I need help on math .... can someone help me?

Question

The first of three numbers exceeds twice the second number by 4, while the third number is twice the first. If the sum of the three numbers is 54, find the largest number

Answer 1

Write the numbers as algebraic expressions

 

2x + 4

x

2(2x + 4)

 

Now, add all three numbers together to get a sum of 54 and solve for x

 

2x + 4 + x + 2(2x + 4) = 54

2x + 4 + x + 4x + 8 = 54

7x + 12 = 54

7x = 42

x = 6

 

Because x = 6, our three numbers are:

 

2x + 4 = 2(6) + 4 = 12 + 4 = 16

x = 6

2(2x + 4) = 2(12 + 4) = 32

 

The largest number is 32.

Answer 2

It the 2nd number is x, the first number would be 2x + 4, and the 3rd number is 2(2x + 4). Adding the 3 numbers gives a sum of 54, so: 2x + 4 + x + 2(2x + 4) = 54, 2x + 4 + x + 4x + 8 = 54, 7x + 12 = 54, 7x = 42, x = 6.

Then substituting 6 for x, the first number is 2(6) + 4 = 12 + 4 = 16. Then, the 3rd number is 2(2 X 6 + 4) = 2(16) = 32.

So the largest number is 32. (Check: 6 + 16 + 32 + 54)

Answer 3

n = the 2nd

2n + 4 = the 1st

2(2n + 4) = the third

 

n + 2n + 4 + 4n + 8 = 54

 

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