x+y+z=35 ( 1)
2x=y+5 (2)
1/3y= 1/2z (3)
X=(y+5)/2 (4) can be deduced from (2).
Z=2/3y (5) can be deduced from (3)
Substitute (4) and (5) into (1) to get y= 15, and then x= 10 can be solved; z= 10
The second question:
Let the hundredth digit be A, the tenth digit be B, and the last digit be C.
a+c=b ①
7a-(b+c)=2 ②
a+b+c= 14 ③
Substitute ① into ② ③ to get 7a-(a+c+c) = ②.
a+a+c+c= 14 ⑤
2 (a+c) = 14 from ⑤, a+c = 7, c = 7-A6.
Substituting 6 into 6 gives 7a-(a+7-a+7-a) = 2 and a = 2⑦.
Substitute ⑦ into ⑦ to get c=5 ⑧.
Substitute ⑦ ⑧ into ⑧ to get b=7.
So this three-digit number is 275.