To solve the given rational equation, we need to perform the following steps:
The given equation is:
`(x+5)/x - 3/(x+5) = 15/(x²+5x)`
Step 1:
To make calculations easier, we can multiply all terms by `x(x + 5)(x² + 5x)`, the least common denominator (LCD).
The equation becomes:
`x(x + 5)(x² + 5x) * (x+5)/x -x(x + 5)(x² +5x)* 3/(x+5) = x(x + 5)(x² + 5x) * 15/(x²+5x)`
Step 2:
Each term will now simplify:
`(x + 5)(x² + 5x) - 3x(x² +5x) = 15x(x + 5)`
Step 3:
Distribute each term:
`x³ + 10x² + 25x - 3x³ -15x² = 15x² + 75x`
Step 4:
Combine like terms to form a polynomial:
`-2x³ - 5x² + 25x = 15x² + 75x`
Step 5:
Set the equation to zero:
`-2x³ - 20x² -50x = 0`
Step 6:
Since our equation is equal to zero, we can divide all terms by -2 to simplify:
`x³ + 10x² + 25x = 0`
Step 7:
Now we can factor the left side:
`x(x² + 10x + 25) = 0`
Step 8:
Then set each factor equal to zero and solve for `x` to get the solutions:
If `x = 0`, we insert it into the original equation, we will get `5/0 - 3/5 = 15/0`, which is undefined. Therefore, `x = 0` is an extraneous solution.
For `x² + 10x + 25 = 0`, we use the quadratic formula to solve for `x`:
`x = [-b ± sqrt(b² - 4ac)] / (2a)`
For `a = 1, b = 10, and c = 25,` the solutions are `x = [-10 ± sqrt((10)² - 4*1*25)] / (2*1) = -5 ± sqrt(0)`
This will simplify to `x = -5 ± 0`.
Therefore, we have one real solution, `x = -5` and we insert into the original equation to check for extraneity.
`(x+5)/x - 3/(x+5) = 15/(x²+5x)`
`(-5+5)/-5 - 3/(-5+5) = 15/((-5)²+5*-5)`
`0 - 3/0 = 15/(25-25)`
This gives an undefined result, therefore `x = -5` is also an extraneous solution.
This leaves us with `x = -2` as the only valid solution.