Question 255598
let b = burl's age.
let s = burl's son's age.


first equation is:


b = 2s


this means burl's age is equal to 2 times his son's age today.


second equation is:


b-10 = 3*(s-10)


this means burl's age 10 years ago was equal to 3 times his son's age 10 years ago.


your system of equations is:


b = 2*s
b-10 = 3*(s-10)


solve simultaneously by substitution.


take b = 2*s from first equation and substitute for b in second equation of:


b-10 = 3*(s-10) to get:


2*s - 10 = 3*(s-10).


simplify to get:


2*s - 10 = 3*s - 30


subtract 2*s from both sides of equation to get:


-10 = 3*s - 2*s - 30


add 30 to both sides of equation to get:


-10 + 30 = 3*s - 2*s


combine like terms to get:


20 = s


burl's son is 20 years old today.


burl must be 40 years old today (2*20 = 40).


ten years ago burl was 30.


ten years ago burl's son was 10.


30/10 = 3 means burl was 3 times as old as his son 10 years ago.


answer is:


burl is 40 years old today.
burl's son is 20 years old today.