Respuesta :
Answer:
b= 2 is the solution for the given equation.
Step-by-step explanation:
Here, the given expression is:
[tex]\frac{7}{(b+3)} + \frac{5}{(b-3)} = \frac{10b -2}{(b^{2} -9)}[/tex]
Simplifying Left side, we get
[tex]\frac{7}{(b+3)} + \frac{5}{(b-3)}[/tex]
= [tex]\frac{7(b-3) + 5(b+3)}{(b+3)(b-3)}[/tex]
Also, by ALGEBARIC IDENTITY:[tex]x^{2} -y^{2} = (x+y)(x-y)[/tex]
So, [tex] (b+3)(b-3) = b^{2} -9 [/tex]
So, LHS becomes [tex]\frac{7(b-3) + 5(b+3)}{b^{2} -9}[/tex]
Compare both Left side, Right side we get
[tex]\frac{7(b-3) + 5(b+3)}{b^{2} -9}[/tex] = [tex]\frac{10b -2}{(b^{2} -9)}[/tex]
or, 7(b-3) + 5(b+3) = 10b -2
⇒ 7b - 21 + 5b + 15 = 10b -2
or, 12b - 10b = 6-2
or, 2b = 4 ⇒ b = 4/2 = 2
⇒ b= 2 is the solution for the given equation.
