Question 1766
assuming your equations are {{{f(x) = (7/x) + (7/(x+4))}}} and {{{f(x) = 1}}}, then ignore the f(x) bit and just put y instead...so y equals the first one and also the second one, so the two equations are equal to each other, ie...

{{{(7/x) + (7/(x+4)) = 1}}}

do fractions on this, to give...

{{{7(x+4) + 7(x) = x(x+4)}}}

then simplify out and collect terms (being careful with negative signs) to give...

{{{x^2 - 10x - 28 = 0}}}

Solve this with the Quadratic formula to give 

{{{x = (10 +- (sqrt(212)))/(2)}}}

{{{sqrt(212)}}} is also {{{(sqrt(4*53))}}}. then take the 4 out, becoming 2 and then cancel the fraction, which gives

{{{x = 5 +- (sqrt(53))}}}

cheers
Jon.