Question 1008638
We write this is as
{{{z = k*sqrt(xy)}}}
Then we plug in to get
{{{3 = k*sqrt(3*12)}}}
{{{3 = 6k}}}
{{{k = 1/2}}}
and we have
{{{z = (1/2)*sqrt(xy)}}}
Then
{{{6 = (1/2)*sqrt(64x)}}}
{{{6 = 4*sqrt(x)}}}
{{{3/2 = sqrt(x)}}}
and
{{{x = 9/4}}}