In the fllowing equation, what is the value of n ?

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In the fllowing equation, what is the value of n ?
$$\frac{3(n+4)−8}{5}=\frac{11−(6−2n)}{2}$$
$$A.\ -\frac{33}{4}$$
$$B.\ -\frac{17}{4}$$
$$C.\ \frac{17}{4}$$
$$D.\ \frac{17}{16}$$
$$E.\ \frac{33}{16}$$

The OA is B.

Please, can any experte assist me with this PS question? I need help to solve it. Thanks.
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by Brent@GMATPrepNow » Thu Nov 02, 2017 8:03 am
swerve wrote:In the fllowing equation, what is the value of n ?
$$\frac{3(n+4)−8}{5}=\frac{11−(6−2n)}{2}$$
$$A.\ -\frac{33}{4}$$
$$B.\ -\frac{17}{4}$$
$$C.\ \frac{17}{4}$$
$$D.\ \frac{17}{16}$$
$$E.\ \frac{33}{16}$$
First expand and simplify the NUMERATORS.
We get: (3n + 4)/5 = (5 + 2n)/2

Next, we can eliminate the fractions by multiplying both sides by 10, the least common multiple of 5 and 2.
We get: (10)(3n + 4)/5 = (10)(5 + 2n)/2
Simplify: (2)(3n + 4) = (5)(5 + 2n)
Expand both sides: 6n + 8 = 25 + 10n
Subtract 6n from both sides: 8 = 25 + 4n
Subtract 25 from both sides: -17 = 4n
Divide both sides by 4 to get: -17/4 = n

Answer: B

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Brent
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by Scott@TargetTestPrep » Fri Nov 01, 2019 6:56 pm
swerve wrote:In the fllowing equation, what is the value of n ?
$$\frac{3(n+4)−8}{5}=\frac{11−(6−2n)}{2}$$
$$A.\ -\frac{33}{4}$$
$$B.\ -\frac{17}{4}$$
$$C.\ \frac{17}{4}$$
$$D.\ \frac{17}{16}$$
$$E.\ \frac{33}{16}$$

The OA is B.

Please, can any experte assist me with this PS question? I need help to solve it. Thanks.
Cross multiplying, we have:

2[3(n + 4) - 8] = 5[11 - (6 - 2n)]

6(n + 4) - 16 = 5(11 - 6 + 2n)

6n + 24 - 16 = 5(5 + 2n)

6n + 8 = 25 + 10n

-17 = 4n

-17/4 = n

Answer: B

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