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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