Question

The following rational equation has denominators that contain variables. for this equation, A. write the value or values of the variable that makes the denominator zero. These are the restrictions on the variable. B.
keeping the restrictions in mind, solve the equation.
3/x+4 + 2/x-4 = 16/(x+4)(x-4)
A. what is/ Are The value our values of the variable that make(s) the denominators zero?
x=

Answer 1

Answer:

A. -4 and 4

B. No solution.

Step-by-step explanation:

The given equation is

$\dfrac{3}{x+4}+\dfrac{2}{x-4}=\dfrac{16}{(x+4)(x-4)}$

A.

Equate the denominators equal to 0 to find the restrictions on the variable.

$x+4=0\Rightarrow x=-4$

$x-4=0\Rightarrow x=4$

Therefore, $x\neq -4,4$.

B.

We have,

$\dfrac{3}{x+4}+\dfrac{2}{x-4}=\dfrac{16}{(x+4)(x-4)}$

$\dfrac{3(x-4)+2(x+4)}{(x+4)(x-4)}=\dfrac{16}{(x+4)(x-4)}$

Multiply both sides by (x-4)(x+4).

$3x-12+2x+8=16$

$5x-4=16$

Add 4 on both sides.

$5x=16+4$

$5x=20$

Divide both sides by 5.

$x=4$

Here the solution is x=4 but it is the restricted value.

Therefore, the given equation has no solution.