How to solve (2/x+1)+(3/5x+5)=0

I have some algebra homework and I'm having trouble with fractional equations.

(2/[x + 1]) + (3/[5x + 5]) = 0
10x + 10 + 3x + 3 = 0
13x = - 13
x = - 1

Answer: x = - 1

Check:
(2/[- 1 + 1]) + (3/[5{- 1} + 5]) = 0
2/0 + (3/[- 5 + 5]) = 0
0 + 3/0 = 0
0 + 0 = 0

Assuming you're solving for x; since it's addition, you don't really need the parentheses. So it's just 2/x+1+3/5x+5=0 and you can add the fractions: 2/x+3/5x=13/5x and add 5+1=6
So you have: 13/5x+6=0
subtract 6 from each side: 13/5x= -6
multiply both sides by 5x: 13= -30x
now divide by by -30: x= -13/30= -.43333

(2/x + 1) + (3/5x + 5) = 0
combine terms:
2/x + 3/5x + 6 = 0
subtract 6 from both sides:
2/x + 3/5x = -6
multiply 2/x by 5/5:
10/5x + 3/5x = -6
add thr fractions:
13/5x = -6
multiply both sides by 5x:
13 = -30x
divide both sides by -30:
x = -13/30

- .--

(2/x+1)+(3/5x+5)=0
2/x+3/5x+6=0
(amplyfy the first fraction)
10/5x+3/5x=-6
13/5x=-6
1/x=-30/13
x=-13/30

multiply the whole equation by x:

2 + x + 3/5 + 5x = 0

6x = -13/5

x= -13/30

multiply the whole equation by x:

2 + x + 3/5 + 5x = 0

6x = -13/5

x= -13/30

Do you mean 2/(x + 1) + 3/(5x + 5)??

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