Question 86294
{{{4/(3x)}}} + {{{1/x}}} = {{{14/(3x^2)}}}
Multiply everything by the LCD of the fractions, which is {{{3x^2}}}:
{{{3x^2}}}{{{4/(3x)}}} + {{{3x^2}}}{{{1/x}}} = {{{3x^2}}}{{{14/(3x^2)}}}
Let the denominators cancel out:
4x + 3x = 14
Solve in the usual manner:
7x = 14
{{{(7x)/7}}} = {{{14/7}}}
x = 2 [potentially]
Must check for extraneous solutions/zero denominators:
3x = 3(2) which isn't = 0 so OK
{{{3x^2}}} = 3(2)^2 = 12 which isn't = 0 so OK
x = 2 is a valid solution