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The sum of three consecutive integers is 53 more than the least of the integers. Find the integers.
2006-10-23 15:35:05 · 5 answers · asked by Anonymous in Education & Reference ➔ Homework Help
letthe integers be x,x+1,x+2
the sum
3x+3=53+x
2x=50
x=25
the integers are 25,26 and 27
2006-10-23 15:37:00 · answer #1 · answered by raj 7 · 0⤊ 0⤋
I'm not going to provide the complete answer, but I'll help with setting up the equation. You have 3 consecutive integers. x, x+1, x+2. The sum of the three is 53 more then the least of the integers.
x+(x+1)+(x+2)=x+53 . Then solve for x.
If you need more help, send your work back showing where you get stuck, and I'll help more.
2006-10-23 15:39:41 · answer #2 · answered by Snoopy 5 · 0⤊ 0⤋
the first integer is x
the second integer is x+1
the third integer is x+2
53 more than the last integer would be 53 + x
so (x)+(x+1)+(X+2)= (53+x)
3x+3 = 53 + x
2x = 50
x = 25 then substitute into the other equations you have
25,26,27
2006-10-23 15:39:54 · answer #3 · answered by mackistan 2 · 0⤊ 0⤋
x + x+1 + x+2 = 53 + x
3x + 3 = 53+ x
2x = 50
x = 25
so the three numbers are 25, 26, 27
2006-10-23 15:39:01 · answer #4 · answered by Up_In_Smoke 2 · 0⤊ 0⤋
(x + x+1 + x+2) -x = 53
(3x+3)-x =53
2x =50
x=25
integers are 25,26,27
2006-10-23 15:40:40 · answer #5 · answered by ace 2 · 0⤊ 0⤋