Question

Without actually solving the simultaneous equations given below, decide which simultaneous equations have unique solutions, no solution or infinitely many solutions.

$\left(\mathrm{v}\right)\frac{x-2y}{3}=1;2x-4y=\frac{9}{2}$

Answer 1

$$\begin{gathered} \frac{x-2y}{3}=1 \\ 2x-4y=\frac{9}{2} \end{gathered}$$

These equations can be written as follows:

$$\begin{gathered} x-2y=3 \\ 2x-4y=\frac{9}{2} \end{gathered}$$

Comparing the above equations with the equations $a_{1}x+b_{1}y=c_1\mathrm{and}{a}_{2}x+b_{2}y=c_2,\mathrm{we}\mathrm{get}$:

$$\begin{gathered} a_1=1b_1=-2c_1=3 \\ {a}_2=2b_2=-4c_2=\frac{9}{2} \\ \therefore \frac{a_1}{a_2}=\frac{1}{2};\frac{b_1}{b_2}=\frac{-2}{-4}=\frac{1}{2}\mathrm{and}\frac{c_1}{c_2}=\frac{3}{{\displaystyle \raisebox{1ex}{$9$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}=\frac{2}{3} \end{gathered}$$

$$\begin{gathered} \mathrm{i}.\mathrm{e}.,\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2} \\ \therefore \mathrm{The}\mathrm{given}\mathrm{equations}\mathrm{have}\mathrm{no}\mathrm{solutions}. \end{gathered}$$