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2d=8-3c

C is less than 1.

Just to check the boundary take c as -1.

2d=8-(-1) .......
2d=11
D=5.5

But c is less than -1.

Since c has a negetive sign in above equation a smaller value of c will give a bigger value to d.

Only 6.5 is bigger than 5.5.

Hence option e

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In this question, although c<-1, a smart method would be to assume c=-1 and simplify the equation. This will give you a boundary value for d.
Once you have the boundary value, bring back the inequality and check which option satisfies the inequality.

So, if c = -1, then 3(-1) + 2d = 8. Solving this equation, we have,

2d = 8 + 3 = 11 or d = 5.5.

But, c<-1 in actual.

So, in reality, the number getting added to 8 (in the equation above), will be more than 3. This is because 3c<-3 and when you shift this to the RHS, you will have a number more than 3.

Therefore, the RHS will be more than 11 and so, the value of d will be more than 5.5. The only possible value is 6.5.
The correct answer option is E.

Note that the question itself is giving you some elbow space to by asking you what COULD be the value of c. This is because of the inequality, c<-1. With this inequality, you will never be able to find a unique value for d. Hence, the could be clause.

An alternative approach would be to backsolve, starting with option C.
If d = 4.5, 3c + 9 = 8. This means, c = \(\frac{1}{3}\), which is not less than -1. If d= 4.5 can give us a value of c that is positive, the values in option A and B will also do the same. Options A, B and C can all be eliminated.

If d = 5.5, 3c + 11 = 8. This means, c = -1. Not acceptable because c has to be less than -1 according to the question. Option D can be eliminated.

Answer option E HAS to be the answer.

Hope this helps!
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Bunuel
If 3c + 2d = 8 and c < –1, which one of the following could be a value for d?

(A) 2.5

(B) 3.5

(C) 4.5

(D) 5.5

(E) 6.5

Initially I want to plug in values of c but then I realized I am already provided 5 limited values of d so why not use those.
Working out d as 2.5 results in c being 1 which is not <-1
Trying out other options quickly, we find out E works. Hence E.
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