Respuesta :
ANSWER
The solution to the equation is
[tex]h = 7[/tex]
EXPLANATION
The equation given to us is
[tex] \frac{1}{h - 5} + \frac{2}{h + 5} = \frac{16}{ {h}^{2} - 25} [/tex]
The least common multiple of the denominators is,
[tex] {h}^{2} - 25 = (h - 5)(h + 5)[/tex]
We multiply through with the LCM to obtain,
[tex](h + 5)(h - 5) \times \frac{1}{h - 5} + (h + 5)(h - 5) \times \frac{2}{h + 5} = \frac{16}{ {h}^{2} - 25} \times (h + 5)(h - 5)[/tex]
We cancel out common factors to obtain,
[tex]1(h + 5) + 2(h - 5) = 16[/tex]
We expand the brackets to obtain,
[tex]h + 5 + 2h - 10 = 16[/tex]
We group like terms to get,
[tex]h + 2h = 16 - 5 + 10[/tex]
This simplifies to,
[tex]3h = 21[/tex]
We now divide through by 3 to obtain,
[tex]h = 7[/tex]
Therefore the solution is 7.
The solution to the equation is
[tex]h = 7[/tex]
EXPLANATION
The equation given to us is
[tex] \frac{1}{h - 5} + \frac{2}{h + 5} = \frac{16}{ {h}^{2} - 25} [/tex]
The least common multiple of the denominators is,
[tex] {h}^{2} - 25 = (h - 5)(h + 5)[/tex]
We multiply through with the LCM to obtain,
[tex](h + 5)(h - 5) \times \frac{1}{h - 5} + (h + 5)(h - 5) \times \frac{2}{h + 5} = \frac{16}{ {h}^{2} - 25} \times (h + 5)(h - 5)[/tex]
We cancel out common factors to obtain,
[tex]1(h + 5) + 2(h - 5) = 16[/tex]
We expand the brackets to obtain,
[tex]h + 5 + 2h - 10 = 16[/tex]
We group like terms to get,
[tex]h + 2h = 16 - 5 + 10[/tex]
This simplifies to,
[tex]3h = 21[/tex]
We now divide through by 3 to obtain,
[tex]h = 7[/tex]
Therefore the solution is 7.