find three consecutive odd integers such that the sum of three times the first and the third is 72
odd numbers are 2 apart so if we call the first one x,
then the next two are x+2 and x+4
translate to math:
"the sum of three times the first and the third is 72"
3x + x+4 = 72
solve for x, etc
thank you
Actually, I was wrong
To find three consecutive odd integers, let's assume the first odd integer as 'x'.
The second odd integer will be 'x + 2' because it is the next odd number after 'x'.
The third odd integer will be 'x + 4' because it is the odd number after the second one.
Now, we can form the equation as given:
3 times the first integer (3x) plus the third integer (x + 4) equals 72.
Simplifying this equation, we get:
3x + (x + 4) = 72
Combining like terms:
4x + 4 = 72
Subtracting 4 from both sides of the equation:
4x = 68
Dividing both sides of the equation by 4:
x = 17
So the first odd integer is 17.
The second odd integer is 17 + 2 = 19.
The third odd integer is 17 + 4 = 21.
Therefore, the three consecutive odd integers that satisfy the given conditions are 17, 19, and 21.