Could you help me with this Algebra question?

Question

Solve the following problem. Be sure to show all steps (V.E.S.T.) and work in order to receive full credit.

The sum of three numbers is 26. The second number is twice the first and the third number is 6 more than the second. Find the numbers.

Answer 1

 V.E.S.T. is: 

V. variable

E. equation

S. solve it

T. test it

 

 

 

 

Answer 2

a + b + c = 26

b = 2a

c = b + 6 = 2a + 6

a + (2a) + (2a + 6) = 26

5a + 6 = 26

5a = 20

a =4

a = 4

b = 2a = 2(4) = 8

c = b + 6 = (8) + 6 = 14

4 + 8 + 14 = 26

Answer 3

4+8+14=26

The first number is 4

The second is 4*2=8

The third is 8+6=14

Hope this helps! :)

Answer 4

"The sum of three numbers is 26."

x+y+z = 26

"the second number is twice the first"

y = 2x

"the third number is 6 more than the second"

z = 6+y

You have the system of equations

x + y + z = 26

2x - y = 0

-y + z = 6

Solve using Gauss-Jordan Elimination:

x+y+z = 26

-3y-2z = -52

-y+z = 6

x+y+z = 26

y+(2/3)z = 52/3

-y+z = 6

x+(1/3)z = 26/3

y+(2/3)z = 52/3

(5/3)z = 70/3

x+(1/3)z = 26/3

y+(2/3)z = 52/3

z = 14

x = 4

y = 8

z = 14

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What is V.E.S.T.?

Answer 5

4,8,14

Answer 6

first number is x

second is 2x

third is 2x + 6

so x + 2x + 2x + 6 =26, 5x = 20; x = 4

first number = 4, second 8, third 14 - like the other person said