find three consecutive odd integers such that the sum of the middle and largest integer is 21 more than the smallest integer
middle number --- x
1st number ---- x-2
3rd number --- x+2
middle + larger > smaller by 21
middle + larger = smaller + 21
x + x+2 = x-2 + 21
x = 17
the three numbers are 15, 17, 19
check:
sum of middle + larger = 17+19 = 36
the smaller = 15
is 36>15 by 21 ? , looks good
To solve this problem, let's break it down into steps:
Step 1: Let's represent the first odd integer as "x".
Step 2: Since we need three consecutive odd integers, the second odd integer would be "x + 2" (adding 2 to the first odd integer gives us the next odd integer).
Step 3: Similarly, the third odd integer would be "x + 4" (adding 2 to the second odd integer gives us the next odd integer).
Now, we need to set up the equation based on the given information:
"The sum of the middle and largest integer is 21 more than the smallest integer."
The smallest integer is "x".
The middle integer is "x + 2".
The largest integer is "x + 4".
According to the given information, we can write the equation as:
(x + 2) + (x + 4) = x + 21
Now, we can simplify and solve for "x":
2x + 6 = x + 21
Subtracting "x" from both sides:
2x - x + 6 = x - x + 21
x + 6 = 21
Subtracting 6 from both sides:
x + 6 - 6 = 21 - 6
x = 15
Therefore, the first odd integer is 15.
To find the other two consecutive odd integers, we can substitute the value of "x" into our previous representations:
First odd integer = x = 15
Second odd integer = x + 2 = 15 + 2 = 17
Third odd integer = x + 4 = 15 + 4 = 19
So, the three consecutive odd integers are 15, 17, and 19.