solve for x and express solution set in set builder notation: 2x/(x+5) + 10/(x^2+2x-15) = x/(x - 3)

When we try to add or subtract fractions, all ratios have
to have the same denominator.

The denominator of the ratio
10/(x^2+2x-15) is a quadratic.

We'll check if the quadratic
has real roots. If it has, we can re-write it as a product of linear
factors.

To verify if it has real roots, we'll apply the
quadratic formula:

x1 = [-b+sqrt(b^2 -
4ac)]/2a

2x/(x+5)  +  10/(x^2+2x-15)   =  x/(x -
3)

 We'll identify
a,b,c:

a=1

b=2

c=-15

x1
= [-2+sqrt(4+60)]/2

x1 =
(-2+8)/2

x1 = 3

x2 =
-5

We'll re-write the
quadratic:

x^2+2x-15 =
(x-3)(x+5)

We'll re-write the
expression:

2x/(x+5)  +  10/(x-3)(x+5)   =  x/(x -
3)

In order to find out x, we'll multiply the first ratio
by (x-3) and the third ratio by (x+5):

2x(x-3) + 10 =
x(x+5)

We'll remove the
brackets:

2x^2 - 6x + 10 = x^2 +
5x

We'll subtract x^2 + 5x both
sides:

x^2 - 11x + 10 =
0

We'll apply the quadratic formula,
again:

x1 =
[11+sqrt(121-40)]/2

x1 =
[11+sqrt81]/2

x1 = (11+9)/2

x1
= 9

x2 = (11-9)/2

x2 =
1

The real roots of the given expression are:
{1 ; 9}.