Find two solutions for each of the following:
(i) 3x + 4y = 12
(ii) 3x + 5y = 0
(iii) 4y + 5 = 0

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Solution

(i) $3 x + 4 y = 12$
Or, $y = \frac{12 - 3 x}{4}$

When $x = 0$ , $y = \frac{12}{4} = 3$ .
When $x = 4$ , $y = \frac{0}{4} = 0$ .
Hence , the two solutions are $(0, 3)$ and $(4, 0)$ .

(ii) $3 x + 5 y = 0$
Or, $y = - \frac{3}{5} x$
When $x = 5$ , $y = - \frac{3}{5} \times 5 = - 3$ .

When $x = 10$ , $y = - \frac{3}{5} \times 10 = - 6$ .
Hence, the two solution are $(5, - 3)$ and $(10, - 6)$ .

(iii) $4 y + 5 = 0$
Or, $y = - \frac{5}{4}$
It is independent of $x$ .
When $x = 1$ , $y = - \frac{5}{4}$ .
When $x = 2$ , $y = - \frac{5}{4}$ .
Hence, the two solutions are $(1, - \frac{5}{4})$ and $(2, - \frac{5}{4})$ .

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