math問題

1.
The sum of 3 consecutive integers is 24. Find the smallest integer.

let the smallest integer be y
y+(y+1)+(y+2)=24
3y+1+2=24
3y=21
y=7
Therefore, the smallest integer is 7

2.
John is 7 years older than his brother. 9 years ago, he was twice as old as his brother.

let John is y years old now
so, his bother is y-7 years old now

y-9=(y-7-9)*2
y-9=2y-32
-9+32=2y-y
y=23
Therefore, John is 23 years old now
so, his bother is 23-7=16 years old now

1. Let x be the smaller number

x+(x+1)+(x+2)=24
3x+3=24
x=(24-3)/3
x=7

They are 7, 8 & 9


2. Let x be John year old(now), y be his brober

a)x-7=y
b)x-9=2(y-9)

Sub a) into b)
x-9=2[(x-7)-9]
x-9=2(x-16)
x-9=2x-32
-9+32=2x-x
23=x

John is 23 year old now, his brother is 16

1. Let the smallest integer be n. Then we have:
n+(n+1)+(n+2)=24
3n=21
Therefore, n=7
The smallest integer is 7.
2. Let the age of John be n now. Then the age of his brother now is (n-7).
We have:
n-9=2(n-7-9)
n=23
Therefore, the age of John now is 23.