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Under the math problem.
Problem 1: solution: let a, b and c be x, y and z respectively.

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.