Question

(1) 5 years ago a man's age was 5 times the age of his daughter's. 3 years hence, twice

Answer 1

Answer: Daughter = 13,   Man = 45

Step-by-step explanation:

                     Today           5 yrs ago          3 years hence (future)

Man                    x                  x - 5                     x + 3

Daughter           y                  y - 5                      y + 3

5 years ago a man's age was 5 times the age of his daughter's

EQ1:  x - 5 = 5(y - 5)    →     x - 5 = 5y - 25        →      x - 5y = -20

3 years hence, twice the man's age will be equal to 6 times his daughter's age.

EQ2: 2(x + 3) = 6(y + 3)     →    2x + 6 = 6y + 18      →     2x - 6y = 12

Solve the system of equations using the Elimination method:

EQ1:  x - 5y = -20  →  -2(x - 5y = -20)   →   -2x + 10y = 40

EQ2: 2x - 6y = 12  →    1(2x - 6y = 12)   →    2x -  6y = 12  

                                                                             4y =  52

                                                                           ÷4     ÷4  

                                                                              y =  13

Substitute y = 13 into either equation to solve for x:

EQ1: x - 5y = -20

       x         = 5y - 20

       x         = 5(13) - 20

       x         =  65   - 20

              x = 45