four times the sum of three consecutives odd integers is the same as 221 more than 5 times the largest integer. Find the integers
that's better.
consecutive odd integers differ by 2.
so, if the integers are x, x+2, x+4 then you just have to solve
4(x + x+2 + x+4) = 221+5(x+4)
217?
217,219,221
Hmmm. How did you arrive at that answer?
I think you went astray at the step
7x = 217
31,33,35
To solve this problem, we'll break it down into steps and use algebra to find the solution.
Let's start by assigning variables to the unknowns. Let's say our three consecutive odd integers are x, x+2, and x+4.
According to the problem, "four times the sum of three consecutive odd integers is the same as 221 more than 5 times the largest integer." We can translate this sentence into an equation:
4(x + (x + 2) + (x + 4)) = 5(x + 4) + 221
Now, let's simplify the equation and solve for x:
4(3x + 6) = 5x + 20 + 221
12x + 24 = 5x + 241
12x - 5x = 241 - 24
7x = 217
x = 31
So, the first odd integer is 31, the second is 33, and the third is 35.
Therefore, the three consecutive odd integers are 31, 33, and 35.