Question

Solve the following equation. Remember to check for extraneous solutions 7/(b+3) + 5/(b-3) = (10b-2)/(b²-9).

Answer 1

Answer:

b= 2 is the solution for the given equation.

Step-by-step explanation:

Here, the given expression is:

$\frac{7}{(b+3)} + \frac{5}{(b-3)} = \frac{10b -2}{(b^{2} -9)}$

Simplifying Left side, we get

$\frac{7}{(b+3)} + \frac{5}{(b-3)}$

= $\frac{7(b-3) + 5(b+3)}{(b+3)(b-3)}$

Also, by ALGEBARIC IDENTITY:$x^{2} -y^{2} = (x+y)(x-y)$

So, $(b+3)(b-3) = b^{2} -9$

So, LHS becomes $\frac{7(b-3) + 5(b+3)}{b^{2} -9}$

Compare both Left side, Right side we get

$\frac{7(b-3) + 5(b+3)}{b^{2} -9}$ =   $\frac{10b -2}{(b^{2} -9)}$

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.