Question 92635
 {{{(1/(x-3))+(1/(x+3))}}}={{{(10/(x^2-9))}}}
:
Note that x^2 - 9 is the difference of squares and can be factored
{{{(1/(x-3))+(1/(x+3))}}}={{{(10/(x-3)(x+3))}}}
:
Multiply the equation thru by (x-3)(x+3), results:
1(x+3) + 1(x-3) = 10
:
x + 3 + x - 3 = 10
:
2x = 10
:
x = 10/2
x = +5
:
:
Check the solution in original equation
{{{(1/(5-3))+(1/(5+3))}}}={{{(10/(25-9))}}}
{{{(1/(2))+(1/(8))}}}={{{(10/(16))}}}
{{{(8/(16))+(2/(16))}}}={{{(10/(16))}}}
:
Could you follow this OK? Any questions?