SOLUTION: 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 numbers.
please help
Question 666555: 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 numbers.
please help me. Answer by mananth(16946) (Show Source):
You can put this solution on YOUR website! second number be x
first number = 2x+4
third number =2(2x+4)
sum =54
x+2x+4+2(2x+4)=54
x+2x+4+4x+8=54
7x+12=54
7x=54-12
7x=42
x=6
First number = 6
second number = 2x+4 ------> 2*6+4 = 16
Third number = 2*16=32
6,16,32
add them up you get 54
