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i) \(2x + 3y = 9\overline{.35}\)

ii)\( x - \frac{y}{5} - 10 = 0 \)

iii) -2x + 3y = 6

iv) x = 3y

v) 2x = -5y

vi) 3x + 2 = 0

vii) y - 2 = 0

viii)5 = 2x

Answer :

i)\(2x + 3y = 9\overline{.35}\)

Rearranging the terms we get,

\(2x + 3y - 9\overline{.35} = 0\)

On comparing with ax + by + c = 0, then the values of a = 2, b = 3 and c = -\(9\overline{.35}\)

ii)\( x - \frac{y}{5} - 10 = 0 \)

Rearranging the terms we get,

\( x - \frac{y}{5} - 10 = 0 \)

On comparing with ax + by + c = 0, then the values of a = 1, b = \( \frac{-1}{5} \) and c = -10

iii)-2x + 3y = 6

Rearranging the terms we get,

-2x + 3y - 6 = 0

On comparing with ax + by + c = 0, then the values of a = -2, b = 3 and c = -6

iv)x = 3y

Rearranging the terms we get,

x - 3y + 0= 0

On comparing with ax + by + c = 0, then the values of a = 1, b = -3 and c = 0

v)2x = -5y

Rearranging the terms we get,

2x + 5y + 0 = 0

On comparing with ax + by + c = 0, then the values of a = 2, b = 5 and c = 0

vi)3x + 2 = 0

Rearranging the terms we get,

3x + (0)y + 2 = 0

On comparing with ax + by + c = 0, then the values of a = 3, b = 0 and c = 2

vii)y - 2 = 0

Rearranging the terms we get,

(0)x + y - 2 = 0

On comparing with ax + by + c = 0, then the values of a = 0, b = 1 and c = -2

viii)5 = 2x

Rearranging the terms we get,

2x + (0)y - 5 = 0

On comparing with ax + by + c = 0, then the values of a = 2, b = 0 and c = -5

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