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Math problem in the fifth grade of primary school: the age of father and son this year is 6 1 year. Seven years later, the father is twice as old as his son. How old are the father and son this year?
Solution: Suppose the son is X this year and the father is (6 1-x) years old. (The total age this year is 6 1, the son is X, and the father is naturally 6 1-x)

(6 1-x)+7 ÷ (x+7)=2

(Father's age after seven years) (Son's age after seven years) (Father's age after seven years is twice that of his son, so it is equal to two)

Think of the following (x+7)=2 as divisor and quotient, so

(6 1-x)+7= (x+7)×2 (using the distribution law)

6 1+7-x=2x+ 14

68-x=2x+ 14

68-x-2x= 14

X= 18 (this year's son is 18 because the age of this year's son is set).

6 1- 18=43 (years old) (here is the father's age)

Note: distinguish the letter x from the multiplication sign X.

Seven years later, my father's age has risen again, and so has my son.