If x represents the encouragement points given to a punctual student and y represents the points given for irregularity of a student, then find how many points are fixed for each, if x and y are connected by the relation, $\frac{(1 + i) x - 2 i}{3 + i} + \frac{(2 - 3 i) y + i}{3 - i} = i ?$

Students getting y point are lacking which value system ?

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Solution

Only one point at x = 3 and y = -1

Students getting y point are lacking regularity in their work.

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Similar questions

If $\frac{(1 + i) x - 2 i}{3 + i} + \frac{(2 - 3 i) y + i}{3 - i} = i$ , then the real values of x and y are given by :

x

y

\frac{(1 + i) x - 2 i}{3 + i} + \frac{(2 - 3 i) y + i}{3 - i} = i

x

y

\frac{(1 + i) x - 2 i}{3 + i} + \frac{(2 - 3 i) y + i}{3 - i} = i

\frac{(1 + i) x - 2 i}{3 + i} + \frac{(2 - 3 i) y + i}{3 - i} = i

(x, y) =

\frac{(1 + i) x - 2 i}{3 + i} + \frac{(2 - 3 i) y + i}{3 - i} = i

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