Find two solutions for each of the following:
(i) 3x + 4y = 12 (ii) 3x + 5y = 0
(iii) 4y + 5 = 0
(i) 3x + 4y = 12
⇒ 4y = 12 – 3x
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Let x = 4,
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⇒ y = 0
Therefore, (4, 0) is a solution
Let x = – 4,
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⇒ y = 6
Therefore, ( – 4, 6) is a solution
(ii) 3x + 5y = 0
⇒ 5y = 0 – 3x
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Let x = 5,
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⇒ y = – 3
Therefore, (5, – 3) is a solution
Let x = – 5,
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⇒ y = 3
Therefore, ( – 5, 3) is a solution
(iii) 4y + 5 = 0
⇒ 4y = 0 – 5
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Let x = 1,
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Therefore,
is a solution.
Let x = – 1,
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Therefore,
is a solution.
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