)

$\frac{a_{1}}{a_{2}}=\frac{3}{6}=\frac{1}{2}, \frac{b_{1}}{b_{2}}=\frac{-5}{-10}=\frac{1}{2}, \frac{c_{1}}{c_{2}}=\frac{20}{40}=\frac{1}{2}$

$\therefore \quad \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$

Hence, infinite solutions

(iv) x – 3y – 7 = 0 ........(i)

3x – 3y – 15 = 0 ........(ii)

$\frac{a_{1}}{a_{2}}=\frac{1}{3}, \frac{b_{1}}{b_{2}}=1, \frac{c_{1}}{c_{2}}=\frac{7}{15}$

$\therefore \quad \frac{\mathbf{a}_{1}}{\mathbf{a}_{2}} \neq \frac{\mathbf{h}_{1}}{\mathbf{h}_{2}}$

Hence, unique solution

$\frac{x}{(-3)(-15)-(-3)(-7)}=\frac{y}{3 \times(-7)-1 \times(-15)}$

$=\frac{1}{1 \times(-3)-3(-3)}$

$\Rightarrow \frac{x}{45-21}=\frac{y}{-21+15}=\frac{1}{-3+9}$

$\Rightarrow \frac{x}{24}=\frac{y}{-6}=\frac{1}{6}$

$x=\frac{\mathbf{2 4}}{\mathbf{6}}=4, \quad y=\frac{-\mathbf{6}}{\mathbf{6}}=-1$