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.