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

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

To solve for x, we notice that the denominators are x+5 , x^2+2x-15 and x-3.

Also notice that (x+5)(x-3) = x^2+(5-3)x+5*-3 = x^2+2x-15.

So the least common multiple of the denominators (LCM) is x^2+2x-15 or (x+5)(x-2)

Now we multiply the equation by the LCM in order to get rid off the denominators.

2x(x+5)(x-3)/(x+5) +10(x^2+2x-15)/(x^2+2x-15) = x(x+5)(x-3)/(x-3). The following are the process of simplifications.

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

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

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

Now we bring all terms to left retaining zero on the right.

2x^2-x^2 -5x +10 = 0.

x^2-11x+10 = 0.

Now this is a quadratic equation'

We split the middle term and rewrite so that we can factor the left side.

x^2-10x-1x +10 = 0

x(x-10) -1(x-10) = 0

(x-310)(x-1) = 0.

x-10 = 0 or x-1 = 0.

So x= 10 or x= 1.

Therefore the solution set is S = { x : x = 1 or x = 10}.

We have to solve: 2x/(x+5) + 10/(x^2+2x-15) = x/(x - 3)

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

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

=> 2x/(x+5) + 10/[x(x+5) -3(x+15)] = x/(x - 3)

=> 2x/(x+5) + 10/[(x-3)(x+5)]= x/(x - 3)

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

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

=> x^2 - 11x +10 =0

=> x^2 - 10x - x +10 =0

=> x(x-10) -1(x-10) =0

=> (x-1)(x+10) =0

=> x= 1 and x = 10

**Therefore x = ( 1, 10)**

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}.**