Question

question #7 solve the equation. remember to check for extraneous solutions. $\frac{k + 3}{6}+\frac{k + 2}{6}=\frac{k^{2}-4}{6k}$
$k=-1$
$k=-1$ and $k = 1$
$k=-3$ and $k = 3$
$k=-1$ and $k=-4$

Explanation:

Step1: Combine left - hand side fractions

Since the denominators of $\frac{k + 3}{6}$ and $\frac{k+2}{6}$ are the same, we add the numerators: $\frac{(k + 3)+(k + 2)}{6}=\frac{2k+5}{6}$. So the equation becomes $\frac{2k + 5}{6}=\frac{k^{2}-4}{6k}$.

Step2: Cross - multiply

Cross - multiplying gives us $(2k + 5)\times6k=6\times(k^{2}-4)$. Expand both sides: $12k^{2}+30k = 6k^{2}-24$.

Step3: Rearrange to form a quadratic equation

Subtract $6k^{2}-24$ from both sides: $12k^{2}+30k-(6k^{2}-24)=0$, which simplifies to $12k^{2}+30k - 6k^{2}+24 = 0$, and further to $6k^{2}+30k + 24 = 0$. Divide through by 6: $k^{2}+5k + 4 = 0$.

Step4: Factor the quadratic equation

Factor $k^{2}+5k + 4$ as $(k + 1)(k + 4)=0$.

Step5: Solve for k

Setting each factor equal to zero gives $k+1 = 0$ or $k + 4=0$, so $k=-1$ or $k=-4$.

Step6: Check for extraneous solutions

The original equation has a denominator of $6k$. When $k=-1$, the denominators $6$ and $6k=-6$ are non - zero. When $k=-4$, the denominators $6$ and $6k=-24$ are non - zero.

Answer:

$k=-1$ and $k=-4$