Question about finding age

Question

Two years ago, my age was four times that of my son. Eight years ago, my age was ten times that of my son. Find the age of my son now.

Answer 1

Hi Rei,

 

Two years ago, my age was four times that of my son. Eight years ago, my age was ten times that of my son. Find the age of my son now.

 

x = My current age
y = son's current age

Two years ago...
x-2
y-2

My age was 4 times son's age:
(x-2) = 4(y-2)

Eight years ago...
x-8
y-8
My age was 10 times son's age:
(x-8) = 10(y-8)

So, we have two equations and two unknowns.  A system of linear equations can be used to solve them. 

Before we try to solve the system, it is a good idea to simplify each equation.

 

(x-2) = 4(y-2)                    (x-8) = 10(y-8)

x - 2 = 4y -8                       x - 8 = 10y -80
x+ 6 = 4y                            x +72= 10y

 

This system can be solved using Elimination:

To eliminate x multiply the 2nd equation by -1 and add:

x+ 6 = 4y                      x+ 6 = 4y               x +   6 = 4y

x +72= 10y               -1[x +72= 10y]          -x + -72= -10y

                                                                -----------------

                                                                  0  +  -66=-6y

                                                                          -66=-6y

Divide by -6:                                                         11=y



ANSWER:  The son is now 11 years old.