Let the mean and variance of 8 numbers x, y, 10, 12, 6, 12, 4, 8 be...

Given, the data $x, y, 10, 12, 4, 6, 8, 12$.

So,

Mean = \dfrac{x + y + 10 + 12 + 4 + 6 + 8 + 12}{8}$</p><p>$⇒ 9 = \dfrac{x + y + 52}{8}$</p><p>$⇒ x + y + 52 = 72$</p><p>$⇒ x + y = 20$</p><p>And,</p><p>Variance = \left(\dfrac{\mathop{∑}_{}{x}_{i}^{2}}{n}\right) - {\left(\dfrac{\mathop{∑}_{}{x}_{i}}{n}\right)}^{2}$

$⇒ 9 . 25 = \left(\dfrac{{x}^{2} + {y}^{2} + 100 + 144 + 16 + 36 + 64 + 144}{8}\right) - {\left(9\right)}^{2}$

$⇒ {x}^{2} + {y}^{2} = 218$

$⇒ {\left(x + y\right)}^{2} - 2xy = 218$

$⇒ {20}^{2} - 2xy = 218$

$⇒ 2xy = 182$

$⇒ xy = 91$

Now solving $x + y = 20 \text{ and } xy = 91$ we get,

$x = 13, y = 7 ⇒ 3x - 2y = 25$