The sum of 3 consecutive numbers is 147
so, X (first interger)
X+1 (second)
X+2 (third)
*combine like terms
X+X+1+X+2=147
3x+3=147
-3 -3
3X=144
/3 /3
X=48
(substitute 48 for x)
48+1=49
48+2=50
So the intergers are 48, 49, and 50.
so, X (first interger)
X+1 (second)
X+2 (third)
*combine like terms
X+X+1+X+2=147
3x+3=147
-3 -3
3X=144
/3 /3
X=48
(substitute 48 for x)
48+1=49
48+2=50
So the intergers are 48, 49, and 50.