Question

question
solve for all values of x.
1 + \frac{7}{x - 4}=\frac{2}{x^{2}-16}
answer
attempt 1 out of 2
additional solution no solution
x =

Explanation:

Step1: Factor the denominator

Note that $x^{2}-16=(x + 4)(x - 4)$. The given equation $1+\frac{7}{x - 4}=\frac{2}{x^{2}-16}$ can be rewritten as $1+\frac{7}{x - 4}=\frac{2}{(x + 4)(x - 4)}$.

Step2: Multiply through by the common - denominator

Multiply each term by $(x + 4)(x - 4)$:
$(x + 4)(x - 4)\times1+(x + 4)(x - 4)\times\frac{7}{x - 4}=(x + 4)(x - 4)\times\frac{2}{(x + 4)(x - 4)}$
$(x + 4)(x - 4)+7(x + 4)=2$.

Step3: Expand the expressions

Expand $(x + 4)(x - 4)$ using the difference - of - squares formula $(a + b)(a - b)=a^{2}-b^{2}$, so $(x + 4)(x - 4)=x^{2}-16$. Expand $7(x + 4)=7x+28$.
The equation becomes $x^{2}-16 + 7x+28 = 2$.

Step4: Simplify the equation

Combine like terms: $x^{2}+7x + 12 = 2$.
Subtract 2 from both sides to get a quadratic equation in standard form: $x^{2}+7x+10 = 0$.

Step5: Factor the quadratic equation

Factor $x^{2}+7x + 10$ as $(x + 2)(x+5)=0$.

Step6: Solve for x

Set each factor equal to zero:
If $x + 2 = 0$, then $x=-2$.
If $x + 5 = 0$, then $x=-5$.
We need to check for extraneous solutions by substituting these values back into the original equation. The original equation has denominators $x - 4$ and $x^{2}-16=(x + 4)(x - 4)$. When $x=-2$, $x-4=-6
eq0$ and $x^{2}-16=4 - 16=-12
eq0$. When $x=-5$, $x - 4=-9
eq0$ and $x^{2}-16 = 25-16 = 9
eq0$.

Answer:

$x=-2,x=-5$