Question 392281
Three consecutive odd integers that equal 63.

Since 1, 3, 5... each number has to add 2 to itself to equal the next.
1 plus 2 equals 3 and so on.

So the three unknown, consecutive odd numbers algebraically would be set up as:
{{{x+(x+2)+(x+4) = 63}}}
Since all the numbers are being added, you can drop the parentheses.
{{{x+x+2+x+4 = 63}}}
You, then, combine like terms.
There are 3 x's, and a 4 and a 2.
So you have:
3x+6 = 63
You subtract 6 from both sides of the equation.
3x = 57
Then you divide both sides of the equation by 3.
x = 19.
Since x is 19, you plug that in the original formula we created.
19+(19+2)+(19+4) = 63
You solve the problem.
19+21+23 = 63
63 = 63.
So your answer is:
The three consecutive odd integers that equal 63 are 19, 21, and 23.