In a bag there are a number of red and blue balls. The ratio of red to blue balls is 3:7.

Question

I add 10 balls and the ratio goes to 6:7 red to blue balls.
How many blue balls are there?

3n + 7n = 10n
(3n+10) + 7n = 10n + 10
6n + 7n = 10n + 10
13n = 10n + 10
3n = 10
N = 3.333
7n = 23.333 ( number of blue balls)

Answer 1

original red --- 3x
original blue --- 7x

number of reds added -- a
number of blues added -- 10-a

(3x+a)/(7x + 10-a) = 6/7
42x + 60 - 6a = 21x + 7a
21x - 13a = -60

I think you have something missing in your problem, maybe a typo
We have to know how the 10 balls are distributed by colour, were they all red? or what ?

Your opening equation is true for all values of n
you made an error in your 3rd line and lost one of the 10's , which explains why your answer makes no sense
Also how did your 3n suddenly become 6n ?

Answer 2

I did miss a bit. The 10 additional balls were all red.

Answer 3

then
(3x+10)/(7x) = 6:7
42x = 21x + 70
21x=70
x = 70/21 = 10/3

so the number of reds originally was 3(10/3) = 10
the number of blues originally was 7(10/3) = 70/3 ?????
how can we have partial balls ?

I think there is something very wrong with your question.

notice that "mathematically" my answer is correct

check:
original reds = 10
original blues = 70/3
10 : 70/3
= 30:70
= 3:7

new reds = 20
blues = 70/3
20:70/3
= 60:70
= 6:7