निम्नलिखित समीकरणों में से प्रत्येक समीकरण के दो हल ज्ञात कीजिए

$(i)$ $4 x+3 y=12$

$(ii)$ $2 x+5 y=0$

$(iii)$ $3 y+4=0$

$ (i)$ Taking $x=0,$ we get $3 y=12,$ i.e., $y=4 .$ So, $(0,\,4)$ is a solution of the given equation. Similarly, by taking $y=0,$ we get $x=3 .$ Thus, $(3,\,0) $ is also a solution.

$(ii)$ Taking $x=0,$ we get $5 y=0,$ i.e., $y=0 .$ So $(0,\,0)$ is a solution of the given equation.

Now, if you take $y=0,$ you again get $(0,\,0)$ as a solution, which is the same as the earlier one. To get another solution, take $x=1,$ say. Then you can check that the corresponding value of $y$ is $-\frac{2}{5} \cdot$ So $\left(1,-\frac{2}{5}\right)$ is another solution of $2 x+5 y=0$

$(iii)$ Writing the equation $3 y+4=0$ as $0 . x+3 y+4=0,$ you will find that $y=-\frac{4}{3}$ for any value of $x$. Thus, two solutions can be given as $\left(0,-\frac{4}{3}\right)$ and $\left(1,-\frac{4}{3}\right)$.