One number is 3 more than another and their sum is 41

Question

solve this problem by using a two order system

Answer 1

let one number be x and the other will be x+3

 

so their sum is x+(x+3)=41

After adding you get 2x+3=41

subtract 3 from both sides and you get 2x=38

divide both sides by 2 and x=19

 

 

therefore one number is 19 and if you add 3 to 19 you get the second number: then 19+3=22

 

so the two numbers are 19 and 22

 

Proof

 

19+22=41