Question 13369
find three consecutive intergers that equal out to be 45....

Let x be the first integer; so we have
x; then the second integer would be x+1; then the third would br x+2;

x+(x+1)+(x+2)=45
We add all the x's and the ones together:
3x+3=45
We subtract 3 from both sides;
3x+3-3=45-3
3x=42
Divide 3 from both sides;
{{{3x/3=45/3}}}
x=14
So the first integer would be 14; then the second 14+1=15; and the third 14+2=16

So we have 14+15+16=45

=)