How do you solve the equation: 1/(x+3)+1/(x+5)=1

Multiply by the LCD: (x + 3)(x + 5)

(x + 5) + (x + 3) = (x + 3)(x + 5)
2x + 8 = x^2 + 8x + 15
0 = x^2 + 6x + 7
-7 = x^2 + 6x
-7 + 9 = x^2 + 6x + 9
2 = (x + 3)^2
± √2 = x + 3
x = -3 ± √2

multiply both sides of the equation by (x+3)*(x+5) to obtain the equation

x+5 + x + 3 = (x+3)*(x+5), so 2x + 8 = x^2 + 8x + 15, or, putting everything on one side,

x^2 + 6x + 7 = 0. then you can solve using the quadratic equation to obtain

x = [-6 +/- sqrt(36-28)]/2, which gives x = -3 +sqrt(2) or x = -3 - sqrt(2).

Multiply everything on both sides by (x+3) and (x+5).
You get
(x+5) + (x+3) = (x+3)*(x+5)
which is
2x+8 = x^2 + 8x + 15
or
x^2 + 6x + 7 = 0

Now use the quadratic formula to solve this:
x = (-6 +- squrt(36 - 28))/2
= (-6 +- squrt(8))/2
= (-6 +- 2*squrt(2))/2
= -3 +- squrt(2).
So your answers are squrt(2)-3 and -squrt(2)-3.

1/(x+3) + 1/(x+5) =1
=> (x+5)/[(x+3)(x+5)] + (x+3)/[(x+3)(x+5)] = [(x+3)(x+5)]/[(x+3)(x+5)]
=> x+5 + x+3 = (x+3)(x+5) { with x<> -3 and x<> -5}
=> 2x +8 = x^2 +8x +15
=> x^2 + 6x + 7 = 0
=> x^2 + 2*3x + 9 -2 =0
=> (x+3)^2 =2
=> x+3 = 2^(1/2) or x+3 = -2^(1/2)
=> x = 2^(1/2) - 3 or x = -2^(1/2) -3

Ans x = 2^(1/2) - 3 or x = -2^(1/2) -3

Get a common denominator of (x + 3)(x + 5):

(x + 5 + x + 3)/(x + 5)(x + 3) = 1

2x + 8 = x^2 + 8x + 15

x^2 + 6x + 7 = 0

(x + 3)^2 - 2 = 0

x = +/-√2 - 3

multiply through by (x+3)(x+5)

(x+5)+(x+3)=(x+5)(x+3)

you can take it from there