Find three consecutive integers such that the sum of the first and three times the second is eighteen less than five times the third integers
The numbers are x, x+1, x+2. So,
x + 3(x+1) = 5(x+2)-18
To solve this problem, let's represent the three consecutive integers as x, x+1, and x+2.
According to the given information, the sum of the first integer (x) and three times the second integer (3(x+1)) is eighteen less than five times the third integer (5(x+2)).
We can set up the equation as follows:
x + 3(x+1) = 5(x+2) - 18
Now, let's solve for x:
x + 3x + 3 = 5x + 10 - 18
Combine like terms:
4x + 3 = 5x - 8
Move the variables to one side:
4x - 5x = -8 - 3
Simplify:
-x = -11
Multiply both sides by -1 to isolate x:
x = 11
So, the first integer is 11.
The next consecutive integers would be:
x+1 = 11 + 1 = 12
x+2 = 11 + 2 = 13
Therefore, the three consecutive integers that satisfy the given condition are 11, 12, and 13.