Medium
Find two solutions for each of the following equations:
$(i)$ $4x + 3y = 12$
$(ii)$ $2x + 5y = 0$
$(iii)$ $3y + 4 = 0$
$(i)$ $4x + 3y = 12$
$(ii)$ $2x + 5y = 0$
$(iii)$ $3y + 4 = 0$
(N/A) $(i)$ For $4x + 3y = 12$,if we take $x = 0$,then $3y = 12$,which gives $y = 4$. So,$(0, 4)$ is a solution. If we take $y = 0$,then $4x = 12$,which gives $x = 3$. So,$(3, 0)$ is another solution.
$(ii)$ For $2x + 5y = 0$,if we take $x = 0$,then $5y = 0$,which gives $y = 0$. So,$(0, 0)$ is a solution. If we take $x = 1$,then $2(1) + 5y = 0$,which gives $5y = -2$,so $y = -\frac{2}{5}$. Thus,$(1, -\frac{2}{5})$ is another solution.
$(iii)$ For $3y + 4 = 0$,we can write this as $0x + 3y = -4$. For any value of $x$,$y$ remains $-\frac{4}{3}$. If we take $x = 0$,$y = -\frac{4}{3}$. If we take $x = 1$,$y = -\frac{4}{3}$. Thus,two solutions are $(0, -\frac{4}{3})$ and $(1, -\frac{4}{3})$.
$(ii)$ For $2x + 5y = 0$,if we take $x = 0$,then $5y = 0$,which gives $y = 0$. So,$(0, 0)$ is a solution. If we take $x = 1$,then $2(1) + 5y = 0$,which gives $5y = -2$,so $y = -\frac{2}{5}$. Thus,$(1, -\frac{2}{5})$ is another solution.
$(iii)$ For $3y + 4 = 0$,we can write this as $0x + 3y = -4$. For any value of $x$,$y$ remains $-\frac{4}{3}$. If we take $x = 0$,$y = -\frac{4}{3}$. If we take $x = 1$,$y = -\frac{4}{3}$. Thus,two solutions are $(0, -\frac{4}{3})$ and $(1, -\frac{4}{3})$.
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