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निम्नलिखित समीकरणों में से प्रत्येक समीकरण के दो हल ज्ञात कीजिए
$(i)$ $4 x+3 y=12$
$(ii)$ $2 x+5 y=0$
$(iii)$ $3 y+4=0$
Solution
$ (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 $\left(1,-\frac{4}{3}\right)$.
Similar Questions
निम्नलिखित आलेखों में से प्रत्येक आलेख के लिए दिए गए विकल्पों से सही समीकरण का चयन कीजिए
| आकृति $(i)$ के लिए | आकृति $(ii)$ के लिए |
| $(i)$ $y=x$ | $(i)$ $y=x+2$ |
| $(ii)$ $x+y=0$ | $(ii)$ $y=x-2$ |
| $(iii)$ $y=2 x$ | $(iii)$ $y=-x+2$ |
| $(iv)$ $2+3 y=7 x$ | $(iv)$ $x+2 y=6$ |
