On comparing the ratios $\frac{\mathbf{a}_{1}}{\mathbf{a}_{2}}, \frac{\mathbf{b}_{1}}{\mathbf{b}_{2}}$ and $\frac{\mathbf{c}_{1}}{\mathbf{c}_{\mathbf{z}}}$
Question.
On comparing the ratios $\frac{\mathbf{a}_{1}}{\mathbf{a}_{2}}, \frac{\mathbf{b}_{1}}{\mathbf{b}_{2}}$ and $\frac{\mathbf{c}_{1}}{\mathbf{c}_{2}}$, find out whether the following pairs of linear equations are consistent, or inconsistent.
(i) $3 x+2 y=5 ; 2 x-3 y=7$
(ii) $2 x-3 y=8 ; 4 x-6 y=9$
(iii) $\frac{\mathbf{3}}{\mathbf{z}} \mathrm{x}+\frac{\mathbf{5}}{\mathbf{3}} \mathrm{y}=7 ; 9 \mathrm{x}-10 \mathrm{y}=14$
(iv) $5 x-3 y=11 ;-10 x+6 y=-22$
(v) $\frac{\mathbf{4}}{\mathbf{3}} x+2 y=8 ; 2 x+3 y=12$
On comparing the ratios $\frac{\mathbf{a}_{1}}{\mathbf{a}_{2}}, \frac{\mathbf{b}_{1}}{\mathbf{b}_{2}}$ and $\frac{\mathbf{c}_{1}}{\mathbf{c}_{2}}$, find out whether the following pairs of linear equations are consistent, or inconsistent.
(i) $3 x+2 y=5 ; 2 x-3 y=7$
(ii) $2 x-3 y=8 ; 4 x-6 y=9$
(iii) $\frac{\mathbf{3}}{\mathbf{z}} \mathrm{x}+\frac{\mathbf{5}}{\mathbf{3}} \mathrm{y}=7 ; 9 \mathrm{x}-10 \mathrm{y}=14$
(iv) $5 x-3 y=11 ;-10 x+6 y=-22$
(v) $\frac{\mathbf{4}}{\mathbf{3}} x+2 y=8 ; 2 x+3 y=12$
Solution:
(i) $3 x+2 y-5=0$ ...(i)
$2 x-3 y-7=0$ ...(ii)
$\frac{a_{1}}{a_{2}}=\frac{3}{2} ; \frac{h_{1}}{b_{2}}=\frac{2}{-3}=-\frac{2}{3}$
$\Rightarrow \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$
The equations have a unique solution.
Hence, consistent.
(ii) $2 x-3 y=8$ ...(i)
$4 x-6 y=9$ ...(ii)
$\frac{a_{1}}{a_{2}}=\frac{2}{4}, \frac{b_{1}}{b_{2}}=\frac{-3}{-6}, \frac{c_{1}}{c_{2}}=\frac{8}{9}$
$\Rightarrow \frac{\mathbf{a}_{1}}{\mathbf{a}_{2}}=\frac{\mathbf{b}_{1}}{\mathbf{b}_{2}} \neq \frac{\mathbf{c}_{1}}{\mathbf{c}_{2}}$
The equations have no solution.
Hence inconsistent.
(iii) $\frac{\mathbf{3}}{\mathbf{2}} \mathbf{x}+\frac{\mathbf{5}}{\mathbf{3}} \mathbf{y}=7$ .....(i)
$9 x-10 y=14$ .....(ii)
$\frac{a_{1}}{a_{2}}=\frac{3 / 2}{9}=\frac{1}{6}, \frac{b_{1}}{b_{2}}=\frac{5 / 3}{-10}=\frac{-1}{6}$
$\Rightarrow \frac{\mathbf{a}_{1}}{\mathbf{a}_{2}} \neq \frac{\mathbf{b}_{1}}{\mathbf{b}_{2}}$
The equations have infinite solutions.
Hence, consistent.
(iv) $5 x-3 y=11$ ....(i)
$-10 x+6 y=-22$ ....(ii)
$\frac{a_{1}}{a_{2}}=\frac{5}{-10}=\frac{-1}{2}, \frac{b_{1}}{b_{2}}=\frac{-3}{6}=\frac{-1}{2}$
$\frac{c_{1}}{c_{2}}=\frac{11}{-22}=\frac{-1}{2}$
$\Rightarrow \frac{\mathbf{a}_{1}}{\mathbf{a}_{2}}=\frac{\mathbf{b}_{1}}{\mathbf{b}_{2}}=\frac{\mathbf{c}_{1}}{\mathbf{c}_{2}}$
The equations have infinite solutions.
Hence, consistent.
(v) $\frac{4}{3} \mathbf{x}+\mathbf{2} \mathbf{y}=8$ .....(i)
$2 x+3 y=12$ ....(ii)
$\Rightarrow \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$
The equations have infinite solutions.
Hence, consistent.
(i) $3 x+2 y-5=0$ ...(i)
$2 x-3 y-7=0$ ...(ii)
$\frac{a_{1}}{a_{2}}=\frac{3}{2} ; \frac{h_{1}}{b_{2}}=\frac{2}{-3}=-\frac{2}{3}$
$\Rightarrow \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$
The equations have a unique solution.
Hence, consistent.
(ii) $2 x-3 y=8$ ...(i)
$4 x-6 y=9$ ...(ii)
$\frac{a_{1}}{a_{2}}=\frac{2}{4}, \frac{b_{1}}{b_{2}}=\frac{-3}{-6}, \frac{c_{1}}{c_{2}}=\frac{8}{9}$
$\Rightarrow \frac{\mathbf{a}_{1}}{\mathbf{a}_{2}}=\frac{\mathbf{b}_{1}}{\mathbf{b}_{2}} \neq \frac{\mathbf{c}_{1}}{\mathbf{c}_{2}}$
The equations have no solution.
Hence inconsistent.
(iii) $\frac{\mathbf{3}}{\mathbf{2}} \mathbf{x}+\frac{\mathbf{5}}{\mathbf{3}} \mathbf{y}=7$ .....(i)
$9 x-10 y=14$ .....(ii)
$\frac{a_{1}}{a_{2}}=\frac{3 / 2}{9}=\frac{1}{6}, \frac{b_{1}}{b_{2}}=\frac{5 / 3}{-10}=\frac{-1}{6}$
$\Rightarrow \frac{\mathbf{a}_{1}}{\mathbf{a}_{2}} \neq \frac{\mathbf{b}_{1}}{\mathbf{b}_{2}}$
The equations have infinite solutions.
Hence, consistent.
(iv) $5 x-3 y=11$ ....(i)
$-10 x+6 y=-22$ ....(ii)
$\frac{a_{1}}{a_{2}}=\frac{5}{-10}=\frac{-1}{2}, \frac{b_{1}}{b_{2}}=\frac{-3}{6}=\frac{-1}{2}$
$\frac{c_{1}}{c_{2}}=\frac{11}{-22}=\frac{-1}{2}$
$\Rightarrow \frac{\mathbf{a}_{1}}{\mathbf{a}_{2}}=\frac{\mathbf{b}_{1}}{\mathbf{b}_{2}}=\frac{\mathbf{c}_{1}}{\mathbf{c}_{2}}$
The equations have infinite solutions.
Hence, consistent.
(v) $\frac{4}{3} \mathbf{x}+\mathbf{2} \mathbf{y}=8$ .....(i)
$2 x+3 y=12$ ....(ii)
$\Rightarrow \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$
The equations have infinite solutions.
Hence, consistent.