Solve the following problem. show all steps (v. e.s. t.) 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

We have three numbers:
first number=x
second number=2x
third number=2x+6

we can suggest this equation:
x+2x+(2x+6)=26
x+2x+2x=26-6
5x=20
x=20/5
x=4

First number=x=4
second number=2x=2(4)=8
third number=2x+6=2(4)+6=8+6=14

the numbers are 4, 8 and 14.

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 

10.4 is the answer for you