Question

question
solve for all values of $x$.
$\frac{5}{x - 4}+\frac{1}{x + 1}=\frac{x + 2}{x + 1}$
answer
attempt 1 out of 2
additional solution no solution
$x =$

Explanation:

Step1: Find common denominator

The common denominator of the left - hand side is $(x - 4)(x + 1)$. Rewrite the left - hand side: $\frac{5(x + 1)+(x - 4)}{(x - 4)(x + 1)}=\frac{x + 2}{x + 1}$.

Step2: Simplify the numerator of the left - hand side

Expand and combine like terms: $\frac{5x+5+x - 4}{(x - 4)(x + 1)}=\frac{x + 2}{x + 1}$, so $\frac{6x + 1}{(x - 4)(x + 1)}=\frac{x + 2}{x + 1}$.

Step3: Cross - multiply (note $x

eq - 1$)
$(6x + 1)(x + 1)=(x + 2)(x - 4)(x + 1)$. Since $x
eq - 1$, we can divide both sides by $(x + 1)$ (we will check this value separately later). We get $6x+1=(x + 2)(x - 4)$.

Step4: Expand the right - hand side

$6x+1=x^{2}-4x+2x - 8$, which simplifies to $x^{2}-8x - 9 = 0$.

Step5: Factor the quadratic equation

$x^{2}-8x - 9=(x - 9)(x+1)=0$.

Step6: Solve for $x$

Setting each factor equal to zero gives $x-9 = 0$ or $x + 1=0$. But when $x=-1$, the original equation has undefined terms (denominators of $\frac{1}{x + 1}$ and $\frac{x + 2}{x + 1}$ are zero). So we discard $x=-1$. From $x - 9=0$, we get $x = 9$.

Answer:

$x = 9$