Question

If the average of b and c is 8, and d=3b-4, what is the average of c and d in terms of b?

Answer 1

$\underline{ \huge \mathcal{ Ànswér} } \huge: -$

Average of b and c is 8, that is

$➢ \: \: \dfrac{b + c}{2} = 8$

$➢ \: \: b + c = 16$

$➢ \: \: c = 16 - b$

now let's solve for average of c and d :

$➢ \: \: \dfrac{c + d}{2}$

$➢ \: \: \dfrac{16 - b + 3b - 4}{2}$

$➢ \: \: \dfrac{12 + 2b}{2}$

$➢ \: \: \dfrac{2(6 + b)}{2}$

$➢ \: \: b + 6$

Therefore, the average of c and d, in terms of b is : -

$\large \boxed{ \boxed{b + 6}}$

$\mathrm{✌TeeNForeveR✌}$

Answer 2

Answer:

b+6

Problem:

If the average of b and c is 8, and d=3b-4, what is the average of c and d in terms of b?

Step-by-step explanation:

We are given (b+c)/2=8 and d=3b-4.

We are asked to find (c+d)/2 in terms of variable, b.

We need to first solve (b+c)/2=8 for c.

Multiply both sides by 2: b+c=16.

Subtract b on both sides: c=16-b

Now let's plug in c=16-b and d=3b-4 into (c+d)/2:

([16-b]+[3b-4])/2

Combine like terms:

(12+2b)/2

Divide top and bottom by 2:

(6+1b)/1

Multiplicative identity property applied:

(6+b)/1

Anything divided by 1 is that anything:

(6+b)

6+b

b+6