|
Question 815700: find three consecutive integers such that the sum of the first and three times the second is 44 less than five the third
Answer by algebrahouse.com(1659)   (Show Source):
You can put this solution on YOUR website! x = first consecutive integer
x + 1 = second consecutive integer {consecutive integers increase by 1}
x + 2 = 3rd consecutive integer
x + 3(x + 1) = 5(x + 2) - 44 {sum of 1st and 3 times 2nd is 44 less than 3 times the 3rd}
x + 3x + 3 = 5x + 10 - 44 {used distributive property}
4x + 3 = 5x - 34 {combined like terms}
3 = x - 34 {subtracted 4x from each side}
x = 37 {added 34 to each side}
x + 1 = 38 {substituted 37, in for x, into x + 1}
x + 2 = 39 {substituted 37, in for x, into x + 2}
37, 38, and 39 are the three consecutive integers
For more help from me, visit: www.algebrahouse.com
|
|
|
| |
