Question 190049
Break up the equation
(1) {{{2x = 3y}}}
(2) {{{3y = 6z}}}
From (1)
{{{x = (3/2)*y}}}
From (2)
{{{6z = 3y}}}
{{{z = (1/2)*y}}}
----------------
{{{1/x + 1/y + 1/z}}}
{{{1/((3/2)*y) + 1/y + 1/((1/2)*y)}}}
Multiply by {{{2/2}}}
{{{(2/2)*(1/((3/2)*y) + 1/y + 1/((1/2)*y))}}}
{{{(2/2)*(1/((3/2)*y)) + (2/2)*(1/y) + (2/2)*(1/((1/2)*y))}}}
{{{2/(3y) + 2/(2y) + 2/y }}}
{{{4/(6y) + 6/(6y) + 12/(6y)}}}
{{{22/(6y)}}}
(1) {{{11/(3y)}}}
and since {{{y = 2z}}}
(2) {{{11/(6z)}}}
And since {{{y = (2/3)*x}}}
(3) {{{11/(4x)}}}
(1),(2), and (3) are answers