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3 years ago

11

Help due today will give brainlist

1 answer:

yawa3891 [41]3 years ago

4 0

Answer:

4 fancy and 5 plain

Step-by-step explanation:

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Solving Rational equations. LCD method. Show work. Image attached.

posledela

$(k-2)(k-6)=k^2-2k-6k+12=k^2-8k+12$

So in order to get all the fractions to have a common denominator, we need to multiply $\dfrac k{k-2}$ by $\dfrac{k-6}{k-6}$ , and $\dfrac1{k-6}$ by $\dfrac{k-2}{k-2}$ :

$\dfrac k{k-2}\cdot\dfrac{k-6}{k-6}=\dfrac{k(k-6)}{(k-2)(k-6)}=\dfrac{k^2-6k}{k^2-8k+12}$

$\dfrac1{k-6}\cdot\dfrac{k-2}{k-2}=\dfrac{k-2}{(k-2)(k-6)}=\dfrac{k-2}{k^2-8k+12}$

Now,

$\dfrac4{k^2-8k+12}=\dfrac{(k^2-6k)+(k-2)}{k^2-8k+12}$

As long as $k\neq2$ and $k\neq6$ (which we can't have because otherwise $k^2-8k+12=0$ ), we can cancel $k^2-8k+12$ in the denominators on both sides:

$4=(k^2-6k)+(k-2)$

$4=k^2-5k-2$

$0=k^2-5k-6$

We can factorize the right side:

$0=(k-6)(k+1)$

which tells us that $k=6$ and $k=-1$ are solutions.

4 0

3 years ago