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Urgent! A problem in the guidance of mathematics curriculum in junior two.
Let the two old people do A and B respectively; The two youths are C and D respectively; The two middle-aged couples are E and F respectively.

Then:

Answer? -B? = 195=(A+B)(A-B)

c? -Dee? = 195=(C+D)(C-D)

e? -F? = 195=(E+F)(E-F)

Let's write down all the factors of 195 first:

1,3,5, 13, 15,39,65, 195

Then among them:

1× 195= 195

3×65= 195

5×39= 195

13× 15= 195

And we generally think that people over 60 are called old people.

People around the age of 20 but not over the age of 30 are called young people.

People between the ages of 30 and 40 are called middle-aged people.

That is to say:

A+B≥ 120

C+D≥60

E+F≤60

In this way, we basically determine the number of the above equations, namely:

Answer? -B? = 195 =(A+B)(A-B)= 195× 1

c? -Dee? = 195=(C+D)(C-D)=65×3

e? -F? = 195=(E+F)(E-F)=39×5

Therefore, the equation can be listed by the above formula:

A+B= 195

A-B= 1

C+D=65

C-D=3

E+F=39

E-F=5

Solve separately:

A=98,B=97,C=34,D=3 1,E=27,F=22