It is given that \((\frac {x}{3}+1, \frac {y-2}{3}) = (\frac {5}{3}, \frac {1}{3})\).
Since the ordered pairs are equal, the corresponding elements will also be equal.
\((\frac {x}{3}+1, \frac {y-2}{3}) = (\frac {5}{3}, \frac {1}{3})\)
Therefore,
\(\frac {x}{3}+1=\frac {5}{3}\)
⇒ \(\frac{x}{3} = \frac{5}{3} -1,\frac{y-2}{3} = \frac{1}{3}\)
⇒\(\frac{x}{3} = \frac{2}{3}, y = \frac{1}{3}+\frac{2}{3}\)
⇒ x=2, y=1
Let \(M = \{1, 2, 3, ....., 16\}\), if a relation R defined on set M such that R = \((x, y) : 4y = 5x – 3, x, y (\in) M\). How many elements should be added to R to make it symmetric.
Find the mean deviation about the mean for the data 38, 70, 48, 40, 42, 55, 63, 46, 54, 44.