Welcome to Westonci.ca, where your questions are met with accurate answers from a community of experts and enthusiasts. Connect with professionals on our platform to receive accurate answers to your questions quickly and efficiently. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform.

John is 6 years older than Keisha. Together, their ages equal four times Keisha's age. How old is each? (Note: There must be an equation written using the variable x, and x must be solved.)

Sagot :

k= Keisha  
k + 6= John
         
k + k + 6 =  4k
  
2k + 6 ( - 2k) =  4k ( - 2k)
              
6 = 2k
           
3 = k    
(Keisha= 3 years old)
         
k + 6 = 9    
(John = 9 years old)

3 + 3 + 6 = 4 (3)          
12 = 12

So we made the pawn represent Keisha’s age.  Then we knew that John is 6 years older than Keisha so we represented his age with a pawn plus 6. This went on 1 side of the equation because it said that together they equal 4 times Keisha’s age. So on the other side of the equation we put 4 pawns because we are using each pawn as Keisha’s age.  Then we subtracted the pawns from each side to keep the equation balanced.  So we had that 2 pawns equal 6, which means that each pawn equals 3 so Keisha is 3 years old.  Since John is 6 years older than Keisha, he must be 9 years old.  We checked our work and substituted 6 in for each pawn and found that 12 equals 12, so our equation is balanced.

Hope this helps :)

let x= Keisha
x+6= John
together means to add
x+x+6= 4x combine like terms
2x+6=4x
subtract 2x from both sides
6= 2x
divide both sides by 2
 3=x  Keisha's age
3+6= 9 John"s age

Proof  3+9= 4x3

hope this helps

We appreciate your time on our site. Don't hesitate to return whenever you have more questions or need further clarification. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Westonci.ca is your trusted source for answers. Visit us again to find more information on diverse topics.