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

Question.

Which of the following pairs of linear equations has unique solution, no solution, or infi nitely many solutions. In case there is a unique solution, find it by using cross multiplication method.

(i) x – 3y – 3 = 0

3x – 9y – 2 = 0

(ii) 2x + y = 5

3x + 2y = 8

(iii) 3x – 5y = 20

6x – 10y = 40

(iv) x – 3 y – 7 = 0

3x – 3y – 15 = 0

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

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

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

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

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

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