If x^2 + 4y^2 + 3z^2 + 19/4 = 2√3(x = y+z), then the value of ( x - 4…

$x^2 + 4y^2 + 3z^2 + \frac{19}{4} = 2\sqrt{3}(x = y+z)$,

then the value of ( x - 4y + 3z) = ?

we can find the values of the variables by =

Coefficient of variables on right sides divide by coefficient of same variable on left side along with the signs as given below =

= x = \(\sqrt {3}\)

= y = \(\frac{\sqrt {3}}{4}\)

= z = \(\frac{1}{\sqrt {3}}\)

So the value of  ( x - 4y + 3z) =  ( \(\sqrt {3}\) - 4(\(\frac{\sqrt {3}}{4}\)) + 3( \(\frac{1}{\sqrt {3}}\))) = $\sqrt{3}$